我应该如何处理浮点数，这些数字可能变得很小以至于变成零

Question

所以我只是修复了以下代码中的一个有趣的错误，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，所以我通过简单地保留最小的非零值来解决这个问题，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)

这可行，但对我来说似乎有点笨拙。我不会进行大量此类数值编程，并且我不确定这是否是人们使用的修复方法，或者是否有更好的东西我可以寻求。

Answer 1

既然，math.log(a * b) 等于 math.log(a) + math.log(b)，为什么不取对数之和probabilities 数组的所有成员？

这将避免 p 变得太小而导致下溢的问题。

编辑：这是 numpy 版本，对于大型数据集来说更干净且速度更快：

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