DEEPAK SAINI DEEPAK SAINI - 2 months ago 12
Python Question

Replicating elements in numpy array

I have a numpy array say

a = array([[1, 2, 3],
[4, 5, 6],
[7, 8, 9]])


I have an array 'replication' of the same size where replication[i,j](>=0) denotes how many times a[i][j] should be repeated along the row. Obiviously, replication array follows the invariant that np.sum(replication[i]) have the same value for all i.
For example, if

replication = array([[1, 2, 1],
[1, 1, 2],
[2, 1, 1]])


then the final array after replicating is:

new_a = array([[1, 2, 2, 3],
[4, 5, 6, 6],
[7, 7, 8, 9]])


Presently, I am doing this to create new_a:

##allocate new_a
h = a.shape[0]
w = a.shape[1]
for row in range(h):
ll = [[a[row][j]]*replicate[row][j] for j in range(w)]
new_a[row] = np.array([item for sublist in ll for item in sublist])


However, this seems to be too slow as it involves using lists. Can I do the intended entirely in numpy, without the use of python lists?

Answer

You can flatten out your a and replication arrays, then use the .repeat() method:

import numpy as np

a = array([[1, 2, 3],
           [4, 5, 6],
           [7, 8, 9]])

replication = array([[1, 2, 1],
                     [1, 1, 2],
                     [2, 1, 1]])

new_a = a.ravel().repeat(replication.ravel()).reshape(a.shape[0], -1)

print(repr(new_a))
# array([[1, 2, 2, 3],
#        [4, 5, 6, 6],
#        [7, 7, 8, 9]])
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