Python 的 numpy.exp 函数中的溢出错误
声明：本页面是StackOverFlow热门问题的中英对照翻译，遵循CC BY-SA 4.0协议，如果您需要使用它，必须同样遵循CC BY-SA许可，注明原文地址和作者信息，同时你必须将它归于原作者（不是我）：StackOverFlow
原文地址: http://stackoverflow.com/questions/40726490/
Warning: these are provided under cc-by-sa 4.0 license. You are free to use/share it, But you must attribute it to the original authors (not me):
StackOverFlow
Overflow Error in Python's numpy.exp function
提问by Fcoder
I want to use numpy.explike this:
我想这样使用numpy.exp：
cc = np.array([
[0.120,0.34,-1234.1]
])
print 1/(1+np.exp(-cc))
But this gives me error:
但这给了我错误：
/usr/local/lib/python2.7/site-packages/ipykernel/__main__.py:5: RuntimeWarning: overflow encountered in exp
I can't understand why? How can I fix this? It seems the problem is with third number (-1234.1)
我不明白为什么？我怎样才能解决这个问题？似乎问题出在第三个数字上(-1234.1)
回答by Praveen
As fuglede says, the issue here is that np.float64can't handle a number as large as exp(1234.1). Try using np.float128instead:
正如 fuglede 所说，这里的问题是np.float64无法处理像exp(1234.1). 尝试使用np.float128：
>>> cc = np.array([[0.120,0.34,-1234.1]], dtype=np.float128)
>>> cc
array([[ 0.12, 0.34, -1234.1]], dtype=float128)
>>> 1 / (1 + np.exp(-cc))
array([[ 0.52996405, 0.58419052, 1.0893812e-536]], dtype=float128)
Note however, that there are certain quirks with using extended precision. It may not work on Windows; you don't actually get the full 128 bits of precision; and you might lose the precision whenever the number passes through pure python. You can read more about the details here.
但是请注意，使用扩展精度存在某些怪癖。它可能不适用于 Windows；您实际上并没有获得完整的 128 位精度；每当数字通过纯 python 时，您可能会失去精度。您可以在此处阅读有关详细信息的更多信息。
For most practical purposes, you can probably approximate 1 / (1 + <a large number>)to zero. That is to say, just ignore the warning and move on. Numpy takes care of the approximation for you (when using np.float64):
对于大多数实际用途，您可能可以近似1 / (1 + <a large number>)为零。也就是说，只需忽略警告并继续前进。Numpy 为您处理近似值（使用时np.float64）：
>>> 1 / (1 + np.exp(-cc))
/usr/local/bin/ipython3:1: RuntimeWarning: overflow encountered in exp
#!/usr/local/bin/python3.4
array([[ 0.52996405, 0.58419052, 0. ]])
If you want to suppress the warning, you could use scipy.special.expit, as suggested by WarrenWeckesser in a comment to the question:
如果你想抑制警告，你可以使用scipy.special.expit，正如 WarrenWeckesser 在对问题的评论中所建议的那样：
>>> from scipy.special import expit
>>> expit(cc)
array([[ 0.52996405, 0.58419052, 0. ]])
回答by fuglede
The largest value representable by a numpyfloat is 1.7976931348623157e+308, whose logarithm is about 709.782, so there is no way to represent np.exp(1234.1).
numpy浮点数可表示的最大值为 1.7976931348623157e+308，其对数约为 709.782，因此无法表示np.exp(1234.1)。
In [1]: import numpy as np
In [2]: np.finfo('d').max
Out[2]: 1.7976931348623157e+308
In [3]: np.log(_)
Out[3]: 709.78271289338397
In [4]: np.exp(709)
Out[4]: 8.2184074615549724e+307
In [5]: np.exp(710)
/usr/local/bin/ipython:1: RuntimeWarning: overflow encountered in exp
#!/usr/local/bin/python3.5
Out[5]: inf
回答by jmd_dk
A possible solution is to use the decimalmodule, which lets you work with arbitrary precision floats. Here is an example where a numpyarray of floats with 100 digits precision is used:
一个可能的解决方案是使用该decimal模块，它可以让您使用任意精度的浮点数。下面是一个示例，其中numpy使用了 100 位精度的浮点数数组：
import numpy as np
import decimal
# Precision to use
decimal.getcontext().prec = 100
# Original array
cc = np.array(
[0.120,0.34,-1234.1]
)
# Fails
print(1/(1 + np.exp(-cc)))
# New array with the specified precision
ccd = np.asarray([decimal.Decimal(el) for el in cc], dtype=object)
# Works!
print(1/(1 + np.exp(-ccd)))
回答by user1559897
exp(-1234.1) is too small for 32bit or 64bit floating-point numbers. Since it cannot be represented, numpy produces the correct warning.
exp(-1234.1) 对于 32 位或 64 位浮点数来说太小了。由于无法表示，numpy 会产生正确的警告。
Using IEEE 754 32bit floating-pointnumbers, the smallest positive number it can represent is 2^(-149), which is roughly 1e-45.
使用IEEE 754 32bit floating-point数字，它可以表示的最小正数是2^(-149)，大约是 1e-45。
If you use IEEE 754 64 bit floating-pointnumbers, the smallest positive number is 2^(-1074)which is roughy 1e-327.
如果使用IEEE 754 64 bit floating-point数字，则最小的正数2^(-1074)是粗略的 1e-327。
In either case, it cannot represent a number as small as exp(-1234.1) which is about 1e-535.
在任何一种情况下，它都不能表示像 exp(-1234.1) 这样小的数字，大约为 1e-535。
You should be using the expitfunction from scipy to compute the sigmoid function. This would give you better precision.
您应该使用expitscipy 中的函数来计算 sigmoid 函数。这会给你更好的精度。
For practical purposes, exp(-1234.1) is a very small number. If rounding to zero makes sense in your use case, numpy produces benign results by rounding it to zero.
出于实际目的，exp(-1234.1) 是一个非常小的数字。如果在您的用例中舍入为零是有意义的，则 numpy 通过将其舍入为零来产生良性结果。
