Create an array according to index in another array in Python

Question

I have an array like this:

X= [[1,2,3],
    [3,2,1],
    [2,1,3]]

Now I want to create another array Y. The elements in Y should take value 1 at positions where elements in X equal 2, otherwise they should take value 0. In this example, Y should equal to

Y=[[0,1,0],
   [0,1,0],
   [1,0,0]]

What did you try and what didn't work?

slider
– slider

2018-12-10 22:46:54 +00:00
Commented Dec 10, 2018 at 22:46 — slider
– slider, Commented Dec 10, 2018 at 22:46

sacuL · Accepted Answer · 2018-12-10 22:48:43Z

5

This would be greatly facilitated (and sped up) by using numpy:

import numpy as np

Y = (np.array(X) == 2).astype(int)

>>> Y
array([[0, 1, 0],
       [0, 1, 0],
       [1, 0, 0]])

answered Dec 10, 2018 at 22:48

sacuL

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Comments

blhsing · Accepted Answer · 2018-12-10 22:47:35Z

4

You can use a list comprehension like this:

Y = [[int(i == 2) for i in l] for l in X]

answered Dec 10, 2018 at 22:47

blhsing

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1 Comment

Allen Over a year ago

Thanks. It works, though a bit slow compared with using the numpy module.

enumaris · Accepted Answer · 2018-12-10 22:53:16Z

3

Y = [[1 if i==2 else 0 for i in row] for row in X]

edited Dec 10, 2018 at 22:53

answered Dec 10, 2018 at 22:48

enumaris

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3 Comments

Barmar Over a year ago

You have it backwards

Barmar Over a year ago

Should be 1 if ... else 0

enumaris Over a year ago

Yes, sorry I had the numbers backwards. Thanks for the reminder.

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