floating-point - 我应该如何处理 float ，这些数字可以变得如此之小以至于变成零

Question

所以我只是在下面的代码中修复了一个有趣的错误，但我不确定我采用的最好的方法:

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，我通过简单地保留最小的非零值来解决这个问题，并在 p 变为 0 时使用它。

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)

这有效，但对我来说似乎有点笨拙。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，或者是否有更好的东西我可以去。

Answer 1

从，math.log(a * b)等于 math.log(a) + math.log(b) ，为什么不把probabilities的所有成员的日志求和？大批？

这样就避免了p的问题变得如此之小以至于流量不足。

编辑:这是 numpy 版本，对于大型数据集，它更干净且速度更快:

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