My current function is as follows:
def soft_max(z):
t = np.exp(z)
a = np.exp(z) / np.sum(t, axis=1)
return a
However I get the error: ValueError: operands could not be broadcast together with shapes (20,10) (20,) since np.sum(t, axis=1) isn't a scalar.
I want to have t / the sum of each row but I don't know how to do this.
You want to do something like (see this post)
def softmax(x, axis=None):
x = x - x.max(axis=axis, keepdims=True)
y = np.exp(x)
return y / y.sum(axis=axis, keepdims=True)
As of version 1.2.0, scipy includes softmax as a special function:
https://scipy.github.io/devdocs/generated/scipy.special.softmax.html
Use the axis argument do run it over rows.
from scipy.special import softmax
softmax(arr, axis=0)
suppose your z is a 2 dimensional array, try
def soft_max(z):
t = np.exp(z)
a = np.exp(z) / np.sum(t, axis=1).reshape(-1,1)
return a
np.sum(t, axis=1, keepdims=True) is better than using .reshape. they both are doing the same thing. Overflow may be caused by your input data I think.
keepdims:np.sum(t, axis=1, keepdims=True)?RuntimeWarning: overflow encountered in exp