arr = np.array([[[1,2],[3,4]],[[5,6],[7,8]]])
I'm trying to get the indices of arr when arr==1.
I thought this would work but it doesn't give the expected output:
>>> np.where(arr==1)
(array([0], dtype=int64), array([0], dtype=int64), array([0], dtype=int64))
If you change arr:
arr = np.array([[[1,2],[3,4]],[[1,1],[7,8]]])
and you will get
np.where(arr==1)
# (array([0, 1, 1]), array([0, 0, 0]), array([0, 0, 1]))
it is mean:
arr[0][0][0] == 1
arr[1][0][0] == 1
arr[1][0][1] == 1
When you want display coordinate in one row:
np.array(np.where(arr==1)).T
# array([[0, 0, 0],[1, 0, 0],[1, 0, 1]])
arr[1,0,1] is nicer looking numpy.For multiple numbers, you can mix np.where and np.isin functions:
Here is an example:
import numpy as np
val = np.array([1,2,3])
arr = np.array([[[1,2],[3,4]],[[5,6],[7,8]]])
loc_i, loc_j, loc_k = np.where(np.isin(arr, val))
print('locations:')
[print(f'({loc_i[i]},{loc_j[i]},{loc_k[i]})') for i in range(loc_i.size)]
An interesting option is to use argwhere to get indices of elements in question.
To present a more instructive example, assume that we want indices of your arr where the value is <= 3.
The code to get them is np.argwhere(arr <= 3), getting:
array([[0, 0, 0],
[0, 0, 1],
[0, 1, 0]], dtype=int64)
Meaning that the "wanted" elemens are:
arr[0, 0, 0], arr[0, 0, 1], arr[0, 1, 0],
If you want e.g. to print indices of the "wanted" elements and their values, you can run:
for ind in np.argwhere(arr <= 3):
print(f'{ind}: {arr.__getitem__(tuple(ind))}')
The result is:
[0 0 0]: 1
[0 0 1]: 2
[0 1 0]: 3
Note also that my code works regardless of the number of dimensions in your array, whereas in other solutions the number of dimensions is somehow "fixed" in the code.
argwhere does is np.transpose(np.where(...))
[0][0][0]arr==1 or 2 or 3?arr[ np.where( arr=1 )]should display1.where/nonzerogives a tuple of arrays suitable for indexing. Here its 3 arrays sincearris 3d.