Python:处理大数字

Question

我需要计算困惑，我试着用来计算

def get_perplexity(test_set, model):
perplexity = 1
n = 0
for word in test_set:
n += 1
perplexity = perplexity * 1 / get_prob(model, word)
perplexity = pow(perplexity, 1/float(n))
return perplexity

经过一些步骤后，我的perplexity等于无穷大。我需要得到号码，作为做pow(perplexity, 1/float(n))的最后一步

有什么可以把数字相乘得到结果吗？

3.887311155784627e+243
8.311806360146177e+250
1.7707049372801292e+263
1.690802669602979e+271
3.843294667766984e+278
5.954424789834101e+290
8.859529887856071e+295
7.649470766862909e+306

Answer 1

重复乘法会导致一些棘手的数字不稳定，因为乘法的结果需要越来越多的位来表示。我建议你将其转换为对数空间，并使用求和而不是乘法：

import math
def get_perplexity(test_set, model):
log_perplexity = 0
n = 0
for word in test_set:
n += 1
log_perplexity -= math.log(get_prob(model, word))
log_perplexity /= float(n)
return math.exp(log_perplexity)

这样，你所有的对数都可以用标准位数表示，而且你不会得到数字放大和精度损失。此外，您可以使用decimal模块引入任意精度：

import decimal
def get_perplexity(test_set, model):
with decimal.localcontext() as ctx:
ctx.prec = 100  # set as appropriate
log_perplexity = decimal.Decimal(0)
n = 0
for word in test_set:
n += 1
log_perplexity -= decimal.Decimal(get_prob(model, word))).ln()
log_perplexity /= float(n)
return log_perplexity.exp()

Answer 2

由于e+306仅为10^306，您可以将两部分组成的类

class BigPowerFloat:
POWER_STEP = 10**100
def __init__(self, input_value):
self.value = float(input_value)
self.power = 0
def _move_to_power(self):
while self.value > self.POWER_STEP:
self.value = self.value / self.POWER_STEP
self.power += self.POWER_STEP
# you can add similar for negative values           

def __mul__(self, other):
self.value *= other
self._move_to_power()
# TODO other __calls for /, +, - ...
def __str__(self):
pass
# make your cust to str