numpy where replace with numpy array depending on condition

Question

I have:

import numpy as np

a = np.array([[1,1,1],[-1,-1,-1]])

print (a)

Output:

[[ 1  1  1]
 [-1 -1 -1]]

I can do:

b = np.where(np.mean(a,axis=1) > 0,1,0)

print (b)

Which correctly results in:

[1 0]

But when I do:

b = np.where(np.mean(a,axis=1) > 0,np.array([1,1]),np.array([0,0]))

print (b)

It results in the same:

[1 0]

What I want is:

[[1 1]
 [0 0]]

In verbose, I want to replace elements of ndarray based on mean along axis 1 with array and not single integer. So the output from a 2D array should be a 2D array.

why not just replace 1 with [1, 1] 0 with [0, 0]. value array need to be the same shape with condition array — zuijiang
– zuijiang, Commented May 2, 2021 at 17:10
If you mean np.where(randn.mean(axis=0) > 0,[1,0],[1,0]) that neither works. In fact I want to generate labels from a 2D array. So axis 0 will have a the same length but axis 1 must be different after the operation. — Synthaze
– Synthaze, Commented May 2, 2021 at 17:12

hpaulj · Accepted Answer · 2021-05-02 17:19:59Z

2

In [254]: np.mean(a,axis=1) > 0,np.array([1,1]),np.array([0,0])
Out[254]: (array([ True, False]), array([1, 1]), array([0, 0]))

The 3 arguments are (2,) shaped arrays. They broadcast against each other to return a (2,) array.

The key is, broadcasting. It is picking elements from the 2 arrays based on corresponding elements of the condition. It is't a wholesale pick between the two.

If the condition is (2,1) shaped, that will broadcast against the (2,) to produce a (2,2) result

In [255]: (np.mean(a,axis=1) > 0)[:,None],np.array([1,1]),np.array([0,0])
Out[255]: 
(array([[ True],
        [False]]),
 array([1, 1]),
 array([0, 0]))

In [256]: np.where((np.mean(a,axis=1) > 0)[:,None],np.array([1,1]),np.array([0,0]))
Out[256]: 
array([[1, 1],
       [0, 0]])

answered May 2, 2021 at 17:19

hpaulj

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1 Comment

Synthaze Over a year ago

Thanks for the trick / great explanation!

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