change gaps in numpy array according to gap size

You can use image-processing based binary_closing -

from scipy.ndimage.morphology import binary_closingdef remove_small_nnz(a, W):    K = np.ones(W, dtype=int)    m = a==0    p = binary_closing(m,K)    a[~m & p] = 0    return a

Sample run -

In [97]: aOut[97]: array([1, 3, 1, 0, 0, 1, 8, 3, 0, 8, 2, 4, 7, 0, 0, 4, 1])In [98]: remove_small_nnz(a, W=3)Out[98]: array([1, 3, 1, 0, 0, 1, 8, 3, 0, 8, 2, 4, 7, 0, 0, 4, 1])In [99]: remove_small_nnz(a, W=4)Out[99]: array([1, 3, 1, 0, 0, 0, 0, 0, 0, 8, 2, 4, 7, 0, 0, 4, 1])In [100]: remove_small_nnz(a, W=5)Out[100]: array([1, 3, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 4, 1])

Since you're only looking for nonzeros, you can cast the array to boolean, and look for spots where there is a sequence of however many Trues in a row as you're looking for.

import numpy as npdef orig(fhr, minseq):    p = np.where(fhr>0, 1, 0).astype(int)    s = np.array([1]+ list(np.diff(p)))    sind = np.where(s==1)[0][1:]    for i in range(len(sind) - 1):        s1 = sind[i]        e1 = sind[i+1]        subfhr = np.where(fhr[s1:e1] > 0, 1, 0).sum()        if (subfhr < minseq):            fhr[s1:e1] = 0    return fhrdef update(fhr, minseq):    # convert the sequence to boolean    nonzero = fhr.astype(bool)    # stack the boolean array with lagged copies of itself    seqs = np.stack([nonzero[i:-minseq+i] for i in range(minseq)],                    axis=1)    # find the spots where the sequence is long enough    inseq = np.r_[np.zeros(minseq, np.bool), seqs.sum(axis=1) == minseq]    # the start and end of the series is are assumed to be included in result    inseq[minseq] = True    inseq[-1] = True        # make sure that the full sequence is included.     # There may be a way to vectorize this further    for ind in np.where(inseq)[0]:        inseq[ind-minseq:ind] = True    # Apply the inseq array as a mask    return inseq * fhrfhr = np.array([1, 3, 1, 0, 0, 1, 8, 3, 0, 8, 2, 4, 7, 0,0,4,1])minseq = 4 print(np.all(orig(fhr, minseq) == update(fhr, minseq)))# True