Question

所以我在下面的代码中修复了一个有趣的错误，但我不确定我采取的最佳方法：

p = 1
probabilities = [ ... ] # a (possibly) long list of numbers between 0 and 1
for wp in probabilities:

  if (wp > 0):
    p *= wp

# Take the natural log, this crashes when 'probabilites' is long enough that p ends up
# being zero
try:
    result = math.log(p)

因为结果不需要精确，我通过简单地保持最小的非零值来解决这个问题，并且如果p变为0则使用它。

p = 1
probabilities = [ ... ] # a long list of numbers between 0 and 1
for wp in probabilities:

  if (wp > 0):
    old_p = p
    p *= wp
    if p == 0:
      # we've gotten so small, its just 0, so go back to the smallest
      # non-zero we had
      p = old_p
      break

# Take the natural log, this crashes when 'probabilites' is long enough that p ends up
# being zero
try:
    result = math.log(p)

这很有效，但对我来说似乎有些愚蠢。我不做大量的这种数值编程，我不确定这是否是人们使用的那种修复方式，或者我是否可以选择更好的方法。

Answer 1

由于math.log(a * b)等于math.log(a) + math.log(b)，为什么不取一个probabilities数组所有成员的日志总和？

这样可以避免p在流量不足时出现这么小的问题。

编辑：这是numpy版本，对于大型数据集来说更干净，速度更快：

import numpy
prob = numpy.array([0.1, 0.213, 0.001, 0.98 ... ])
result = sum(numpy.log(prob))

我应该如何处理浮动数字，这些数字可以变得如此之小以至于变为零

1 个答案: