How to retain only the first occurrence of strictly increasing values in each row of a NumPy array?

Below is one elegant vectorized solution using NumPy’s cumulative maximum. The idea is to compute the cumulative maximum along each row and then mark an entry if it is the first time that the maximum changed (with the very first element always kept):

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import numpy as np

arr = np.array([
    [1, 1, 1, 2, 2, 2, 3, 3, 3, 4, 4, 5, 5],
    [1, 1, 2, 2, 2, 3, 2, 2, 3, 3, 3, 4, 4],
    [3, 2, 1, 2, 1, 1, 2, 3, 4, 5, 4, 3, 2]
])

# Compute the cumulative maximum along rows.
cum_max = np.maximum.accumulate(arr, axis=1)

# Create a boolean mask:
# First element in every row is always True.
# For the rest, True only if the cummax is greater than the previous cummax.
mask = np.zeros_like(arr, dtype=bool)
mask[:, 0] = True
mask[:, 1:] = cum_max[:, 1:] > cum_max[:, :-1]

# Build the output array.
out = np.where(mask, arr, 0)
print(out)

Explanation:

1. np.maximum.accumulate(arr, axis=1) computes the running maximum along each row. It increases only when a new, strictly higher element is encountered.

2. The mask sets the first element of each row as True. For the rest, an entry is marked True only if the cumulative maximum increased from the previous column; that is, the value represents the first occurrence of a new maximum.

3. Finally, np.where(mask, arr, 0) keeps only the values corresponding to the first occurrence of an increase and replaces all other elements with zero.

When you run the above code, you get the desired output:

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[[1 0 0 2 0 0 3 0 0 4 0 5 0]
 [1 0 2 0 0 3 0 0 0 0 0 4 0]
 [3 0 0 0 0 0 0 0 4 5 0 0 0]]