Softmax function of a numpy array by row

问题:

I am trying to apply a softmax function to a numpy array. But I am not getting the desired results. This is the code I have tried:

 import numpy as np
 x = np.array([[1001,1002],[3,4]])
 softmax = np.exp(x - np.max(x))/(np.sum(np.exp(x - np.max(x)))
 print softmax

I think the x - np.max(x) code is not subtracting the max of each row. The max needs to be subtracted from x to prevent very large numbers.

This is supposed to output

 np.array([
    [0.26894142, 0.73105858],
    [0.26894142, 0.73105858]])

But I am getting:

np.array([
    [0.26894142, 0.73105858],
    [0, 0]])

回答1:

A convenient way to keep the axes that are consumed by "reduce" operations such as max or sum is the keepdims keyword:

mx = np.max(x, axis=-1, keepdims=True)
mx
# array([[1002],
#        [   4]])
x - mx
# array([[-1,  0],
#        [-1,  0]])
numerator = np.exp(x - mx)
denominator = np.sum(numerator, axis=-1, keepdims=True)
denominator
# array([[ 1.36787944],
#        [ 1.36787944]])
numerator/denominator
# array([[ 0.26894142,  0.73105858],
         [ 0.26894142,  0.73105858]])

回答2:

EDIT. As of version 1.2.0, scipy includes softmax as a special function:

https://scipy.github.io/devdocs/generated/scipy.special.softmax.html

I wrote a very general softmax function operating over an arbitrary axis, including the tricky max subtraction bit. The function is below, and I wrote a blog post about it here.

def softmax(X, theta = 1.0, axis = None):
    """
    Compute the softmax of each element along an axis of X.

    Parameters
    ----------
    X: ND-Array. Probably should be floats. 
    theta (optional): float parameter, used as a multiplier
        prior to exponentiation. Default = 1.0
    axis (optional): axis to compute values along. Default is the 
        first non-singleton axis.

    Returns an array the same size as X. The result will sum to 1
    along the specified axis.
    """

    # make X at least 2d
    y = np.atleast_2d(X)

    # find axis
    if axis is None:
        axis = next(j[0] for j in enumerate(y.shape) if j[1] > 1)

    # multiply y against the theta parameter, 
    y = y * float(theta)

    # subtract the max for numerical stability
    y = y - np.expand_dims(np.max(y, axis = axis), axis)

    # exponentiate y
    y = np.exp(y)

    # take the sum along the specified axis
    ax_sum = np.expand_dims(np.sum(y, axis = axis), axis)

    # finally: divide elementwise
    p = y / ax_sum

    # flatten if X was 1D
    if len(X.shape) == 1: p = p.flatten()

    return p

回答3:

The x - np.max(x) code is not doing row-wise subtraction. Let's do it step-wise. First we will make a 'maxes' array by tiling or making a copy of the column:

maxes = np.tile(np.max(x,1), (2,1)).T

This will create a 2X2 matrix which will correspond to the maxes for each row by making a duplicate column(tile). After this you can do:

 x = np.exp(x - maxes)/(np.sum(np.exp(x - maxes), axis = 1))

You should get your result with this. The axis = 1 is for the row-wise softmax you mentioned in the heading of your answer. Hope this helps.

回答4:

How about this?

For taking max along the rows just specify the argument as axis=1 and then convert the result as a column vector(but a 2D array actually) using np.newaxis/None.

In [40]: x
Out[40]: 
array([[1001, 1002],
       [   3,    4]])

In [41]: z = x - np.max(x, axis=1)[:, np.newaxis]

In [42]: z
Out[42]: 
array([[-1,  0],
       [-1,  0]])

In [44]: softmax = np.exp(z) / np.sum(np.exp(z), axis=1)[:, np.newaxis]

In [45]: softmax
Out[45]: 
array([[ 0.26894142,  0.73105858],
       [ 0.26894142,  0.73105858]])

In the last step, again when you take sum just specify the argument axis=1 to sum it along the rows.

回答5:

My 5-liner (which uses scipy logsumexp for the tricky bits):

def softmax(a, axis=None):
    """
    Computes exp(a)/sumexp(a); relies on scipy logsumexp implementation.
    :param a: ndarray/tensor
    :param axis: axis to sum over; default (None) sums over everything
    """
    from scipy.special import logsumexp
    lse = logsumexp(a, axis=axis)  # this reduces along axis
    if axis is not None:
        lse = np.expand_dims(lse, axis)  # restore that axis for subtraction
    return np.exp(a - lse)

You may have to use from scipy.misc import logsumexp if you have an older scipy version.