DEEPAK SAINI - 4 months ago 24

Python Question

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]])
```