Index of last occurence of True in every row

Question

I have a 2D array:

a = ([[False False False False False  True  True  True  True  True  True  True
   True  True  True  True  True  True  True  True  True  True  True  True
   True False False False]
 [False False False False  True  True  True  True  True  True  True  True
   True  True  True  True  True  True  True  True  True  True  True  True
  False False False False]])

I am trying to get the index of last occurrence of 'True' in every row. So resulting array should be

b = ([24, 23])

To find the occurence of first True, I know i can use the np.argmax().

b = np.argmax(a==True,axis=1)

Is there a function to find from the last? I tried reversing the values of array and then using np.argmax(), but it will give the index of the reversed array.

After you get the index in the reversed array, subtract it from the array length to get the index in the original array. — Barmar, Feb 21 '21 at 17:36

score 2 · Answer 1 · answered Feb 21 '21 at 18:06

2

Process each row in the backwards direction and subtract the result from the row length - 1:

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

Even == True is not needed.

The result is:

array([24, 23], dtype=int64)

answered Feb 21 '21 at 18:06

Valdi_Bo

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score 1 · Answer 2 · answered Feb 21 '21 at 18:24

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

b = a[...,::-1]
c = [len(i) - np.argmax(i) - 1 for i in b]
print(c)

list 'c' will have indices with last True value

Index of last occurence of True in every row

2 Answers2

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