O.rka - 3 months ago 19

Python Question

I wrote a softmax regression function

`def softmax_1(x)`

`m x n`

`x = np.arange(-2.0, 6.0, 0.1)`

scores = np.vstack([x, np.ones_like(x), 0.2 * np.ones_like(x)])

#scores shape is (3, 80)

def softmax_1(x):

"""Compute softmax values for each sets of scores in x."""

return(np.exp(x)/np.sum(np.exp(x),axis=0))

Converting it into a DataFrame I have to transpose

`DF_activation_1 = pd.DataFrame(softmax_1(scores).T,index=x,columns=["x","1.0","0.2"])`

So I wanted to try and make a version of the softmax function that takes in the transposed version and computes the softmax function

`scores_T = scores.T`

#scores_T shape is (80,3)

def softmax_2(y):

return(np.exp(y/np.sum(np.exp(y),axis=1)))

DF_activation_2 = pd.DataFrame(softmax_2(scores_T),index=x,columns=["x","1.0","0.2"])

Then I get this error:

`Traceback (most recent call last):`

File "softmax.py", line 22, in <module>

DF_activation_2 = pd.DataFrame(softmax_2(scores_T),index=x,columns=["x","1.0","0.2"])

File "softmax.py", line 18, in softmax_2

return(np.exp(y/np.sum(np.exp(y),axis=1)))

ValueError: operands could not be broadcast together with shapes (80,3) (80,)

`np.sum`

Answer

Change

```
np.exp(y/np.sum(np.exp(y),axis=1))
```

to

```
np.exp(y)/np.sum(np.exp(y),axis=1, keepdims=True)
```

This will mean that `np.sum`

will return an array of shape `(80, 1)`

rather than `(80,)`

, which will broadcast correctly for the division. Also note the correction to the bracket closing.

Source (Stackoverflow)

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