Change Numpy array values in-place

Question

Say when we have a randomly generated 2D 3x2 Numpy array a = np.array(3,2) and I want to change the value of the element on the first row & column (ie a[0,0]) to 10. If I do

a[0][0] = 10

then it works and a[0,0] is changed to 10. But if I do

a[np.arange(1)][0] = 10

then nothing is changed. Why is this?

I want to change some columns values of a selected list of rows (that is indicated by a Numpy array) to some other values (like a[row_indices][:,0] = 10 ) but it doesn't work as I'm passing in an array (or list) that indicates rows.

Answer 1

a[x][y] is wrong . It happens to work in the first case, a[0][0] = 10 because a[0] returns a view , hence doing resul[y] = whatever modifies the original array. However, in the second case, a[np.arange(1)][0] = 10 , a[np.arange(1)] returns a copy (because you are using array indexing).

You should be using a[0, 0] = 10 or a[np.arange(1), 0] = 10

Answer 2

Advanced indexing always returns a copy as a view cannot be guaranteed.

Advanced indexing always returns a copy of the data (contrast with basic slicing that returns a view).

If you replace np.arange(1) with something that returns a view (or equivalent slicing) then you get back to basic indexing, and hence when you chain two views, the change is reflected into the original array.

For example:

import numpy as np


import numpy as np


arr = np.arange(2 * 3).reshape((2, 3))
arr[0][0] = 10
print(arr)
# [[10  1  2]
#  [ 3  4  5]]

arr = np.arange(2 * 3).reshape((2, 3))
arr[:1][0] = 10
print(arr)
# [[10 10 10]
#  [ 3  4  5]]

arr = np.arange(2 * 3).reshape((2, 3))
arr[0][:1] = 10
print(arr)
# [[10  1  2]
#  [ 3  4  5]]

etc.

---

If you have some row indices you want to use, to modify the array you can just use them, but you cannot chain the indexing, eg:

arr = np.arange(5 * 3).reshape((5, 3))
row_indices = (0, 2)
arr[row_indices, 0] = 10
print(arr)
# [[10  1  2]
#  [ 3  4  5]
#  [10  7  8]
#  [ 9 10 11]
#  [12 13 14]]