Making new Numpy array with indexes of another Numpy array

Question

I have an 10 x N array as follows:

[[ 0. 1. 0. ..., 0. 0. 0.]
 [ 0. 0. 0. ..., 1. 0. 0.]
 [ 1. 0. 0. ..., 0. 0. 0.]
 ..., 
 [ 0. 0. 0. ..., 1. 0. 0.]
 [ 0. 0. 0. ..., 0. 0. 1.]
 [ 0. 0. 0. ..., 0. 0. 1.]]

I want a Numpy array of the following 1 x N format, where each element in the new array is the value of the index filled with '1' in the 10 x N array.

For example, the process would convert the above into the array:

[[ 1. 7. 0. ..., 7, 9. 9.]]

I have had some success with using the function:

np.where(array > 0)[0][0]

This gives me a value for my final array but my attempts to fill the array in the required format have not worked. Furthermore, my implementations have not been very pythonic. Is there a pythonic solution to the above question?

Answer 1

Setup

a = np.array([[0, 0, 0, 0, 0, 1, 0, 0, 0, 0],
 [1, 0, 0, 0, 0, 0, 0, 0, 0, 0],
 [0, 0, 1, 0, 0, 0, 0, 0, 0, 0],
 [0, 0, 0, 0, 1, 0, 0, 0, 0, 0],
 [0, 0, 1, 0, 0, 0, 0, 0, 0, 0],
 [0, 0, 0, 0, 0, 0, 1, 0, 0, 0]])

---

You only care about the column, since your second array is 1D, right now you are grabbing the row from numpy.where

If you can guarantee that there is only one 1 per row, just grab the columns from the output of numpy.where:

np.where(a==1)[1]

array([5, 0, 2, 4, 2, 6], dtype=int64)

• Jack Rolph: Perfect! Thank you! :D