CODEWITHSUNDEEP
pythonnumpymultidimensional-array
user15300490
user15300490

Reputation:

How to get index of numpy multidimensional array in reverse order?

For example I have this np.array:

[[True, True, False, False]
[True, False, True, False]
[False, True, False, True]
[False, False, True, True]]

I want to get the the first index that is True in each row but counting from the back of the row. So expected output is a (4,) array corresponding to each row:

[1,  # First index that is True is at index 1
 2,  # First index that is True is at index 2
 3,  # First index that is True is at index 3
 3]  # First index that is True is at index 3

Upvotes: 1

Views: 501

Answers (2)

Reti43
Reti43

Reputation: 9796

a = np.array(
    [[True, True, False, False],
     [True, False, True, False],
     [False, True, False, True],
     [False, False, True, True]]
)

idx = a.shape[1] - 1 - np.argmax(a[:,::-1], axis=1)

np.argmax() will find the index of the highest value (for each row with axis=1). Since True is equivalent to 1 and False to 0, it'll record the index of the first True value. Simply reverse the rows so you find the "last" one and then substract this from the row length to account for the fact that you're counting backwards.

Upvotes: 1

Shoaib Mirzaei
Shoaib Mirzaei

Reputation: 522

you can use python to reversea row and find an element: row.reverse() and row.find(True). in numpy you can use numpy.flip(row) to reverse a row and numpy.where(row == True) to find an element in a row.

import numpy as np

x = np.array([[True, True, False, False],
[True, False, True, False],
[False, True, False, True],
[False, False, True, True]])

result = []
for row in x:
    row = np.flip(row)
    index = np.where(row == True)
    result.append(index[0][0])
print(result)

Upvotes: 0

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