Replace all but the first 1 in an array with 0

I am trying to find a way to replace all of the duplicate 1 with 0. As an example:

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

Should become:

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

I found a similar problem, however the solution does not seem to work https://stackoverflow.com/questions/37909364/numpy-setting-duplicate-values-in-a-row-to-0

Assume array contains only zeros and ones, you can find the max value per row using numpy.argmax and then use advanced indexing to reassign the values on the index to a zeros array.

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

res = np.zeros_like(arr)
idx = (np.arange(len(res)), np.argmax(arr, axis=1))
res[idx] = arr[idx]

res
array([[0, 1, 0, 0, 0],
       [1, 0, 0, 0, 0],
       [1, 0, 0, 0, 0]])

Try looping through each row of the grid

In each row, find all the 1s. In particular you want their indices (positions within the row). You can do this with a list comprehension and enumerate, which automatically gives an index for each element.

Then, still within that row, go through every 1 except for the first, and set it to zero.

grid = [[0, 1, 0, 1, 0], [1, 0, 0, 1, 0], [1, 1, 1, 0, 1]]

for row in grid:
    ones = [i for i, element in enumerate(row) if element==1]
    for i in ones[1:]:
        row[i] = 0

print(grid)

Gives: [[0, 1, 0, 0, 0], [1, 0, 0, 0, 0], [1, 0, 0, 0, 0]]

You can use cumsum:

(arr.cumsum(axis=1).cumsum(axis=1) == 1) * 1

this will create a cummulative sum, by then checking if a value is 1 you can find the first 1s