How to set first value in numpy array that meets a condition to 1 but not the rest

Question

I want to set the max value in a numpy array equal to 1, and the rest of the values to 0, so that there is only one value equal to 1 in the new array.

Right now I'm doing this with:

new_arr = np.where(arr == np.max(arr), 1, 0)

However, if there are multiple values in arr that are equal to np.max(arr) then there will be multiple values in new_arr equal to 1. How do I make it so that there is only one value in new_arr equal to 1 (the first value equal to np.max(arr) seems like a fine option but not necessary).

Answer 1

You can use:

new_arr = np.zeros(shape=arr.shape)
new_arr[np.unravel_index(np.argmax(arr),shape=arr.shape)] = 1

This also works for multi-dimensional arrays. np.argmax gives the flattened index of the first instance of the max element, and np.unravel_index converts the flat index to the index based on the array shape.

Answer 2

You almost have it.

This will give you the index of the last occurrence of the max value

np.where(arr == np.max(arr))[0][-1]

and if you want the first occurence of the maximum value, it is :

np.where(arr == np.max(arr))[0][0]

Example

import numpy as np

arr = np.array(np.random.choice([1, 2, 3, 4], size=10))
print(arr)

Output : [4 1 2 4 1 1 4 2 4 3]

Then:

np.where(arr == np.max(arr))[0][-1] # last index of max value

Output: 8

or

np.where(arr == np.max(arr))[0][0] # first index of max value

Output: 0

You can then proceed to replace by index.