How to modify the value of numpy array based on another array?

Question

I have a numpy array. I want to modify one array index by the chosen elements of another array. For example:

import numpy as np
t1 = np.ones((10,3))
t2 = np.arange(10)
t1[np.where(t2>5)][:,2] = 10
print(t1)

What I want t1 is:

array([[1., 1., 1.],
       [1., 1., 1.],
       [1., 1., 1.],
       [1., 1., 1.],
       [1., 1., 1.],
       [1., 1., 1.],
       [1., 1., 10.],
       [1., 1., 10.],
       [1., 1., 10.],
       [1., 1., 10.]])

But the output of t1 is:

array([[1., 1., 1.],
       [1., 1., 1.],
       [1., 1., 1.],
       [1., 1., 1.],
       [1., 1., 1.],
       [1., 1., 1.],
       [1., 1., 1.],
       [1., 1., 1.],
       [1., 1., 1.],
       [1., 1., 1.]])

What the problem there?

oppressionslayer · Accepted Answer · 2019-12-06 02:50:23Z

3

It's backwards, it should be:

t1[:,2][np.where(t2>5)] = 10

output:

array([[ 1.,  1.,  1.],
       [ 1.,  1.,  1.],
       [ 1.,  1.,  1.],
       [ 1.,  1.,  1.],
       [ 1.,  1.,  1.],
       [ 1.,  1.,  1.],
       [ 1.,  1., 10.],
       [ 1.,  1., 10.],
       [ 1.,  1., 10.],
       [ 1.,  1., 10.]])

answered Dec 6, 2019 at 2:50

oppressionslayer

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Comments

stardust · Accepted Answer · 2021-06-28 08:09:46Z

2

The most pythonic way to do this is probably

t1[:, 2] = np.where(t2 > 5,       # where t2 > 5
                    10,           # put 10
                    t1[:, 2])     # into the third column of t1

Besides added clarity, for very large objects this will have a time benefit, as there is no creation of an intermediate indexing array np.where(t2 > 5) and no resulting intermediate callbacks to that python object - everything is done in-place with c-compiled code.

edited Jun 28, 2021 at 8:09

stardust

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answered Dec 6, 2019 at 8:23

Daniel F

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1 Comment

MoralCode Over a year ago

where() for the multidimensional case simply decides which array to pull from (second arg or third arg) based on the value of the first arg, so the into the third column of t1 comment probably should point to the left side of the = on the first line

Austin · Accepted Answer · 2019-12-06 02:56:32Z

1

You can do:

t1[np.where(t2>5), 2] = 10

Syntax: array[<row>, <col>]

answered Dec 6, 2019 at 2:56

Austin

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