Numercially stable softmax

问题:

Is there a numerically stable way to compute softmax function below? I am getting values that becomes Nans in Neural network code.

np.exp(x)/np.sum(np.exp(y))

回答1:

The softmax exp(x)/sum(exp(x)) is actually numerically well-behaved. It has only positive terms, so we needn't worry about loss of significance, and the denominator is at least as large as the numerator, so the result is guaranteed to fall between 0 and 1.

The only accident that might happen is over- or under-flow in the exponentials. Overflow of a single or underflow of all elements of x will render the output more or less useless.

But it is easy to guard against that by using the identity softmax(x) = softmax(x + c) which holds for any scalar c: Subtracting max(x) from x leaves a vector that has only non-positive entries, ruling out overflow and at least one element that is zero ruling out a vanishing denominator (underflow in some but not all entries is harmless).

Footnote: theoretically, catastrophic accidents in the sum are possible, but you'd need a ridiculous number of terms. For example, even using 16 bit floats which can only resolve 3 decimals---compared to 15 decimals of a "normal" 64 bit float---we'd need between 2^1431 (~6 x 10^431) and 2^1432 to get a sum that is off by a factor of two.

回答2:

Softmax function is prone to two issues: overflow and underflow

Overflow: It occurs when very large numbers are approximated as infinity

Underflow: It occurs when very small numbers (near zero in the number line) are approximated (i.e. rounded to) as zero

To combat these issues when doing softmax computation, a common trick is to shift the input vector by subtracting the maximum element in it from all elements. For the input vector x, define z such that:

z = x-max(x)

And then take the softmax of the new (stable) vector z

---

Example:

In [266]: def stable_softmax(x):
     ...:     z = x - max(x)
     ...:     numerator = np.exp(z)
     ...:     denominator = np.sum(numerator)
     ...:     softmax = numerator/denominator
     ...:     return softmax
     ...: 

In [267]: vec = np.array([1, 2, 3, 4, 5])

In [268]: stable_softmax(vec)
Out[268]: array([ 0.01165623,  0.03168492,  0.08612854,  0.23412166,  0.63640865])

In [269]: vec = np.array([12345, 67890, 99999999])

In [270]: stable_softmax(vec)
Out[270]: array([ 0.,  0.,  1.])

---

For more details, see chapter Numerical Computation in deep learning book.

回答3:

Thank Paul Panzer's explanation, but I am wondering why we need to subtract max(x). Therefore, I found more detailed information and hope it will be helpful to the people who has the same question as me. See the section, "What’s up with that max subtraction?", in the following link's article.

https://nolanbconaway.github.io/blog/2017/softmax-numpy

回答4:

There is nothing wrong with calculating the softmax function as it is in your case. The problem seems to come from exploding gradient or this sort of issues with your training methods. Focus on those matters with either "clipping values" or "choosing the right initial distribution of weights".

回答5:

Extending @kmario23's answer to support 1 or 2 dimensional numpy arrays or lists (common if you're passing a batch of results through the softmax function):

import numpy as np


def stable_softmax(x):
    z = x - np.max(x, axis=-1, keepdims=True)
    numerator = np.exp(z)
    denominator = np.sum(numerator, axis=-1, keepdims=True)
    softmax = numerator / denominator
    return softmax


test1 = np.array([12345, 67890, 99999999])  # 1D
test2 = np.array([[12345, 67890, 99999999], [123, 678, 88888888]])  # 2D
test3 = [12345, 67890, 999999999]
test4 = [[12345, 67890, 999999999]]

print(stable_softmax(test1))
print(stable_softmax(test2))
print(stable_softmax(test3))
print(stable_softmax(test4))

 [0. 0. 1.]

[[0. 0. 1.]
 [0. 0. 1.]]

 [0. 0. 1.]

[[0. 0. 1.]]