Inverse of bincount

mediumnumpy_manipulation

Implement repeat_counts(counts).

Given a 1D non-negative integer array counts, return a 1D array in which index i appears counts[i] times, in order. For example counts = [2, 0, 3][0, 0, 2, 2, 2]. (This is the inverse of np.bincount.) No for/while loops.

Constraints
  • no scipy, sklearn
  • no for/while loops

Your solution

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Hints

Hint 1

np.repeat(np.arange(len(counts)), counts) repeats each index by its count.