Update 2D numpy array using pairs of indices without iteration

Question

Reproducible setup

I have an array of n pairs of indices:

indexarray=\
np.array([[0,2,4,6],
          [1,3,5,7]])

I also have a 2D array:

zeros = np.zeros((10,9))

and a list of n values:

values = [1,2,3,4]

I would like to add each kth element from the values list to the element in the zeros list having indeces equal to the kth indices pair

A solution

# works, but for loop is not suitable for real-world use-case
for index, (row, col) in enumerate(indexarray.T):
    zeros[row, col] = values[index]

Visualize what I get:

plt.imshow(zeros)

Results as expected.

How can I do this without iteration?

Similar but different questions:

Index 2D numpy array by a 2D array of indices without loops : here the output is an array of pairs of indices, not a 2D array, as here.
Fill 2D numpy array from three 1D numpy arrays : setup is more complicated, they create a 2D array from 1D arrays, unlike here, where we already have a 2D array
numpy arrays: filling and extracting data quickly : way more complicated setup, do not start from 2D array
How to convert List of Lists of Tuples- pairs (index,value) into 2D numpy array : aim is different and for loop is not avoided

rows, cols = indexarray[0], indexarray[1]; zeros[rows, cols] = values? — Dani Mesejo
– Dani Mesejo, Commented Nov 12, 2021 at 10:58
Are you going to have repeated indices? Do you need to add them together? — Dani Mesejo
– Dani Mesejo, Commented Nov 12, 2021 at 11:01
Personally I don't need that, but future users finding this question might; feel free to add an answer covering that case as well if you feel like it - my problem at hand is solved. — zabop
– zabop, Commented Nov 12, 2021 at 11:03

Dani Mesejo · Accepted Answer · 2021-11-12 11:06:34Z

Simply use:

import numpy as np

indexarray = np.array([[0, 2, 4, 6],
                       [1, 3, 5, 7]])
values = [1, 2, 3, 4]

rows, cols = indexarray[0], indexarray[1]

zeros = np.zeros((10, 9))
zeros[rows, cols] = values

print(zeros)

Output

[[0. 1. 0. 0. 0. 0. 0. 0. 0.]
 [0. 0. 0. 0. 0. 0. 0. 0. 0.]
 [0. 0. 0. 2. 0. 0. 0. 0. 0.]
 [0. 0. 0. 0. 0. 0. 0. 0. 0.]
 [0. 0. 0. 0. 0. 3. 0. 0. 0.]
 [0. 0. 0. 0. 0. 0. 0. 0. 0.]
 [0. 0. 0. 0. 0. 0. 0. 4. 0.]
 [0. 0. 0. 0. 0. 0. 0. 0. 0.]
 [0. 0. 0. 0. 0. 0. 0. 0. 0.]
 [0. 0. 0. 0. 0. 0. 0. 0. 0.]]

An alternative that will add together repeating coordinates, is to use add.at:

np.add.at(zeros, (rows, cols), values)

A second alternative is to use a sparse matrix constructor, for example:

from scipy.sparse import csr_matrix
rows, cols = indexarray[0], indexarray[1]
zeros = csr_matrix((values, (rows, cols)), shape=(10, 9)).toarray()

Output

[[0 1 0 0 0 0 0 0 0]
 [0 0 0 0 0 0 0 0 0]
 [0 0 0 2 0 0 0 0 0]
 [0 0 0 0 0 0 0 0 0]
 [0 0 0 0 0 3 0 0 0]
 [0 0 0 0 0 0 0 0 0]
 [0 0 0 0 0 0 0 4 0]
 [0 0 0 0 0 0 0 0 0]
 [0 0 0 0 0 0 0 0 0]
 [0 0 0 0 0 0 0 0 0]]

Ch3steR · Accepted Answer · 2021-11-12 10:59:44Z

1

You can directly use indexarray in indexing.

r, c = indexarray
zeros[r, c] = values

answered Nov 12, 2021 at 10:59

Ch3steR

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Comments

mozway · Accepted Answer · 2021-11-12 10:59:59Z

1

You can simply use:

zeros[indexarray[0], indexarray[1]] = values

answered Nov 12, 2021 at 10:59

mozway

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