[FIXED] Identifying all consecutive positive triplets in a 1D numpy array

Issue

Consider the 1D array arr shown below, and assume n = 3. I want to identify all 'islands' holding >= n consecutive positive values.

The following code succesfully finds the FIRST set of 3 consecutive positive numbers by determining the initial index, but it does not find all such sets.

import numpy as np

arr = np.array([1, -1, 5, 6, 3, -4, 2, 5, 9, 2, 1, -6, 8])

def find_consec_pos(arr, n):
    mask = np.convolve(np.greater(arr,0), np.ones(n, dtype=int)) >= n
    if mask.any():
        return mask.argmax() - n + 1
    else:
        return None
    

find_consec_pos(arr, 3)

This gives output 2, the index of the 1st triplet of consecutive positive values.
I want to know how to modify the code to get the output [2, 6, 7, 8], identifying all consecutive positive triples.

Solution

You could use sliding_window_view:

In [1]: from numpy.lib.stride_tricks import sliding_window_view

In [2]: sliding_window_view(arr, 3) > 0
Out[2]:
array([[ True, False,  True],
       [False,  True,  True],
       [ True,  True,  True],
       [ True,  True, False],
       [ True, False,  True],
       [False,  True,  True],
       [ True,  True,  True],
       [ True,  True,  True],
       [ True,  True,  True],
       [ True,  True, False],
       [ True, False,  True]])

Turning this into your desired function (and assuming you want a list as output):

def find_consec_pos(arr, n):
    all_n_positive = np.all(sliding_window_view(arr > 0, n), axis=1)
    return np.argwhere(all_n_positive).flatten().tolist()

Demo of some different "window" sizes:

In [4]: arr
Out[4]: array([ 1, -1,  5,  6,  3, -4,  2,  5,  9,  2,  1, -6,  8])

In [5]: find_consec_pos(arr, 3)
Out[5]: [2, 6, 7, 8]

In [6]: find_consec_pos(arr, 4)
Out[6]: [6, 7]

In [7]: find_consec_pos(arr, 2)
Out[7]: [2, 3, 6, 7, 8, 9]

Answered By - ddejohn