Tutorial¶
Introduction¶
Class vegas.Integrator
gives Monte Carlo estimates of arbitrary
multidimensional integrals using the vegas algorithm
(G. P. Lepage, J. Comput. Phys. 27 (1978) 192 and J. Comput. Phys. 439 (2021) 110386).
The algorithm has two components.
First an automatic transformation is applied to to the integration variables
in an attempt to flatten the integrand. Then a Monte Carlo estimate of the
integral is made using the transformed variables. Flattening the integrand
makes the integral easier and improves the estimate. The transformation
applied to the integration variables is optimized
over several iterations of the algorithm: information about the integrand that
is collected during one iteration is used to improve the transformation used
in the next iteration.
Monte Carlo integration makes few assumptions about the integrand — it needn’t be analytic nor even continuous. This makes Monte Carlo integration unusually robust. It also makes it well suited for adaptive integration. Adaptive strategies are essential for multidimensional integration, especially in high dimensions, because multidimensional space is large, with lots of corners, making it easy to lose important features in the integrand.
Monte Carlo integration also provides efficient and reliable methods for estimating the accuracy of its results. In particular, each Monte Carlo estimate of an integral is a random number from a distribution whose mean is the correct value of the integral. This distribution is Gaussian or normal provided the number of integrand samples is sufficiently large. In practice we generate multiple estimates of the integral in order to verify that the distribution is indeed Gaussian. Error analysis is straightforward if the integral estimates are Gaussian.
The vegas
algorithm has been in use for decades and implementations are
available in many programming languages, including Fortran (the original
version), C and C++. The algorithm used here is significantly improved over
the original implementation, and that used in most other implementations.
It uses two adaptive strategies: importance sampling, as in the original
implementation, and adaptive stratified sampling, which is new. The
new algorithm is described in G. P. Lepage, arXiv_2009.05112
(J. Comput. Phys. 439 (2021) 110386).
This module is written in Cython, so it is almost as fast as compiled Fortran or C, particularly when the integrand is also coded in Cython (or some other compiled language), as discussed below.
The following sections describe how to use vegas
. Almost every
example shown is a complete code, which can be copied into a file
and run with Python. It is worthwhile playing with the parameters to see how
things change.
About Printing: The examples in this tutorial use the print function as it is used in Python 3. Drop the outermost parenthesis in each print statement if using Python 2, or add
from __future__ import print_function
at the start of your file.
Basic Integrals¶
Here we illustrate the use of vegas
by estimating the integral
where constant is chosen so that the exact integral is 1. The following code shows how this can be done:
import vegas
import math
def f(x):
dx2 = 0
for d in range(4):
dx2 += (x[d] - 0.5) ** 2
return math.exp(-dx2 * 100.) * 1013.2118364296088
integ = vegas.Integrator([[-1, 1], [0, 1], [0, 1], [0, 1]])
result = integ(f, nitn=10, neval=1000)
print(result.summary())
print('result = %s Q = %.2f' % (result, result.Q))
First we define the integrand f(x)
where x[d]
specifies a point in the
4-dimensional space. We then create an integrator, integ
, which is an
integration operator that can be applied to any 4-dimensional function. It is
where we specify the integration volume.
Finally we apply integ
to our integrand f(x)
,
telling the integrator to estimate the integral using nitn=10
iterations
of the vegas
algorithm, each of which uses no more than neval=1000
evaluations of the integrand. Each iteration produces an independent
estimate of the integral. The final estimate is the weighted average of
the results from all 10 iterations, and is returned by integ(f ...)
.
The call result.summary()
returns
a summary of results from each iteration.
This code produces the following output:
itn integral wgt average chi2/dof Q
-------------------------------------------------------
1 2.6(1.4) 2.6(1.4) 0.00 1.00
2 1.32(25) 1.36(25) 0.75 0.39
3 0.909(96) 0.968(89) 1.79 0.17
4 1.039(69) 1.012(55) 1.32 0.26
5 0.929(34) 0.952(29) 1.41 0.23
6 1.003(26) 0.980(19) 1.47 0.20
7 0.994(18) 0.988(13) 1.27 0.27
8 0.998(14) 0.9922(98) 1.13 0.34
9 1.020(12) 1.0035(75) 1.39 0.20
10 1.011(12) 1.0057(64) 1.27 0.25
result = 1.0057(64) Q = 0.25
There are several things to note here:
Adaptation: Integration estimates are shown for each of the 10 iterations, giving both the estimate from just that iteration, and the weighted average of results from all iterations up to that point. The estimates from the first two iterations are not accurate at all, with errors equal to 25–140% of the final result.
vegas
initially has no information about the integrand and so does a relatively poor job of estimating the integral. It uses information from the samples in one iteration, however, to remap the integration variables for subsequent iterations, concentrating samples where the function is largest and reducing errors. As a result, the per-iteration error is reduced to 3.4% by the fifth iteration, and almost to 1% by the end — an improvement by a factor of more than 100 from the start. Eventually the per-iteration error stops decreasing becausevegas
has found the optimal remapping, at which point it is fully adapted to the integrand.Weighted Average: The final result, 1.0057 ± 0.0064, is obtained from a weighted average of the separate results from each iteration: estimates are weighted by the inverse variance, thereby giving much less weight to the early iterations, where the errors are largest. The individual estimates are statistical: each is a random number drawn from a distribution whose mean equals the correct value of the integral, and the errors quoted are estimates of the standard deviations of those distributions. The distributions are Gaussian provided the number of integrand evaluations per iteration (
neval
) is sufficiently large, in which case the standard deviation is a reliable estimate of the error. The weighted average minimizeswhere are the estimates from individual iterations. If the are Gaussian, should be of order the number of degrees of freedom (plus or minus the square root of double that number); here the number of degrees of freedom is the number of iterations minus 1.
The distributions are likely non-Gaussian, and error estimates unreliable, if is much larger than the number of iterations. This criterion is quantified by the Q or p-value of the , which is the probability that a larger could result from random (Gaussian) fluctuations. A very small Q (less than 0.05-0.1) indicates that the is too large to be accounted for by statistical fluctuations — that is, the estimates of the integral from different iterations do not agree with each other to within errors. This means that
neval
is not sufficiently large to guarantee Gaussian behavior, and must be increased if the error estimates are to be trusted.
integ(f...)
returns a weighted-average object, of typevegas.RAvg
, that has the following attributes:
result.mean
— weighted average of all estimates of the integral;
result.sdev
— standard deviation of the weighted average;
result.chi2
— of the weighted average;
result.dof
— number of degrees of freedom;
result.Q
— Q or p-value of the weighted average’s ;
result.itn_results
— list of the integral estimates from each iteration;
result.sum_neval
— total number of integrand evaluations used.
result.avg_neval
— average number of integrand evaluations per iterationIn this example the final Q is 0.25, indicating that the for this average is not particularly unlikely and thus the error estimate is likely reliable.
Precision: The precision of
vegas
estimates is determined bynitn
, the number of iterations of thevegas
algorithm, and byneval
, the maximum number of integrand evaluations made per iteration. The computing cost is typically proportional to the product ofnitn
andneval
. The number of integrand evaluations per iteration varies from iteration to iteration, here between 860 and 960. Typicallyvegas
needs more integration points in early iterations, before it has fully adapted to the integrand.We can increase precision by increasing either
nitn
orneval
, but it is generally far better to increaseneval
. For example, adding the following lines to the code aboveresult = integ(f, nitn=100, neval=1000) print('larger nitn => %s Q = %.2f' % (result, result.Q)) result = integ(f, nitn=10, neval=1e4) print('larger neval => %s Q = %.2f' % (result, result.Q))generates the following results:
larger nitn => 1.0003(13) Q = 0.79 larger neval => 0.99981(53) Q = 0.28The total number of integrand evaluations,
nitn * neval
, is about the same in both cases, but increasingneval
is more than twice as accurate as increasingnitn
. Typically you want to use no more than 10 or 20 iterations beyond the point wherevegas
has fully adapted. You want some number of iterations so that you can verify Gaussian behavior by checking the and Q, but not too many.It is also generally useful to compare two or more results from values of
neval
that differ by a significant factor (4–10, say). These should agree within errors. If they do not, it could be due to non-Gaussian artifacts caused by a smallneval
.vegas
estimates have two sources of error. One is the statistical error, which is what is quoted byvegas
. The other is a systematic error due to residual non-Gaussian effects. The systematic error vanishes like1/neval
or faster, and so becomes negligible compared with the statistical error asneval
increases. The systematic error can bias the Monte Carlo estimate, however, ifneval
is insufficiently large. This usually results in a large (and small Q), but a more reliable check is to compare results that use significantly different values ofneval
. The systematic errors due to non-Gaussian behavior are likely negligible if the different estimates agree to within the statistical errors.The possibility of systematic biases is another reason for increasing
neval
rather thannitn
to obtain more precision. Makingneval
larger and larger is guaranteed to improve the Monte Carlo estimate, as the statistical error decreases and the systematic error decreases even more quickly. Makingnitn
larger and larger, on the other hand, is guaranteed eventually to give the wrong answer. This is because at some point the statistical error (which falls assqrt(1/nitn)
) will no longer mask the systematic error (which is unaffected bynitn
). The systematic error for the integral above (withneval=1000
) is about -0.0008, which is negligible compared to the statistical error unlessnitn
is of order 1500 or larger — so systematic errors aren’t a problem withnitn=10
.Early Iterations: Integral estimates from early iterations, before
vegas
has adapted, can be quite crude. With very peaky integrands, these are often far from the correct answer with highly unreliable error estimates. For example, the integral above becomes more difficult if we double the length of each side of the integration volume by redefininginteg
as:integ = vegas.Integrator([[-2, 2], [0, 2], [0, 2], [0., 2]])The code above then gives:
itn integral wgt average chi2/dof Q ------------------------------------------------------- 1 0.0011(10) 0.0011(10) 0.00 1.00 2 0.074(56) 0.0011(10) 1.71 0.19 3 0.250(59) 0.0012(10) 9.65 0.00 4 0.93(14) 0.0013(10) 21.40 0.00 5 0.874(70) 0.0015(10) 54.87 0.00 6 0.949(39) 0.0021(10) 162.08 0.00 7 0.949(30) 0.0033(10) 301.18 0.00 8 0.985(25) 0.0050(10) 484.50 0.00 9 0.967(19) 0.0078(10) 738.53 0.00 10 0.988(15) 0.0125(10) 1131.46 0.00 result = 0.0125(10) Q = 0.00
vegas
misses the peak completely in the first iteration, giving an estimate that is completely wrong (by 1000 standard deviations!). Some of its samples hit the peak’s shoulders, sovegas
is eventually able to find the peak (by iterations 5–6), but the integrand estimates are wildly non-Gaussian before that point. This results in a nonsensical final result, as indicated by theQ = 0.00
.It is common practice in using
vegas
to discard estimates from the first several iterations, before the algorithm has adapted, in order to avoid ruining the final result in this way. This is done by replacing the single call tointeg(f...)
in the original code with two calls:# step 1 -- adapt to f; discard results integ(f, nitn=10, neval=1000) # step 2 -- integ has adapted to f; keep results result = integ(f, nitn=10, neval=1000) print(result.summary()) print('result = %s Q = %.2f' % (result, result.Q))The integrator is trained in the first step, as it adapts to the integrand, and so is more or less fully adapted from the start in the second step, which yields:
itn integral wgt average chi2/dof Q ------------------------------------------------------- 1 0.993(17) 0.993(17) 0.00 1.00 2 1.062(48) 1.001(16) 1.83 0.18 3 0.964(20) 0.987(13) 1.91 0.15 4 0.974(16) 0.9817(99) 1.40 0.24 5 0.990(15) 0.9843(82) 1.10 0.35 6 1.012(16) 0.9899(73) 1.34 0.25 7 0.999(15) 0.9917(65) 1.16 0.32 8 1.008(12) 0.9953(58) 1.20 0.30 9 1.013(15) 0.9977(54) 1.20 0.29 10 0.983(14) 0.9958(50) 1.17 0.31 result = 0.9958(50) Q = 0.31The final result is now reliable.
Other Integrands: Once
integ
has been trained onf(x)
, it can be usefully applied to other functions with similar structure. For example, adding the following at the end of the original code,def g(x): return x[0] * f(x) result = integ(g, nitn=10, neval=1000) print(result.summary()) print('result = %s Q = %.2f' % (result, result.Q))gives the following new output:
itn integral wgt average chi2/dof Q ------------------------------------------------------- 1 0.4933(61) 0.4933(61) 0.00 1.00 2 0.5017(54) 0.4980(40) 1.04 0.31 3 0.4975(64) 0.4979(34) 0.52 0.59 4 0.5059(60) 0.4998(30) 0.80 0.49 5 0.5075(64) 0.5012(27) 0.90 0.46 6 0.4907(66) 0.4997(25) 1.15 0.33 7 0.5009(47) 0.5000(22) 0.97 0.45 8 0.5082(58) 0.5010(21) 1.08 0.38 9 0.5016(63) 0.5010(20) 0.94 0.48 10 0.4934(76) 0.5006(19) 0.94 0.49 result = 0.5006(19) Q = 0.49Again the grid is almost optimal for
g(x)
from the start, becauseg(x)
peaks in the same region asf(x)
. The exact value for this integral is very close to 0.5.Non-Rectangular Volumes:
vegas
can integrate over volumes of non-rectangular shape. For example, we can replace integrandf(x)
above by the same Gaussian, but restricted to a 4-sphere of radius 0.2, centered on the Gaussian:import vegas import math def f_sph(x): dx2 = 0 for d in range(4): dx2 += (x[d] - 0.5) ** 2 if dx2 < 0.2 ** 2: return math.exp(-dx2 * 100.) * 1115.3539360527281318 else: return 0.0 integ = vegas.Integrator([[-1, 1], [0, 1], [0, 1], [0, 1]]) integ(f_sph, nitn=10, neval=1000) # adapt the grid result = integ(f_sph, nitn=10, neval=1000) # estimate the integral print(result.summary()) print('result = %s Q = %.2f' % (result, result.Q))The normalization is adjusted to again make the exact integral equal 1. Integrating as before gives:
itn integral wgt average chi2/dof Q ------------------------------------------------------- 1 0.992(20) 0.992(20) 0.00 1.00 2 0.993(19) 0.992(14) 0.00 0.97 3 1.002(18) 0.996(11) 0.09 0.91 4 1.004(22) 0.9973(98) 0.10 0.96 5 1.026(30) 1.0001(93) 0.28 0.89 6 1.053(92) 1.0007(93) 0.29 0.92 7 1.035(30) 1.0038(89) 0.45 0.85 8 0.991(19) 1.0014(80) 0.44 0.88 9 0.968(18) 0.9956(73) 0.76 0.64 10 1.022(37) 0.9966(72) 0.73 0.68 result = 0.9966(72) Q = 0.68It is a good idea to make the actual integration volume as large a fraction as possible of the total volume used by
vegas
— by choosing integration variables properly — sovegas
doesn’t spend lots of effort on regions where the integrand is exactly 0. Also, it can be challenging forvegas
to find the region of non-zero integrand in high dimensions: integratingf_sph(x)
in 20 dimensions instead of 4, for example, would requireneval=1e16
integrand evaluations per iteration to have any chance of finding the region of non-zero integrand, because the volume of the 20-dimensional sphere is a tiny fraction of the total integration volume. The final error in the example above would have been cut in half had we used the integration volume4 * [[0.3, 0.7]]
instead of[[-1, 1], [0, 1], [0, 1], [0, 1]]
.Note, finally, that integration to infinity is also possible: map the relevant variable into a different variable of finite range. For example, an integral over from 0 to infinity is easily re-expressed as an integral over from 0 to 1, where the transformation emphasizes the region in of order free parameter .
Damping: The result in the previous section can be improved somewhat by slowing down
vegas
’s adaptation:... integ(f_sph, nitn=10, neval=1000, alpha=0.1) result = integ(f_sph, nitn=10, neval=1000, alpha=0.1) ...Parameter
alpha
controls the speed with whichvegas
adapts, with smalleralpha
s giving slower adaptation. Here we reducealpha
to 0.1, from its default value of 0.5, and get the following output:itn integral wgt average chi2/dof Q ------------------------------------------------------- 1 1.008(26) 1.008(26) 0.00 1.00 2 0.993(23) 0.999(17) 0.19 0.66 3 1.005(21) 1.002(13) 0.11 0.89 4 1.016(20) 1.006(11) 0.19 0.91 5 0.973(18) 0.9967(95) 0.73 0.57 6 1.016(18) 1.0009(84) 0.77 0.57 7 1.008(18) 1.0023(76) 0.66 0.68 8 0.990(17) 1.0002(69) 0.63 0.73 9 1.008(17) 1.0012(64) 0.58 0.80 10 0.958(17) 0.9959(60) 1.12 0.34 result = 0.9959(60) Q = 0.34Notice how the errors fluctuate less from iteration to iteration with the smaller
alpha
in this case. Persistent, large fluctuations in the size of the per-iteration errors is often a signal thatalpha
should be reduced. With largeralpha
s,vegas
can over-react to random fluctuations it encounters as it samples the integrand.In general, we want
alpha
to be large enough so thatvegas
adapts quickly to the integrand, but not so large that it has difficulty holding on to the optimal tuning once it has found it. The best value depends upon the integrand.adapt=False: Adaptation can be turned off completely by setting parameter
adapt=False
. There are three reasons one might do this. The first is ifvegas
is exhibiting the kind of instability discussed in the previous section — one might use the following code, instead of that presented there:... integ(f_sph, nitn=10, neval=1000, alpha=0.1) result = integ(f_sph, nitn=10, neval=1000, adapt=False) ...The second reason is that
vegas
runs slightly faster when it is no longer adapting to the integrand. The difference is not signficant for complicated integrands, but is noticable in simpler cases.The third reason for turning off adaptation is that
vegas
uses unweighted averages, rather than weighted averages, to combine results from different iterations whenadapt=False
. Unweighted averages are not biased. They have no systematic error of the sort discussed above, and so give correct results even for very large numbers of iterations,nitn
.The lack of systematic biases is not a strong reason for turning off adaptation, however, since the biases are usually negligible (see above). The most important reason is the first: stability.
adapt=False
is particularly useful when the number of integrand evaluationsneval
is small for the integrand, leading to large fluctuations in the errors from iteration to iteration. For example, the following output is from an estimate (withneval=2.5e4
) of an eight-dimensional integral with three sharp peaks along the diagonal (Eq. (45) in arXiv_2009.05112, normalized so that the correct answer equals 1):itn integral wgt average chi2/dof Q ------------------------------------------------------- 1 0.75(43) 0.75(43) 0.00 1.00 2 0.506(58) 0.510(58) 0.32 0.57 3 0.80(21) 0.530(56) 1.02 0.36 4 0.76(11) 0.576(50) 1.81 0.14 5 1.27(29) 0.596(49) 2.74 0.03 6 1.10(19) 0.629(48) 3.56 0.00 7 0.802(73) 0.681(40) 3.63 0.00 8 2.8(2.0) 0.681(40) 3.27 0.00 9 0.907(90) 0.719(36) 3.52 0.00 10 1.07(16) 0.736(35) 3.65 0.00 itn integral average chi2/dof Q ------------------------------------------------------- 1 1.13(14) 1.13(14) 0.00 1.00 2 1.064(96) 1.095(86) 0.13 0.72 3 1.03(10) 1.072(67) 0.19 0.83 4 0.924(94) 1.035(55) 0.58 0.63 5 0.858(71) 1.000(46) 1.08 0.37 6 0.97(11) 0.995(43) 0.84 0.52 7 0.924(69) 0.985(38) 0.84 0.54 8 1.19(16) 1.010(39) 1.01 0.42 9 1.74(73) 1.092(88) 1.01 0.42 10 0.942(89) 1.077(80) 1.02 0.42The first 10 iterations are used to train the
vegas
map; their results are discarded. The next 10 iterations, withadapt=False
, have uncertainties that fluctuate in size by an order of magnitude, but still give a reliable estimate for the integral (1.08(8)). Allowingvegas
to continue adapting in the the second set of iterations gives results like 0.887(25), which is 4.5 standard deviations too low; the real uncertainty is larger than ±0.025.Training the integrator and then setting
adapt=False
for the final results works best if the number of evaluations per iteration (neval
) is the same in both steps. This is because the second ofvegas
’s adaptation strategies (Adaptive Stratified Sampling) is usually reinitialized whenneval
changes, and so is not used at all whenneval
is changed at the same timeadapt=False
is set.
Multiple Integrands Simultaneously¶
vegas
can be used to integrate multiple integrands simultaneously, using
the same integration points for each of the integrands. This is useful
in situations where the integrands have similar structure, with peaks in
the same locations. There can be signficant advantages in sampling
different integrands at precisely the same points in x space, because
then Monte Carlo estimates for the different integrals are correlated.
If the integrands are very similar to each other, the correlations can be
very strong. This leads to greatly reduced errors in ratios or differences
of the resulting integrals as the fluctuations cancel.
Consider a simple example. We want to compute the normalization and first two moments of a sharply peaked probability distribution:
From these integrals we determine the mean and width of the distribution projected onto one of the axes:
This can be done using the following code:
import vegas
import math
import gvar as gv
def f(x):
dx2 = 0.0
for d in range(4):
dx2 += (x[d] - 0.5) ** 2
f = math.exp(-200 * dx2)
return [f, f * x[0], f * x[0] ** 2]
integ = vegas.Integrator(4 * [[0, 1]])
# adapt grid
training = integ(f, nitn=10, neval=2000)
# final analysis
result = integ(f, nitn=10, neval=10000)
print('I[0] =', result[0], ' I[1] =', result[1], ' I[2] =', result[2])
print('Q = %.2f\n' % result.Q)
print('<x> =', result[1] / result[0])
print(
'sigma_x**2 = <x**2> - <x>**2 =',
result[2] / result[0] - (result[1] / result[0]) ** 2
)
print('\ncorrelation matrix:\n', gv.evalcorr(result))
The code is very similar to that used in the previous section. The
main difference is that the integrand function and vegas
return arrays of results — in
both cases, one result for each of the three integrals. vegas
always adapts to
the first integrand in the array. The Q
value is for all three
of the integrals, taken together.
The code produces the following output:
I[0] = 0.00024682(12) I[1] = 0.000123417(61) I[2] = 0.000062327(33)
Q = 0.93
<x> = 0.500017(49)
sigma_x**2 = <x**2> - <x>**2 = 0.0024983(73)
correlation matrix:
[[1. 0.98002885 0.92558296]
[0.98002885 1. 0.98157932]
[0.92558296 0.98157932 1. ]]
The estimates for the individual integrals are separately accurate to about ±0.05%, but the estimate for is accurate to ±0.01%. This is almost an order of magnitude (8x) more accurate than we would obtain absent correlations. The correlation matrix shows that there is 98% correlation between the statistical fluctuations in estimates for and , and so the bulk of these fluctuations cancel in the ratio. The estimate for the variance is 48x more accurate than we would have obtained had the integrals been evaluated separately. Both estimates are correct to within the quoted errors.
The individual results are objects of type gvar.GVar
, which
represent Gaussian random variables. Such objects have means
(result[i].mean
) and standard deviations (result[i].sdev
), but
also can be statistically correlated with other gvar.GVar
s.
Such correlations are handled automatically by gvar
when
gvar.GVar
s are combined with each other or with numbers in
arithmetical expressions. (Documentation for gvar
can be found
at https://gvar.readthedocs.io or with the source code
at https://github.com/gplepage/gvar.git.)
Dictionaries¶
Integrands can return dictionaries instead of arrays. The example in the previous section, for example, can be rewritten as
import vegas
import math
import gvar as gv
def f(x):
dx2 = 0.0
for d in range(4):
dx2 += (x[d] - 0.5) ** 2
f = math.exp(-200 * dx2)
return {'1':f, 'x':f * x[0], 'x**2':f * x[0] ** 2}
integ = vegas.Integrator(4 * [[0, 1]])
# adapt grid
training = integ(f, nitn=10, neval=2000)
# final analysis
result = integ(f, nitn=10, neval=10000)
print(result)
print('Q = %.2f\n' % result.Q)
print('<x> =', result['x'] / result['1'])
print(
'sigma_x**2 = <x**2> - <x>**2 =',
result['x**2'] / result['1'] - (result['x'] / result['1']) ** 2
)
which returns the following output:
{'1': 0.00024682(12),'x': 0.000123417(61),'x**2': 0.000062327(33)}
Q = 0.93
<x> = 0.500017(49)
sigma_x**2 = <x**2> - <x>**2 = 0.0024983(73)
The result returned by vegas
is a dictionary using the same keys as the
dictionary returned by the integrand. Using a dictionary with descriptive
keys, instead of an array, can often make code more intelligible, and,
therefore, easier to write and maintain. Here the values in the integrand’s
dictionary are all numbers; in general, values can be either numbers or
arrays (of any shape).
Dictionaries can also be used for the integration variables. For example, the following code calculates the volume of a unit sphere using spherical coordinates:
import numpy as np
import vegas
def f(xd):
r = xd['r']
theta = xd['theta']
phi = xd['phi']
return r ** 2 * np.sin(theta)
integ = vegas.Integrator(dict(r=(0,1), theta=(0, np.pi), phi=(0, 2 * np.pi)))
volume = integ(f, neval=1000, nitn=10)
print(volume)
Running this code gives a result of 4.1852(44) which agrees well (0.1%) with the correct
result . This can be generalized to DIM
dimensions,
where now xd['phi']
is an array of variables: e.g.,
import numpy as np
import vegas
DIM = 5
def f(xd):
r = xd['r']
phi = xd['phi']
# construct Euclidean coordinates, Jacobian
x = np.zeros(DIM, float)
x[:] = r
x[1:] *= np.cumprod(np.sin(phi), axis=0)
jac = np.prod(x[1:-1], axis=0) * r
x[:-1] *= np.cos(phi)
# calculate contribution to sphere's volume
return jac
integ = vegas.Integrator(dict(
r=(0,1),
phi=(DIM - 2) * [(0, np.pi)] + [(0, 2 * np.pi)]
))
warmup = integ(f, neval=1000, nitn=10)
volume = integ(f, neval=1000, nitn=10)
print(integ.settings(), '\n')
print(f'volume(dim={DIM}) = {volume}')
This code generates the following output:
Integrator Settings:
1000 (approx) integrand evaluations in each of 10 iterations
number of: strata/axis = [3 3 3 2 2]
increments/axis = [ 99 99 99 100 100]
h-cubes = 108 processors = 1
evaluations/batch >= 5e+04
2 <= evaluations/h-cube <= 5e+04
minimize_mem = False adapt_to_errors = False adapt = True
accuracy: relative = 0 absolute = 0
damping: alpha = 0.5 beta= 0.75
key/index axis integration limits
---------------------------------------------
r 0 (0.0, 1.0)
phi 0 1 (0.0, 3.141592653589793)
1 2 (0.0, 3.141592653589793)
2 3 (0.0, 3.141592653589793)
3 4 (0.0, 6.283185307179586)
volume(dim=5) = 5.254(12)
Calculating Distributions¶
vegas
is often used to calculate distributions. The following
code, for example, evaluates both an integral I
and the contributions
dI
to the integral coming from each of five different intervals dr
in the radius measured
from the center of the integration volume. The normalized contributions
dI/I
are then tabulated:
import vegas
import numpy as np
RMAX = (2 * 0.5**2) ** 0.5
def fcn(x):
dx2 = 0.0
for d in range(2):
dx2 += (x[d] - 0.5) ** 2
I = np.exp(-dx2)
# add I to appropriate bin in dI
dI = np.zeros(5, dtype=float)
dr = RMAX / len(dI)
j = int(dx2 ** 0.5 / dr)
dI[j] = I
return dict(I=I, dI=dI)
integ = vegas.Integrator(2 * [(0,1)])
# results returned in a dictionary
result = integ(fcn)
print(result.summary())
print(' I =', result['I'])
print('dI/I =', result['dI'] / result['I'])
print('sum(dI/I) =', sum(result['dI']) / result['I'])
Note the check at the end, to verify that the sum of the
dI[i]
s equals the original integral. Running this script gives
the following output:
itn integral wgt average chi2/dof Q
-------------------------------------------------------
1 0.85040(55) 0.85040(55) 0.00 1.00
2 0.85039(45) 0.85040(35) 0.76 0.60
3 0.85085(39) 0.85061(26) 0.63 0.82
4 0.85105(33) 0.85079(20) 0.52 0.95
5 0.85105(30) 0.85087(17) 0.60 0.94
6 0.85097(24) 0.85091(14) 0.55 0.98
7 0.85099(21) 0.85096(11) 0.72 0.90
8 0.85112(17) 0.851013(93) 0.66 0.95
9 0.85114(15) 0.851053(79) 0.70 0.94
10 0.85101(13) 0.851041(67) 0.68 0.96
I = 0.851041(67)
dI/I = [0.0759(12) 0.2091(23) 0.3217(27) 0.3209(23) 0.0723(12)]
sum(dI/I) = 0.999999999996(26)
The integrator adapts to the full integral I
but also gives
accurate results for the distribution dI
(though not quite as
accurate). Note that sum(dI/I)
is much more accurate than
any individual dI/I
, because of correlations between
the different dI/I
values. (The uncertainty on sum(dI/I)
would
be exactly zero absent roundoff errors.)
Often one has more than five bins in a distribution. Increasing the number to 100 in the example above reveals a problem:
itn integral wgt average chi2/dof Q
-------------------------------------------------------
1 0.85040(55) 0.85040(55) 0.00 1.00
2 0.85039(45) 0.85079(39) 0.75 0.97
3 0.85085(39) 0.820(49) 0.73 1.00
4 0.85105(33) 0.813(40) 0.81 0.99
5 0.85105(30) 0.837(24) 0.96 0.73
6 0.85097(24) 0.85236(75) 1.10 0.06
7 0.85099(21) 0.85242(25) 1.19 0.00
8 0.85112(17) 0.85242(23) 1.22 0.00
9 0.85114(15) 0.85234(21) 1.22 0.00
10 0.85101(13) 0.85179(11) 1.25 0.00
I = 0.85179(11)
dI/I = [0.00012(87) 0.00177(62) 0.00016(83) 0.00067(51) 0.00103(53)...]
sum(dI/I) = 1.0000117(35)
Something is going wrong with the weighted averages of the results from different iterations, as is clear by the third iteration. The weights used in the weighted average are obtained from the inverse of the covariance matrix for the different components of the integral — here a 101×101 matrix. This matrix becomes quite singular as it grows, and therefore is quite sensitive to even small errors in the covariance matrix. These errors, particularly in early iterations, can introduce large errors in the weighted averages.
This problem can be addressed by increasing the number of integrand
evaluations per iteration neval
, which increases the accuracy
of the Monte Carlo estimate of the covariance matrix. A more efficient
solution in this case, however, is to break the integration into two parts: one
where the integrator is adapted to the integrand, but the results
are discarded; and a second step where adaptation is turned off
with adapt=False
to obtain the final result. As discussed above,
vegas
does not use a weighted average when adapt=False
and
so the inversion of the covariance matrices is unnecessary.
To implement this strategy in the code above, replace the line
result = integ(fcn)
with
discard = integ(fcn) # adapt to grid
result = integ(fcn, adapt=False) # no further adaptation
This gives the following output:
itn integral average chi2/dof Q ------------------------------------------------------- 1 0.85113(11) 0.85113(11) 0.00 1.00 2 0.85113(12) 0.851130(81) 0.99 0.50 3 0.85124(11) 0.851166(65) 1.08 0.21 4 0.85129(11) 0.851197(56) 1.04 0.31 5 0.85105(12) 0.851168(51) 0.99 0.54 6 0.85120(11) 0.851173(46) 0.99 0.57 7 0.85114(11) 0.851169(43) 0.98 0.63 8 0.85107(11) 0.851156(40) 1.03 0.29 9 0.85114(11) 0.851154(37) 1.04 0.19 10 0.85110(11) 0.851149(36) 1.02 0.31 I = 0.851149(36) dI/I = [0.00034(24) 0.00067(34) 0.00067(34) 0.00185(49) 0.00101(39)...] sum(dI/I) = 1.000000000000(37)
The correct result for I
is 0.85112.
PDF Integrals¶
The vegas
module has a special-purpose integrator for
evaluating averages over probability distributions.
In its simplest form,
g_ev = vegas.PDFIntegrator(param=g)
creates an integrator optimized for the multi-dimensional Gaussian
probability distribution corresponding to g
. (g
is
an array of Gaussian random variables of type gvar.GVar
,
or a dictionary whose
values are gvar.GVar
s or arrays of gvar.GVar
s.)
Then g_ev(f)
evaluates the expectation value of a
function f(p)
with respect to this distribution,
where p
is a point in the distribution’s parameter
space.
More generally
pdf_ev = vegas.PDFIntegrator(param=g, pdf=pdf)
creates an integrator which calculates expectation values with
respect to an arbitrary probability density function pdf(p)
.
The parameter space for points p
is again defined by
and optimized for the PDF corresponding to g
, but
expectation values are calculated with pdf(p)
. Typically
g
’s means and covariance would be chosen to emphasize
the regions where pdf(p)
is large (e.g., g
might be
set equal to the prior in a Bayesian analysis).
In general, the integrator uses param
to define and optimize the
internal integration parameters. It re-expresses integrals in
terms of variables
that diagonalize param
’s correlation matrix and are centered at
its mean value. This greatly facilitates integration over these
variables using vegas
, making integrals over
many parameters feasible, even when the parameters are highly
correlated. vegas.PDFIntegrator
also pre-adapts the integrator
to param
’s PDF so it is often unnecessary to discard early iterations.
vegas.PDFIntegrator
evaluates the
integrals of both pdf(p) * f(p)
and pdf(p)
. The expectation
value is the ratio of the two integrals, so the PDF need not be
normalized. Note also that Monte Carlo uncertainties in
the two integrals are often highly correlated, in which case
the uncertainties are significantly reduced in the ratio.
A simple illustration of vegas.PDFIntegrator
is given by the following
code, where g
determines both the parameterization of the integrals
and the PDF used for expectation values:
import vegas
import gvar as gv
# multi-dimensional distribution
g = gv.BufferDict()
g['a'] = gv.gvar([2., 1.], [[1., 0.99], [0.99, 1.]])
g['fb(b)'] = gv.BufferDict.uniform('fb', 0.0, 2.0)
# integrator for expectation values in distribution g
g_ev = vegas.PDFIntegrator(g)
# adapt integrator to the PDF
g_ev(neval=10_000, nitn=5)
# want expectation value of [fp, fp**2]
def f_f2(p):
a = p['a']
b = p['b']
fp = a[0] * a[1] + b
return [fp, fp ** 2]
# <f_f2> in distribution g
r = g_ev(f_f2, adapt=False)
print(r.summary())
print('results =', r, '\n')
# mean and standard deviation of fp's distribution
fmean = r[0]
fsdev = gv.sqrt(r[1] - r[0] ** 2)
print ('fp.mean =', fmean, ' fp.sdev =', fsdev)
print ("Gaussian approx'n for fp =", f_f2(g)[0], '\n')
# g's pdf norm
print('PDF norm =', r.pdfnorm)
Here the distribution g
describes two highly correlated Gaussian
variables, a[0]
and a[1]
, and a third uncorrelated variable b
that is uniformly distributed on the interval [0,2] (see the gvar
documentation for more information).
We use the integrator to calculated the expectation value of
fp = a[0] * a[1] + b
and fp**2
, so we can compute the
mean and standard
deviation of the fp
distribution. The output from this code
shows that the Gaussian approximation 3.0(3.1) for the mean and
standard deviation is not particularly
close to the correct value 4.0(3.4):
itn integral average chi2/dof Q
-------------------------------------------------------
1 1.0003(10) 1.0003(10) 0.00 1.00
2 1.0001(11) 1.00022(76) 0.19 0.90
3 1.0004(10) 1.00028(61) 0.24 0.96
4 0.9995(11) 1.00009(54) 0.21 0.99
5 1.0006(10) 1.00020(48) 0.30 0.99
results = [3.9957(52) 27.33(13)]
fp.mean = 3.9957(52) fp.sdev = 3.371(13)
Gaussian approx'n for fp = 3.0(3.1)
PDF norm = 1.00020(48)
In general the function f(p)
in g_ev(f)
can return a number,
or an array of
numbers, or a dictionary whose values are numbers or arrays of numbers.
This allows multiple expectation values to be evaluated simultaneously.
The example above can be coded much more simply using the
PDFIntegrator.stats()
method to evaluate the
expectation value of function f(p)
(rather
than f_f2(p)
above):
# want expectation value of f(p)
def f(p):
a = p['a']
b = p['b']
fp = a[0] * a[1] + b
return dict(a=a, b=b, fp=fp)
r = g_ev.stats(f)
print('results =', r)
print (' f(g) =', f(g))
print('\ncorrelation matrix:')
print(gv.evalcorr([r['a'][0], r['a'][1], r['b'], r['fp']]))
stats(f)
calculates both the mean values and the standard deviations of
each component of f(p)
, combining them into gvar.GVar
objects.
(The standard deviations include the uncertainties coming from the integration
added in quadrature with the uncertainties coming from the distribution.)
The output from this code compares the actual means and standard deviations
from g_ev.stats(f)
with what is obtained from the Gaussian
approximation (f(g)
):
results = {'a': array([2.0(1.0), 1.0(1.0)], dtype=object), 'b': 1.00(58), 'fp': 4.0(3.4)}
f(g) = {'a': array([2.0(1.0), 1.0(1.0)], dtype=object), 'b': 1.00(80), 'fp': 3.0(3.1)}
correlation matrix:
[[ 1.00000000e+00 9.90054625e-01 3.22078905e-04 8.86097896e-01]
[ 9.90054625e-01 1.00000000e+00 -2.59235850e-04 8.88957532e-01]
[ 3.22078905e-04 -2.59235850e-04 1.00000000e+00 1.70539216e-01]
[ 8.86097896e-01 8.88957532e-01 1.70539216e-01 1.00000000e+00]]
This shows that the results for f['a']
agree with the inputs (as expected because
they are Gaussian), but this is less true for r['b']
and r['fp']
whose
distributions are less well approximated by a Gaussian.
Additional information can be obtained by setting the keyword arguments moments
and/or histograms
equal to True
in PDFIntegrator.stats()
. The code
r = g_ev.stats(f, moments=True, histograms=True)
print('Statistics for fp:')
print(r.stats['fp'])
r.stats['fp'].plot_histogram(show=True)
shows the statistical analysis for fp
:
Statistics for fp:
mean = 3.9920(59) sdev = 3.369(15) skew = 1.655(23) ex_kurt = 3.99(16)
split-normal: 0.5988(22) +/- 4.839(12)/0.4289(47)
median: 3.0411(66) +/- 4.009(17)/1.9274(76)
g_ev.stats(f, moments=True, histograms=True)
calculates
moments for each output quantity, and uses the moments to determines
the mean, standard deviation, skewness, and excess kurtosis of each distribution. It also
calculates a histogram for each distribution and fits the histogram with two
two-sided Gaussian models: one that is continuous (split-normal), and the other
that is discontinuous centered on the median. The uncertainties shown for
each quantity come from the vegas
integrations.
The last line in
the code above displays the histogram
for fp
, which confirms that it is not particularly Gaussian:
The discussion in Case Study: Bayesian Curve Fitting illustrates how
vegas.PDFIntegrator
can be used with a non-Gaussian PDF in two
examples, one with 4 parameters
and the other with 22 parameters. It also shows how to
use vegas.PDFIntegrator.sample()
to create (weighted)
random samples of parameter points whose density is proportional
to the integrator’s PDF.
Finally, note that the lsqfit
Python module uses vegas.PDFIntegrator
to implement a least-squares
fitter vegas_fit
that uses Bayesian integration (rather than minimization).
Faster Integrands¶
The computational cost of a realistic multidimensional integral
comes mostly from
the cost of evaluating the integrand at the Monte Carlo sample
points. Integrands written in pure Python are probably fast
enough for problems where neval=1e3
or neval=1e4
gives
enough precision. Some problems, however, require
hundreds of thousands or millions of function evaluations, or more.
We can significantly reduce the cost of evaluating the integrand
by using vegas
’s batch mode. For example, replacing
import vegas
import math
def f(x):
dim = len(x)
norm = 1013.2118364296088 ** (dim / 4.)
dx2 = 0.0
for d in range(dim):
dx2 += (x[d] - 0.5) ** 2
return math.exp(-100. * dx2) * norm
integ = vegas.Integrator(4 * [[0, 1]])
integ(f, nitn=10, neval=2e5)
result = integ(f, nitn=10, neval=2e5)
print('result = %s Q = %.2f' % (result, result.Q))
by
import vegas
import numpy as np
@vegas.llbatchintegrand
def f_batch(x):
# evaluate integrand at multiple points simultaneously
dim = x.shape[1]
norm = 1013.2118364296088 ** (dim / 4.)
dx2 = 0.0
for d in range(dim):
dx2 += (x[:, d] - 0.5) ** 2
return np.exp(-100. * dx2) * norm
integ = vegas.Integrator(4 * [[0, 1]])
integ(f_batch, nitn=10, neval=2e5)
result = integ(f_batch, nitn=10, neval=2e5)
print('result = %s Q = %.2f' % (result, result.Q))
reduces the cost of the integral by an order of magnitude. Internally vegas
processes integration points in batches. (vegas
parameter min_neval_batch
determines the number of integration
points per batch (typically 10,000s).) In batch mode,
vegas
presents integration points to the integrand in batches
rather than offering them one at a
time. Here, for example, function f_batch(x)
accepts an array of integration
points — x[i, d]
where i=0...
labels the integration point and
d=0...
the direction — and returns an array of integrand values
corresponding to those points. The decorator
vegas.lbatchintegrand()
tells vegas
that it should send
integration points to f(x)
in batches.
An alternative to a function decorated with vegas.lbatchintegrand()
is
a class that behaves like a batch integrand:
import vegas
import numpy as np
@vegas.lbatchintegrand
class f_batch:
def __init__(self, dim):
self.dim = dim
self.norm = 1013.2118364296088 ** (dim / 4.)
def __call__(self, x):
# evaluate integrand at multiple points simultaneously
dx2 = 0.0
for d in range(self.dim):
dx2 += (x[:, d] - 0.5) ** 2
return np.exp(-100. * dx2) * self.norm
f = f_batch(dim=4)
integ = vegas.Integrator(f.dim * [[0, 1]])
integ(f, nitn=10, neval=2e5)
result = integ(f, nitn=10, neval=2e5)
print('result = %s Q = %.2f' % (result, result.Q))
This version is as fast as the previous batch integrand, but is
potentially more flexible because it is built around a class rather
than a function. (Some classes won’t allow decorators. An alternative
to the decorator is to derive the class from vegas.LBatchIntegrand
.)
The batch integrands here are fast because they are expressed in terms
numpy
operators that act on entire arrays — they evaluate the
integrand for all integration points in a batch at the same time.
That optimization is not always possible or simple.
It is unnecessary if we write the integrand in Cython, which
is a compiled hybrid of Python and C. The Cython version
of the (batch) integrand is:
# file: cython_integrand.pyx
import numpy as np
# use exp from C
from libc.math cimport exp
def f_batch(double[:, ::1] x):
cdef int i # labels integration point
cdef int d # labels direction
cdef int dim = x.shape[1]
cdef double norm = 1013.2118364296088 ** (dim / 4.)
cdef double dx2
cdef double[::1] ans = np.empty(x.shape[0], float)
for i in range(x.shape[0]):
# integrand for integration point x[i]
dx2 = 0.0
for d in range(dim):
dx2 += (x[i, d] - 0.5) ** 2
ans[i] = exp(-100. * dx2) * norm
return ans
We put this in a separate file called, say,
cython_integrand.pyx
, and rewrite the main code as:
import numpy as np
import pyximport
pyximport.install(inplace=True)
import vegas
from cython_integrand import f_batch
f = vegas.lbatchintegrand(f_batch)
integ = vegas.Integrator(4 * [[0, 1]])
integ(f, nitn=10, neval=2e5)
result = integ(f, nitn=10, neval=2e5)
print('result = %s Q = %.2f' % (result, result.Q))
Module pyximport
is used here to cause the Cython
module cython_integrand.pyx
to be compiled the first time
it is imported. The compiled code is used in subsequent
imports, so compilation occurs only once.
Batch mode is also useful for array-valued integrands. The code from the previous section could have been written as:
import vegas
import gvar as gv
import numpy as np
dim = 4
@vegas.lbatchintegrand
def f(x):
ans = np.empty((x.shape[0], 3), float)
dx2 = 0.0
for d in range(dim):
dx2 += (x[:, d] - 0.5) ** 2
ans[:, 0] = np.exp(-200 * dx2)
ans[:, 1] = x[:, 0] * ans[:, 0]
ans[:, 2] = x[:, 0] ** 2 * ans[:, 0]
return ans
integ = vegas.Integrator(4 * [[0, 1]])
# adapt grid
training = integ(f, nitn=10, neval=2000)
# final analysis
result = integ(f, nitn=10, neval=10000)
print('I[0] =', result[0], ' I[1] =', result[1], ' I[2] =', result[2])
print('Q = %.2f\n' % result.Q)
print('<x> =', result[1] / result[0])
print(
'sigma_x**2 = <x**2> - <x>**2 =',
result[2] / result[0] - (result[1] / result[0]) ** 2
)
print('\ncorrelation matrix:\n', gv.evalcorr(result))
Note that the batch index (here :
) always comes first. An extra
(first) index is also added to each value in the dictionary returned
by a dictionary-valued batch integrand: e.g.,
dim = 4
@vegas.lbatchintegrand
def f(x):
ans = {}
dx2 = 0.0
for d in range(dim):
dx2 += (x[:, d] - 0.5) ** 2
ans['1'] = np.exp(-200 * dx2)
ans['x'] = x[:, 0] * ans['1']
ans['x**2'] = x[:, 0] ** 2 * ans['1']
return ans
It is sometimes more convenient to have the batch index be the last index
(the rightmost) rather than the first. Then @vegas.lbatchintegrand
is
replaced by @vegas.rbatchintegrand
, and vegas.LBatchIntegrand
by
vegas.RBatchIntegrand
. (Note that @vegas.batchintegrand
and @vegas.lbatchintegrand
are the same, as are vegas.BatchIntegrand
and vegas.LBatchIntegrand
.)
Multiple Processors¶
vegas
supports parallel evaluation of integrands
on multiple processors. This
can shorten execution time substantially when the integrand is
costly to evaluate. The following code, for example,
runs more than five times faster
when using nproc=8
processors instead of the
default nproc=1
(on a 2019 laptop):
import vegas
import numpy as np
# Integrand: ridge of N Gaussians spread along part of the diagonal
def ridge(x):
N = 10000
x0 = np.linspace(0.4, 0.6, N)
dx2 = 0.0
for xd in x:
dx2 += (xd - x0) ** 2
return np.average(np.exp(-100. * dx2)) * (100. / np.pi) ** (len(x) / 2.)
def main():
integ = vegas.Integrator(4 * [[0, 1]], nproc=8) # 8 processors
# adapt
integ(ridge, nitn=10, neval=1e4)
# final results
result = integ(ridge, nitn=10, neval=1e4)
print('result = %s Q = %.2f' % (result, result.Q))
if __name__ == '__main__':
main()
The code doesn’t run eight times faster because it takes time to
initiate the nproc
processes, and to feed data and
collect results from them.
Parallel processing only becomes useful when integrands are
sufficiently costly that such overheads become negligible.
Parallel processing is managed by Python’s
multiprocessing
module.
The if __name__ == '__main__'
construct at the end of
this code is essential when running on Windows or MacOS (in its
default mode) as it prevents additional processes being launched
when the main module is imported as part of spawning the nproc
processes; see the multiprocessing
documentation for more
details. This is not an issue for Linux/Unix. It is also important
that the integrand and its return values can be pickled using
Python’s pickle
module. This is the case for most pure
Python integrands.
The code above will generate an AttributeError
when run in some
interactive environments (as opposed to running from the command line)
on some platforms. This can usually be fixed by putting the integrand
function ridge(x)
into a file and importing it into the script.
vegas
also supports multi-processor evaluation of integrands using MPI
(via the Python module mpi4py
which must be installed separately).
Placing the code above in a file ridge.py
, with mpi=True
set
in the integrator (and with nproc=1
or unset) —
integ = vegas.Integrator(4 * [[0, 1]], mpi=True)
— ridge.py
can be run on 8 processors using
mpirun -np 8 python ridge.py
The speedup is similar to that from using the
multiprocessing
module, above.
Note that the random number generator used by vegas
must be
synchronized so that it
produces the same random numbers on the different processors. This
happens automatically for the default random-number generator.
vegas
’s batch mode makes it possible to implement other strategies
for distributing integrand evaluations across multiple processors.
For example, we can create a class parallelintegrand
whose function is similar to decorator vegas.batchintegrand()
,
but where Python’s
multiprocessing
module provides parallel processing:
import multiprocessing
import numpy as np
import vegas
class parallelintegrand(vegas.BatchIntegrand):
""" Convert (batch) integrand into multiprocessor integrand.
Integrand should return a numpy array.
"""
def __init__(self, fcn, nproc=4):
" Save integrand; create pool of nproc processes. "
super().__init__()
self.fcn = fcn
self.nproc = nproc
self.pool = multiprocessing.Pool(processes=nproc)
def __del__(self):
" Standard cleanup. "
self.pool.close()
self.pool.join()
def __call__(self, x):
" Divide x into self.nproc chunks, feeding one to each process. "
nx = x.shape[0] // self.nproc + 1
# launch evaluation of self.fcn for each chunk, in parallel
results = self.pool.map(
self.fcn,
[x[i*nx : (i+1)*nx] for i in range(self.nproc)],
1,
)
# convert list of results into a single numpy array
return np.concatenate(results)
Then fparallel = parallelintegrand(f, 4)
, for example, will create a
new integrand fparallel(x)
that uses 4 CPU cores.
Sums with vegas
¶
The code in the previous sections is inefficient in the way it
handles the sum over 10,000 Gaussians. It is not necessary to include every
term in the sum for every integration point. Rather we can sample the sum,
using vegas
to do the sampling. The trick is to replace the sum with
an equivalent integral:
where is the largest
integer smaller than . The
resulting integral can then be handed to vegas
. Using this trick,
the integral in the previous section can be re-cast as a 5-dimensional
integral:
import vegas
import numpy as np
# Integrand: ridge of N Gaussians spread evenly along the diagonal
def ridge(x):
N = 10000
dim = 4
x0 = 0.4 + 0.2 * np.floor(x[-1] * N) / (N - 1.)
dx2 = 0.0
for xd in x[:-1]:
dx2 += (xd - x0) ** 2
return np.exp(-100. * dx2) * (100. / np.pi) ** (dim / 2.)
def main():
integ = vegas.Integrator(5 * [[0, 1]])
# adapt
integ(ridge, nitn=10, neval=5e4)
# final results
result = integ(ridge, nitn=10, neval=5e4)
print('result = %s Q = %.2f' % (result, result.Q))
if __name__ == '__main__':
main()
This code gives a result with the same precision, but is 5x faster
than the code in the previous section (with nproc=1
; it is another
3x faster when nproc=8
).
The same trick can be generalized to sums over multiple indices, including sums
to infinity. vegas
will provide Monte Carlo estimates of the sums, emphasizing
the more important terms.
Saving Results Automatically¶
Results returned by a vegas
integrator can be pickled for later use using pickle.dump/load
(or
gvar.dump/load
) in the usual way. Results can also be saved automatically using the
save
keyword to specify a file name for the pickled result: for example, running
import vegas
import math
def f(x):
dx2 = 0
for d in range(4):
dx2 += (x[d] - 0.5) ** 2
return math.exp(-dx2 * 100.) * 1013.2118364296088
integ = vegas.Integrator([[-2, 2], [0, 2], [0, 2], [0, 2]])
result = integ(f, nitn=10, neval=1000, save='save.pkl')
print(result.summary())
print('result = %s Q = %.2f' % (result, result.Q))
prints out
itn integral wgt average chi2/dof Q
-------------------------------------------------------
1 0.00050(45) 0.00050(45) 0.00 1.00
2 0.115(59) 0.00051(45) 3.77 0.05
3 0.66(22) 0.00051(45) 6.41 0.00
4 0.72(15) 0.00052(45) 12.27 0.00
5 0.97(12) 0.00053(45) 26.17 0.00
6 1.045(61) 0.00059(45) 79.46 0.00
7 1.016(45) 0.00069(45) 152.45 0.00
8 1.023(29) 0.00093(45) 307.32 0.00
9 0.999(20) 0.00141(45) 573.82 0.00
10 1.025(17) 0.00208(45) 896.20 0.00
result = 0.00208(45) Q = 0.00
but also stores result
in file save.pkl
. The result can be retrieved
later using, for example,
import pickle
with open('save.pkl', 'rb') as ifile:
result = pickle.load(ifile)
print(result.summary())
print('result = %s Q = %.2f' % (result, result.Q))
which gives exactly the same output.
This feature is most useful for expensive integrations, ones taking minutes or hours
to complete. This is because the pickled file is updated after every vegas
iteration. This means that a short script like the one above can be used to
monitor progress before the integration is finished. It also means that results
up through the most recent iteration are saved even if the integration is
terminated early or crashes.
Saved results are also useful because they can be fixed after the code has
finished running.
The early iterations in the output above are clearly wrong
and badly distort the weighted average. The problem is that vegas
isn’t well
adapted to the integrand until around the fifth or sixth iteration. We
can discard the first five iterations (from the saved result)
by using function vegas.ravg()
to redo the weighted average:
import pickle
import vegas
with open('save.pkl', 'rb') as ifile:
result = pickle.load(ifile)
result = vegas.ravg(result.itn_results[5:])
print(result.summary())
print('result = %s Q = %.2f' % (result, result.Q))
This gives the following output
itn integral wgt average chi2/dof Q
-------------------------------------------------------
1 1.045(61) 1.045(61) 0.00 1.00
2 1.016(45) 1.026(36) 0.14 0.71
3 1.023(29) 1.024(23) 0.07 0.93
4 0.999(20) 1.010(15) 0.28 0.84
5 1.025(17) 1.017(11) 0.31 0.87
result = 1.017(11) Q = 0.87
which is greatly improved over the original.
It is also possible to save an adapted integrator using pickle.dump/load
(or gvar.dump/load
). This can also be done automatically, by
replacing, for example, save='save.pkl'
with saveall='saveall.pkl'
in the script above. The pickled file then returns a tuple containing
the most recent result
and integ
. Having the (adapted) integrator,
it is possible to further refine a result later: for example, running
import pickle
def f(x):
dx2 = 0
for d in range(4):
dx2 += (x[d] - 0.5) ** 2
return math.exp(-dx2 * 100.) * 1013.2118364296088
with open('saveall.pkl', 'rb') as ifile:
result, integ = pickle.load(ifile)
result = vegas.ravg(result.itn_results[5:])
new_result = integ(f, nitn=5)
print('\nNew results:')
print(new_result.summary())
print('\nCombined results:')
result.extend(new_result)
print(result.summary())
print('Combined result = %s Q = %.2f' % (result, result.Q))
significantly improves the final result by adding 5 additional
iterations
to what was done earlier. The new iterations are in new_result
and tabulated under “New Results” in the output:
New results:
itn integral wgt average chi2/dof Q
-------------------------------------------------------
1 1.007(17) 1.007(17) 0.00 1.00
2 1.006(14) 1.007(11) 0.00 0.98
3 0.997(13) 1.0029(82) 0.16 0.85
4 1.015(21) 1.0044(76) 0.20 0.90
5 0.990(13) 1.0007(66) 0.39 0.82
Combined results:
itn integral wgt average chi2/dof Q
-------------------------------------------------------
1 1.045(61) 1.045(61) 0.00 1.00
2 1.016(45) 1.026(36) 0.14 0.71
3 1.023(29) 1.024(23) 0.07 0.93
4 0.999(20) 1.010(15) 0.28 0.84
5 1.025(17) 1.017(11) 0.31 0.87
6 1.007(17) 1.0136(94) 0.30 0.91
7 1.006(14) 1.0114(78) 0.28 0.95
8 0.997(13) 1.0076(66) 0.36 0.92
9 1.015(21) 1.0082(63) 0.33 0.95
10 0.990(13) 1.0047(57) 0.48 0.89
Combined result = 1.0047(57) Q = 0.89
The new results are merged onto the end of the original results using
result.extend(new_result)
to obtain the combined results from all
10 iterations (old plus new).
Saving integrators is again useful for costly
integrations that might need to be restarted later since the saved integrator
remembers the variable transformations made to minimize errors, and
so need not be readapted to the integrand when used again. The resulting
pickle file can be large, however, particularly if neval
is large.
The (adapted) vegas.AdaptiveMap
integ.map
can also
be pickled by itself and results in a smaller file.
vegas
Maps and Preconditioning vegas
¶
vegas
adapts by remapping the integration variables (see Importance Sampling). It
is possible to precondition this map, before creating a vegas.Integrator
.
Preconditioned maps can improve vegas
results when
much is known about the integrand ahead of time. Consider,
for example, the integral
which has high, narrow peaks at
Given the locations of the peaks we can create a vegas
map before integrating
that emphasizes the regions around them:
import vegas
import numpy as np
@vegas.batchintegrand
def f(x):
ans = 0
for c in [0.45, 0.7]:
dx2 = np.sum((x - c) ** 2, axis=1)
ans += np.exp(-50 * np.sqrt(dx2))
return ans * 247366.171
dim = 5
map = vegas.AdaptiveMap(dim * [[0, 1]]) # uniform map
x = np.concatenate([ # 2000 points near peaks
np.random.normal(loc=0.45, scale=3/50, size=(1000, dim)),
np.random.normal(loc=0.7, scale=3/50, size=(1000, dim)),
])
map.adapt_to_samples(x, f(x), nitn=5) # precondition map
integ = vegas.Integrator(map, alpha=0.)
r = integ(f, neval=1e4, nitn=5)
print(r.summary())
vegas
maps are objects of type vegas.AdaptiveMap
. Here we
create a uniform map object map
. We then
generate 2000 random points x
from normal distributions centered around
the peak locations. map.adapt_to_samples(x, f(x), nitn=5)
optimizes
the map for integrating f(x)
based on information about the integrand
at the random
points x
. As a result, vegas
is almost fully adapted to the
integrand already in its first iteration:
itn integral wgt average chi2/dof Q
-------------------------------------------------------
1 0.981(25) 0.981(25) 0.00 1.00
2 0.983(11) 0.9825(98) 0.00 0.96
3 1.001(10) 0.9915(71) 0.85 0.43
4 1.000(13) 0.9933(63) 0.67 0.57
5 0.993(11) 0.9931(54) 0.50 0.74
We set alpha=0
in the integrator to prevent
further changes to the pre-adapted map.
These results can be contrasted with what happens without preconditioning, where the integrator is still far from converged by the fifth iteration:
itn integral wgt average chi2/dof Q
-------------------------------------------------------
1 0.74(27) 0.74(27) 0.00 1.00
2 0.93(12) 0.90(11) 0.38 0.54
3 0.94(10) 0.920(74) 0.23 0.79
4 1.091(75) 1.004(53) 1.03 0.38
5 1.063(58) 1.031(39) 0.91 0.46
The exact distribution of random points x
isn’t important; what
matters is that they cover and are concentrated in
the dominant regions contributing to the integral. Should it be needed,
a sample xsample=(x, f(x))
in x
-space is easily converted to the equivalent sample in y
-space using:
x, fx = xsample
y = np.empty(x.shape, float)
jac = np.empty(x.shape[0], float)
map.invmap(x, y, jac)
ysample = (y, jac * fx)
where map.invmap(x, y, jac)
fills array y
with the y
-space
points corresponding to x
, and array jac
with the transformation’s
Jacobian.
Note that vegas
maps can be used with integrators other than vegas
.
The vegas
map maps the integration variables x[d]
into
new variables y[d]
(where 0 < y[d] < 1
) that make the integrand
easier to integrate. The integrator is used
to evaluate the integral in y
-space. To illustrate how this
works, we replace the last three lines of the code at the start of
this section with
the following:
def smc(f, neval, dim):
" integrates f(y) over dim-dimensional unit hypercube "
y = np.random.uniform(0,1, (neval, dim))
fy = f(y)
return (np.average(fy), np.std(fy) / neval ** 0.5)
def g(y):
jac = np.empty(y.shape[0], float)
x = np.empty(y.shape, float)
map.map(y, x, jac)
return jac * f(x)
# with map
r = smc(g, 50_000, dim)
print(' SMC + map:', f'{r[0]:.3f} +- {r[1]:.3f}')
# without map
r = smc(f, 50_000, dim)
print('SMC (no map):', f'{r[0]:.3f} +- {r[1]:.3f}')
Here smc
is a Simple Monte Carlo (SMC) integrator and g(y)
is the
integrand in y
-space that corresponds to f(x)
in x
-space.
The mapping is done by map.map(y, x, jac)
which fills the arrays
x
and jac
with the x
values and Jacobian corresponding
to integration points y
. The result is
SMC + map: 1.013 +- 0.020
SMC (no map): 1.703 +- 0.535
where we give results from SMC with and without using
the vegas
map. The vegas
map greatly improves
the SMC result, which, not surprisingly, is significantly
less accurate the the vegas
result above.
Maps for use with other integrators can be built
directly, as above, or they can be built for a particular
integrand by running several iterations of vegas
with
the integrand and using the vegas
integrator’s map: integ.map
.
vegas
Stratifications¶
Having mapped the integration variables x[d]
to new variables y[d]
,
vegas
evaluates the integral in y
-space using stratified Monte
Carlo sampling (see Adaptive Stratified Sampling). This is
done by dividing each axis d
into nstrat[d]
equal stratifications,
which divide the D
-dimensional integration volume into
prod(nstrat)
sub-volumes or (rectangular) hypercubes. vegas
does a
separate integral
in each hypercube, adjusting the number of integrand evaluations
used in each one to minimize errors. By default, the number of
stratifications is set automatically based on the number neval
of integrand evaluations per iteration: nstrat[d]
is set
equal to for the first directions
and for
the remaining directions, where
and are chosen
to maximize the number stratifications consistent with neval
.
It is also possible, however, to specify the number of
stratifications nstrat[d]
for each direction separately.
Requiring (approximately) the same number of stratifications in each direction
greatly limits the number of stratifications in very high
dimensions, since the product of the nstrat[d]
must be
less than neval/2
(so there are at least two integrand
samples in each hypercube). This restricts vegas
’s ability to
adapt. Often there is a subset of integration directions
that are more challenging than the others. In high
dimensions (and possibly lower dimensions)
it is worthwhile using larger values
for nstrat[d]
for those directions.
An example is the 20-dimensional integral
which has high, narrow peaks at
These peaks are aligned along the diagonal of the integration volume for the first five directions, but are on top of each other in the remaining directions. This makes the integrals over the first five directions much more challenging than the remaining integrals.
To integrate this function, we specify nstrat[d]
rather than neval
.
We set nstrat[d]=10
for the first five directions, and nstrat[d]=1
for all the rest. This fixes the number of function evaluations to be
approximately eight times the number of hypercubes created by
the stratifications (i.e., eight times 100,000), which provides enough
integrand samples for adaptive stratified sampling while also
guaranteeing at least two samples per hypercube.
The following code
import vegas
import numpy as np
dim = 20
r = np.array([
5 * [0.23] + (dim - 5) * [0.45],
5 * [0.39] + (dim - 5) * [0.45],
5 * [0.74] + (dim - 5) * [0.45],
])
@vegas.batchintegrand
def f(x):
ans = 0
for ri in r:
dx2 = np.sum((x - ri[None, :]) ** 2, axis=1)
ans += np.exp(-100 * dx2)
return ans * 356047712484621.56
integ = vegas.Integrator(dim * [[0, 1]], alpha=0.25)
nstrat = 5 * [10] + (dim - 5) * [1]
integ(f, nitn=15, nstrat=nstrat) # warmup
r = integ(f, nitn=5, nstrat=nstrat)
print(r.summary())
print('nstrat =', np.array(integ.nstrat))
gives excellent results:
itn integral wgt average chi2/dof Q
-------------------------------------------------------
1 0.920(50) 0.920(50) 0.00 1.00
2 0.993(30) 0.974(26) 1.60 0.21
3 1.014(30) 0.990(20) 1.30 0.27
4 1.099(64) 0.999(19) 1.73 0.16
5 1.0028(78) 1.0023(72) 1.31 0.26
nstrat = [10 10 10 10 10 1 1 1 1 1 1 1 1 1 1 1 1 1 1 1]
vegas
struggles to converge, however,
when in its default mode with the same number of
integrand evaluations (about 800,000 per iteration):
itn integral wgt average chi2/dof Q
-------------------------------------------------------
1 2.1(1.4) 2.1(1.4) 0.00 1.00
2 0.637(42) 0.638(42) 1.14 0.29
3 0.755(53) 0.684(33) 2.06 0.13
4 0.750(49) 0.704(27) 1.78 0.15
5 0.89(11) 0.716(27) 2.07 0.08
nstrat = [2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 2 1 1 1 1]
vegas
Jacobian¶
The vegas
Jacobian is useful when integrating multiple integrands simultaneously when some of the integrands depend only on a subset of the integration variables.
Consider, for example, the integral
We want to compare this integral for two different cases: a) where is a random number near zero drawn from a distribution proportional to with ; and b) where . The result for the first case is where
The result for the second case is
The first two integrals should have very similar dependence on , while the first and third integrals should have similar dependence on . Given these similarities, we would like to do all three integrals simultaneously; but one integral is two-dimensional and the other two are one-dimensional.
One solution is to turn the one-dimensional integrals into two-dimensional integrals by rewriting as
and similarly for . These are then easily integrated together —
import vegas
import numpy as np
integ = vegas.Integrator(2 * [(0., np.pi/2.)])
@vegas.rbatchintegrand
def f(x):
Ia_num = np.exp(-1e2 * np.sin(x[1])) / ((x[0] - x[1])**2 + 0.01)
Ia_den = np.exp(-1e2 * np.sin(x[1])) / (np.pi/2)
Ib = 1 / (x[0]**2 + 0.01) / (np.pi/2)
return dict(Ia_num=Ia_num, Ia_den=Ia_den, Ib=Ib)
w = integ(f, neval=20000, nitn=10)
r = integ(f, neval=20000, nitn=5)
print(r.summary(True))
print('Ia =', r['Ia_num'] / r['Ia_den'], 'Ia - Ib =', r['Ia_num'] / r['Ia_den'] - r['Ib'])
— but the result is useless:
itn integral wgt average chi2/dof Q
-------------------------------------------------------
1 0.160568(44) 0.160568(44) 0.00 1.00
2 0.160551(40) 0.160559(30) 0.13 0.94
3 0.160533(35) 0.160549(23) 0.36 0.90
4 0.160568(32) 0.160555(18) 0.96 0.47
5 0.160553(29) 0.160554(15) 0.73 0.72
key/index value
-----------------------------
Ia_num 0.160554 (15)
Ia_den 0.0099995 (60)
Ib 14.62 (88)
Ia = 16.0562(99) Ia - Ib = 1.43(88)
is almost 100x less accurate than , despite
having an integrand with no x[1]
dependence. The problem is that the
vegas
map for x[1]
is tailored to , which has
a strong peak at x[1]=0
. That map is highly sub-optimal for integrating
a function, like that in , that is independent of x[1]
.
A much better approach is to rewrite the one-dimensional integrals
as two-dimensional integrals using the vegas
map’s Jacobian in the extra direction .
For example,
This turns the integral into an integral over
with an integrand equal to 1. The Monte Carlo integral over
in vegas
is then exact; it is unaffected by the map for that direction.
This is easily implemented for both one-dimensional integrals by setting
keyword uses_jac=True
in the integrator and modifying the integrand:
@vegas.rbatchintegrand
def f(x, jac):
Ia_num = np.exp(-1e2 * np.sin(x[1])) / ((x[0] - x[1])**2 + 0.01)
Ia_den = np.exp(-1e2 * np.sin(x[1])) / jac[0]
Ib = 1 / (x[0]**2 + 0.01) / jac[1]
return dict(Ia_num=Ia_num, Ia_den=Ia_den, Ib=Ib)
w = integ(f, uses_jac=True, neval=20000, nitn=10)
r = integ(f, uses_jac=True, neval=20000, nitn=5)
Setting uses_jac=True
causes vegas
to call the integrand with
two arguments instead of one: f(x, jac=jac)
where jac[d]
is
the Jacobian for direction d
(array jac
has the same shape
as x
). This gives much better results —
itn integral wgt average chi2/dof Q
-------------------------------------------------------
1 0.160568(44) 0.160568(44) 0.00 1.00
2 0.160551(40) 0.160560(30) 0.28 0.84
3 0.160533(35) 0.160547(22) 0.38 0.89
4 0.160568(32) 0.160554(18) 0.58 0.82
5 0.160553(29) 0.160553(15) 0.58 0.86
key/index value
------------------------------
Ia_num 0.160553 (15)
Ia_den 0.01000103 (59)
Ib 15.0749 (11)
Ia = 16.0536(12) Ia - Ib = 0.97868(38)
— where now and are comparable in precision and the difference is quite accurate. Note that the error on the difference is smaller than either of the separate errors, because of correlations between the two results.
vegas
as a Random Number Generator¶
A vegas
integrator generates random points in its integration volume from a
distribution that is optimized for integrals of whatever function it
was trained on. The integrator
provides low-level access to the random-point generator
through the iterators vegas.Integrator.random()
and
vegas.Integrator.random_batch()
.
To illustrate, the following code snippet estimates the integral of function
f(x)
using integrator integ
:
integral = 0.0
for x, wgt in integ.random():
integral += wgt * f(x)
Here x[d]
is a random point in the integration volume and wgt
is the
weight vegas
assigns to that point in an integration. The iterator generates
integration points and weights corresponding to a single iteration of the
vegas
algorithm. In practice, we would train integ
on a function whose
shape is similar to that of f(x)
before using it to estimate the integral
of f(x)
.
It is usually more efficient to generate and use integration points in
batches. The vegas.Integrator.random_batch()
iterator does just
this:
integral = 0.0
for x, wgt in integ.random_batch():
integral += wgt.dot(batch_f(x))
Here x[i, d]
is an array of integration points, wgt[i]
contains the
corresponding weights, and batch_f(x)
returns an array containing the
corresponding integrand values.
The random points generated by vegas
are stratified into hypercubes: vegas
uses transformed integration variables to improve its Monte Carlo
estimates. It further improves those estimates by subdividing the
integration volume in the transformed variables into a large number of
hypercubes, and doing a Monte Carlo integral in each hypercube separately
(see previous section).
The final result is the sum of the results from all the hypercubes.
To mimic a full vegas
integral estimate using the iterators above, we need
to know which points belong to which hypercubes. The following code
shows how this is done:
import gvar as gv
integral = 0.0
variance = 0.0
for x, wgt, hcube in integ.random_batch(yield_hcube=True):
wgt_fx = wgt * batch_f(x)
# iterate over hypercubes: compute variance for each,
# and accumulate for final result
for i in range(hcube[0], hcube[-1] + 1):
idx = (hcube == i) # select array items for h-cube i
nwf = np.sum(idx) # number of points in h-cube i
wf = wgt_fx[idx]
sum_wf = np.sum(wf) # sum of wgt * f(x) for h-cube i
sum_wf2 = np.sum(wf ** 2) # sum of (wgt * f(x)) ** 2
integral += sum_wf
variance += (sum_wf2 * nwf - sum_wf ** 2) / (nwf - 1.)
# answer = integral; standard deviation = variance ** 0.5
result = gv.gvar(integral, variance ** 0.5)
Here hcube[i]
identifies the hypercube containing x[i, d]
. This example is
easily modified to provide information about the vegas
Jacobian to the integrand, if needed
(see the section vegas Jacobian):
integral = 0.0
variance = 0.0
for x, y, wgt, hcube in integ.random_batch(yield_hcube=True, yield_y=True):
wgt_fx = wgt * batch_f(x, jac=integ.map.jac1d(y))
...
The example is also easily modified for integrands batch_f(xd)
where xd
is a dictionary:
integral = 0.0
variance = 0.0
for x, wgt, hcube in integ.random_batch(yield_hcube=True):
xd = gv.BufferDict(integ.xdict, lbatch_buf=x)
wgt_fx = wgt * batch_f(xd)
...
Here integ.xdict
provides the template dictionary
used to construct the
dictionary xd
.
Implementation Notes¶
This implementation relies upon Cython for its speed and
numpy for array processing. It also uses matplotlib
for graphics, h5py
when minimize_mem=True
, and
mpi4py
for MPI support, but these are all optional.
vegas
also uses the gvar
module (pip install gvar
).
Integration results are returned as objects of type
gvar.GVar
, which is a class representing Gaussian
random variables (i.e., something with a mean and standard
deviation). These objects can be combined with numbers and
with each other in arbitrary arithmetic expressions to
get new gvar.GVar
s with the correct standard
deviations, and properly correlated with other
gvar.GVar
s — that is the tricky part.