我应该如何处理那些可以变得很小以至于变成零的浮动数字 [英] How should I deal with floating numbers that numbers that can get so small that the become zero

问题描述

所以我只是在下面的代码中修复了一个有趣的bug，但是我不确定这个方法是否最好：

 <$ c 
 b = b 
 
 
 
 
 
 
 
 $ b $（如果（wp > 0）：
p * = wp 
 
＃取自然对数，当'probabilites'足够长，p结束
＃为零
尝试：
 result = math.log（p）
  

因为结果不需要确切地说，我解决了这个问题，只需保持最小的非零值，并使用它，如果p永远变成0。

 如果（wp> 0）：
 
 
 
 
 
 
 
 
 
 $ b old_p = p 
p * = wp 
如果p == 0：
＃我们已经变得如此小，它只是0，所以回到最小
＃非零，我们有
p = old_p 
 bre ak 
 
＃以自然记录为例，当'probabilites'足够长以至于p结束时
＃为零
 try：
 result = math.log （p）
  

这个方法可行，但对我来说似乎有点不方便。我不会做这么多的数字编程，我不确定这是否是人们使用的修复方法，或者是有更好的方法可以去做。因为 math.log（a * b）是相等的，所以

解决方案到 math.log（a）+ math.log（b），为什么不把所有 code> array？

这样可以避免 p 变小的问题。

编辑：这是numpy版本，对于大型数据集来说它更干净，速度更快：

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

So I just fixed an interesting bug in the following code, but I'm not sure the approach I took it the best:

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)

Because the result doesn't need to be exact, I solved this by simply keeping the smallest non-zero value, and using that if p ever becomes 0.

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)

This works, but it seems a bit kludgy to me. I don't do a ton of this kind of numerical programming, and I'm not sure if this is the kind of fix people use, or if there is something better I can go for.

解决方案

Since, math.log(a * b) is equal to math.log(a) + math.log(b), why not take a sum of the logs of all members of the probabilities array?

This will avoid the problem of p getting so small it under-flows.

Edit: this is the numpy version, which is cleaner and a lot faster for large data sets:

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

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