numpy 1D array: mask elements that repeat more than n times

Question

Q: given an array of integers like

[1, 1, 2, 2, 2, 3, 3, 3, 3, 4, 4, 4, 5, 5, 5, 5, 5, 5, 5]

I need to mask elements that repeat more than N times. The goal is to retrieve the boolean mask array.

I came up with a rather complicated solution:

import numpy as np

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

N = 3
splits = np.split(bins, np.where(np.diff(bins) != 0)[0]+1)
mask = []
for s in splits:
    if s.shape[0] <= N:
        mask.append(np.ones(s.shape[0]).astype(np.bool_))
    else:
        mask.append(np.append(np.ones(N), np.zeros(s.shape[0]-N)).astype(np.bool_)) 

mask = np.concatenate(mask)

giving e.g.

bins[mask]
Out[90]: array([1, 1, 2, 2, 2, 3, 3, 3, 4, 4, 4, 5, 5, 5])

Is there a nicer way to do this?

---

Wrap-up: Here's a slim version of MSeifert's benchmark plot (thanks for pointing me to simple_benchmark). Showing the four most performant options:

The idea proposed by Florian H, modified by Paul Panzer seems to be a great way of solving this problem as it is pretty straight forward and numpy-only. If you're fine with using numba, MSeifert's solution outperforms the other.

I chose to accept MSeifert's answer as solution as it is the more general answer: It correctly handles arbitrary arrays with (non-unique) blocks of consecutive repeating elements. In case numba is a no-go, Divakar's answer is also worth a look.

Answer 1

Disclaimer: this is just a sounder implementation of @FlorianH's idea:

def f(a,N):
    mask = np.empty(a.size,bool)
    mask[:N] = True
    np.not_equal(a[N:],a[:-N],out=mask[N:])
    return mask

For larger arrays this makes a huge difference:

a = np.arange(1000).repeat(np.random.randint(0,10,1000))
N = 3

print(timeit(lambda:f(a,N),number=1000)*1000,"us")
# 5.443050000394578 us

# compare to
print(timeit(lambda:[True for _ in range(N)] + list(bins[:-N] != bins[N:]),number=1000)*1000,"us")
# 76.18969900067896 us