我正在进行概率计算。我有很多非常小的数字，我想从1中减去所有数字，并且准确地这样做。我可以准确地计算出这些小数的对数。到目前为止我的策略是这样的（使用numpy）：
给定一个小数字x的日志数组，计算：
y = numpy.logaddexp.reduce(x)
现在我想计算1-exp(y)甚至更好log(1-exp(y))之类的内容，但我不确定如何在不失去所有精确度的情况下这样做。
事实上，即使是logaddexp函数也会遇到精度问题。向量x中的值可以在-2到-800之间，甚至更负。上面的向量y基本上会有1e-16左右的整数数字，这是数据类型的eps。因此，例如，准确计算的数据可能如下所示：
In [358]: x
Out[358]:
[-5.2194676211172837,
-3.9050377656308362,
-3.1619783292449615,
-2.71289594096134,
-2.4488395891021639,
-2.3129210706827568,
-2.2709987626652346,
-2.3007776073511259,
-2.3868404149802434,
-2.5180718876609163,
-2.68619816583087,
-2.8849022632856958,
-3.1092603032627686,
-3.3553673369747834,
-3.6200806272462351,
-3.9008385919463073,
-4.1955300857178379,
-4.5023981074719899,
-4.8199676154248081,
-5.1469905756384904,
-5.4824035553480428,
-5.8252945959126876,
-6.174877049340779,
-6.5304687083067563,
-6.8914750074202473,
-7.25737538919104,
-7.6277121540338797,
-8.0020812775389558,
-8.3801247986220773,
-8.7615244716292437,
-9.1459964426584435,
-9.5332867613176404,
-9.9231675781398394,
-10.315433907978701,
-10.709900863130784,
-11.106401278287066,
-11.50478366390567,
-11.904910436107656,
-12.30665638039909,
-12.709907313918777,
-13.114558916892051,
-13.52051570882999,
-13.927690148982549,
-14.336001843810081,
-14.745376846921289,
-15.155747039147968,
-15.567049578271309,
-15.979226409456359,
-16.39222382873956,
-16.805992092998878,
-17.22048507074976,
-17.63565992888303,
-18.051476851117201,
-18.467898784496384,
-18.884891210740903,
-19.302421939667397,
-19.720460922243518,
-20.138980081145718,
-20.557953156947775,
-20.977355568292495,
-21.397164284594595,
-21.817357709992422,
-22.237915577412224,
-22.658818851739369,
-23.080049641202237,
-23.501591116172762,
-23.923427434676114,
-24.345543673975158,
-24.767925767665417,
-25.190560447772668,
-25.61343519140047,
-26.036538171518259,
-26.459858211524278,
-26.883384743252066,
-27.307107768123842,
-27.731017821180984,
-28.155105937748402,
-28.579363622513654,
-29.003782820820732,
-29.428355891997484,
-29.853075584553352,
-30.27793501309668,
-30.702927636836705,
-31.128047239545907,
-31.553287910869187,
-31.978644028878307,
-32.404110243774596,
-32.82968146265631,
-33.255352835270173,
-33.681119740674262,
-34.106977774747804,
-34.532922738484046,
-34.958950627012712,
-35.385057619298891,
-35.811240068471022,
-36.237494492735493,
-36.663817566835519,
-37.090206114019054,
-37.516657098479527,
-37.943167618239784,
-38.369734898447348,
-38.796356285056333,
-39.223029238868548,
-39.64975132991276,
-40.076520232137909,
-40.5033337184027,
-40.930189655741344,
-41.357086000888444,
-41.784020796047173,
-42.210992164885965,
-42.637998308748706,
-43.065037503066776,
-43.492108093959985,
-43.919208495015312,
-44.346337184233221,
-44.773492701130749,
-45.200673643993753,
-45.627878667267964,
-46.055106479082156,
-46.482355838895614,
-46.909625555262096,
-47.336914483704675,
-47.764221524695017,
-48.191545621730768,
-48.618885759506213,
-49.04624096217151,
-49.473610291673936,
-49.900992846179292,
-50.328387758566748,
-50.755794194994508,
-51.183211353532613,
-51.610638462858901,
-52.0380747810147,
-52.46551959421754,
-52.892972215728378,
-53.320431984769073,
-53.747898265489198,
-54.175370445978274,
-54.602847937323247,
-55.030330172705362,
-55.457816606538813,
-55.885306713645889,
-56.312799988467418,
-56.740295944308855,
-57.167794112617116,
-57.59529404228897,
-58.02279529900909,
-58.450297464615232,
-58.877800136490578,
-59.305302926981085,
-59.732805462838542,
-60.160307384683506,
-60.587808346493375,
-61.015308015110463,
-61.442806069768608,
-61.87030220164138,
-62.297796113406662,
-62.725287518829532,
-63.15277614236129,
-63.580261718755196,
-64.007743992695964,
-64.435222718445743,
-64.862697659501919,
-65.290168588270035,
-65.717635285748088,
-66.14509754122389,
-66.572555151982783,
-67.000007923029216,
-67.427455666815376,
-67.854898202982099,
-68.282335358110231,
-68.709766965479957,
-69.137192864839108,
-69.564612902180784,
-69.992026929530198,
-70.419434804735829,
-70.8468363912732,
-71.274231558051156,
-71.701620179229167,
-72.129002134037705,
-72.556377306608397,
-72.983745585807242,
-73.411106865077045,
-73.838461042282461,
-74.265808019561746,
-74.693147703185559,
-75.120480003416901,
-75.547804834380145,
-75.97512211393132,
-76.402431763534764,
-76.829733708143749,
-77.257027876085431,
-77.684314198948414,
-78.111592611476681,
-78.538863051464546,
-78.966125459656723,
-79.393379779652037,
-79.820625957809625,
-80.24786394315754,
-80.675093687306912,
-81.102315144366912]
然后我尝试计算指数的对数总和：
In [359]: np.logaddexp.accumulate(x)
Out[359]:
array([ -5.21946762e+00, -3.66710221e+00, -2.68983273e+00,
-2.00815067e+00, -1.51126604e+00, -1.14067818e+00,
-8.60829425e-01, -6.48188808e-01, -4.86276416e-01,
-3.63085873e-01, -2.69624488e-01, -1.99028599e-01,
-1.45996863e-01, -1.06408884e-01, -7.70565672e-02,
-5.54467248e-02, -3.96506186e-02, -2.81859503e-02,
-1.99225261e-02, -1.40061296e-02, -9.79701394e-03,
-6.82045164e-03, -4.72733966e-03, -3.26317960e-03,
-2.24396350e-03, -1.53767347e-03, -1.05026994e-03,
-7.15209142e-04, -4.85690052e-04, -3.28980607e-04,
-2.22305294e-04, -1.49890553e-04, -1.00858788e-04,
-6.77380054e-05, -4.54139175e-05, -3.03974537e-05,
-2.03154477e-05, -1.35581905e-05, -9.03659252e-06,
-6.01552344e-06, -3.99984336e-06, -2.65671945e-06,
-1.76283376e-06, -1.16860435e-06, -7.73997496e-07,
-5.12213574e-07, -3.38706792e-07, -2.23809375e-07,
-1.47785898e-07, -9.75226648e-08, -6.43149957e-08,
-4.23904687e-08, -2.79246430e-08, -1.83858489e-08,
-1.20995365e-08, -7.95892319e-09, -5.23300609e-09,
-3.43929670e-09, -2.25953475e-09, -1.48391255e-09,
-9.74194956e-10, -6.39351406e-10, -4.19466218e-10,
-2.75121795e-10, -1.80397409e-10, -1.18254918e-10,
-7.74993004e-11, -5.07775611e-11, -3.32619009e-11,
-2.17835737e-11, -1.42634249e-11, -9.33764336e-12,
-6.11190167e-12, -3.99989955e-12, -2.61737204e-12,
-1.71253165e-12, -1.12043465e-12, -7.33052079e-13,
-4.79645919e-13, -3.13905885e-13, -2.05519681e-13,
-1.34650094e-13, -8.83173582e-14, -5.80300378e-14,
-3.82338678e-14, -2.52963381e-14, -1.68421145e-14,
-1.13181549e-14, -7.70918073e-15, -5.35155125e-15,
-3.81152630e-15, -2.80565548e-15, -2.14872312e-15,
-1.71971577e-15, -1.43957518e-15, -1.25665732e-15,
-1.13722927e-15, -1.05925916e-15, -1.00835857e-15,
-9.75131524e-16, -9.53442707e-16, -9.39286186e-16,
-9.30046550e-16, -9.24016349e-16, -9.20080954e-16,
-9.17512772e-16, -9.15836886e-16, -9.14743318e-16,
-9.14029759e-16, -9.13564174e-16, -9.13260398e-16,
-9.13062204e-16, -9.12932898e-16, -9.12848539e-16,
-9.12793505e-16, -9.12757603e-16, -9.12734183e-16,
-9.12718905e-16, -9.12708939e-16, -9.12702438e-16,
-9.12698198e-16, -9.12695432e-16, -9.12693627e-16,
-9.12692451e-16, -9.12691683e-16, -9.12691183e-16,
-9.12690856e-16, -9.12690643e-16, -9.12690504e-16,
-9.12690414e-16, -9.12690355e-16, -9.12690316e-16,
-9.12690291e-16, -9.12690275e-16, -9.12690264e-16,
-9.12690257e-16, -9.12690252e-16, -9.12690249e-16,
-9.12690248e-16, -9.12690246e-16, -9.12690245e-16,
-9.12690245e-16, -9.12690245e-16, -9.12690244e-16,
-9.12690244e-16, -9.12690244e-16, -9.12690244e-16,
-9.12690244e-16, -9.12690244e-16, -9.12690244e-16,
-9.12690244e-16, -9.12690244e-16, -9.12690244e-16,
-9.12690244e-16, -9.12690244e-16, -9.12690244e-16,
-9.12690244e-16, -9.12690244e-16, -9.12690244e-16,
-9.12690244e-16, -9.12690244e-16, -9.12690244e-16,
-9.12690244e-16, -9.12690244e-16, -9.12690244e-16,
-9.12690244e-16, -9.12690244e-16, -9.12690244e-16,
-9.12690244e-16, -9.12690244e-16, -9.12690244e-16,
-9.12690244e-16, -9.12690244e-16, -9.12690244e-16,
-9.12690244e-16, -9.12690244e-16, -9.12690244e-16,
-9.12690244e-16, -9.12690244e-16, -9.12690244e-16,
-9.12690244e-16, -9.12690244e-16, -9.12690244e-16,
-9.12690244e-16, -9.12690244e-16, -9.12690244e-16,
-9.12690244e-16, -9.12690244e-16, -9.12690244e-16,
-9.12690244e-16, -9.12690244e-16, -9.12690244e-16,
-9.12690244e-16, -9.12690244e-16, -9.12690244e-16,
-9.12690244e-16, -9.12690244e-16, -9.12690244e-16,
-9.12690244e-16, -9.12690244e-16, -9.12690244e-16])
最终导致：
In [360]: np.logaddexp.reduce(x)
Out[360]: -9.1269024387687033e-16
所以我的精确度已经消失了。关于如何解决这个问题的任何想法？
答案 0 :(得分：7)
在Python 2.7中，我们为此用例添加了math.expm1()：
>>> from math import exp, expm1
>>> exp(1e-5) - 1 # gives result accurate to 11 places
1.0000050000069649e-05
>>> expm1(1e-5) # result accurate to full precision
1.0000050000166668e-05
此外，总和步骤math.fsum()不会损失精度：
>>> sum([.1, .1, .1, .1, .1, .1, .1, .1, .1, .1])
0.9999999999999999
>>> fsum([.1, .1, .1, .1, .1, .1, .1, .1, .1, .1])
1.0
最后，如果没有别的帮助，你可以使用支持超高精度算术的decimal module：
>>> from decimal import *
>>> getcontext().prec = 200
>>> (1 - 1 / Decimal(7000000)).ln()
Decimal('-1.4285715306122546161332083855139723669559469615692284955124609122046580004888309867906750714454869716398919778588515625689415322789136206397998627088895481989036005482451668027002380442299229191323673E-7')
答案 1 :(得分：4)
我不太了解python，我用Java做大部分工作但在我看来，你最好先同时对所有值执行log-sum-exp技巧，而不是两次 - by-two使用 numpy.logaddexp.accumulate 。
谷歌中的快速搜索显示了python库中的候选者： scipy.misc.logsumexp 。
在任何情况下，自己编程并不困难：
logsumexp(probs) := max(probs) + log(sum[i](exp(probes[i]-max(probs))))
如下所示：
maxValue = -Inf;
for x in probs
if x > maxValue then maxValue = x
expSum = 0
for x in probs
expSum += exp(x - maxValue)
return log(expSum)
返回的单个值，比如 p ，只是 probs 中所有概率总和的对数。请注意，如果输入概率中的最大值和较小值之间存在大的不同，那么较小的值将被忽略，但是与大多数应用中的大数量相比，它们的贡献非常小。
你可以使用更复杂的策略来计算那些小值，以防有大量的小数字，例如probe = 0.5 + 1E-7 + 1E-7 ......数百万次，因此加起来为0.1。您可以做的是将单个和除以几个，其中大致相同比例的值在组合之前首先加在一起。或者你可以使用一些中间的枢轴值而不是最大值，但你必须确保最大值不是太大，因为那时exp（probs [i] - pivot）会导致溢出。
完成此操作后，您仍需要计算 log（1-exp（p））
为此，我发现这篇文档描述了一种方法，可以使用大多数语言数学公共库中可以找到的逻辑函数来尽可能地避免精度损失。
Maechler M, Accurately Computing log(1 − exp(− |a|)) Assessed by the Rmpfr package, 2012
关键是根据输入值 a 使用两种可能的方法之一。
说明：
log1mexp(a) := log(1-exp(a)) ### function that we seek to implement.
log1p(a) := log(1+a) # function provided by common math libraries.
exp1m(a) := exp(a) - 1 # function provided by common math libraries.
使用log1p实现log1mexp有一种显而易见的方法：
log1mexp(a) := log1p(-exp(a))
使用exp1m，您可以：
log1mexp(a) := log(-expm1(a))
然后，当 a＆lt; 时，您应该使用log1p方法。当 a＆gt; = log（.5）时，记录（.5）和expm1。
log1mexp(a) := (a < log(.5)) ? log1p(-exp(a)) : log(-expm1(a)).
有关详细信息，请参阅外部链接。
答案 2 :(得分：3)
我建议相应地将exp（）和log（）替换为0和1附近的Taylor series。通过这种方式，你不会因使用大数字而失去精确度（我的，我只称1个大数字：^）。使用Lagrange remainder公式或仅使用成员的表达式和一些保留来确定差异何时超出精确度。
更新：
Python 2.7的math.expm1（exp(x)-1）和math.log1p（log(1+x)）如果平台的C库的精度（通常为double）足够，那么就为您执行此操作。 （如果没有，你将不得不使用特殊的数学软件（x86的FPU可以扩展精度计算）。）
答案 3 :(得分：2)
我不确定这是否是你想要的
numpy.expm1(x[, out])
Calculate exp(x) - 1 for all elements in the array.
>>> import numpy as np
>>> np.expm1(x).sum()
-200.0
>>> (-np.expm1(x)).sum()
200.0
>>> from scipy import special
>>> (-special.expm1(x)).sum()
200.0
>>> np.log((-special.expm1(x)).sum())
5.2983173665480363
编辑：
很抱歉，我没有意识到这只是雷蒙德·海廷格答案的笨拙版本。
（不是数字问题的答案）
我仍然不确定究竟是什么问题，但是，不是扔十进制或mpmath，也许重新解决问题会有所帮助。例如，如果你在Poisson中加上概率，你最终总是会“接近”1.但对于某些问题，我们可以避免使用生存函数而不是cdf的一些问题。
答案 4 :(得分：1)
答案 5 :(得分：0)
正如Raymond Hettinger所说的那样，但事后你需要乘以-1，因为你想要1 - exp而不是exp - 1