Finding the consecutive zeros in a numpy array

Question

I have the following array

a = [1, 2, 3, 0, 0, 0, 0, 0, 0, 4, 5, 6, 0, 0, 0, 0, 9, 8, 7,0,10,11]

I would like to find the start and the end index of the array where the values are zeros consecutively. For the array above the output would be as follows

[3,8],[12,15],[19]

I want to achieve this as efficiently as possible.

Answer 1

Here's a fairly compact vectorized implementation. I've changed the requirements a bit, so the return value is a bit more "numpythonic": it creates an array with shape (m, 2), where m is the number of "runs" of zeros. The first column is the index of the first 0 in each run, and the second is the index of the first nonzero element after the run. (This indexing pattern matches, for example, how slicing works and how the range function works.)

import numpy as np

def zero_runs(a):
    # Create an array that is 1 where a is 0, and pad each end with an extra 0.
    iszero = np.concatenate(([0], np.equal(a, 0).view(np.int8), [0]))
    absdiff = np.abs(np.diff(iszero))
    # Runs start and end where absdiff is 1.
    ranges = np.where(absdiff == 1)[0].reshape(-1, 2)
    return ranges

For example:

In [236]: a = [1, 2, 3, 0, 0, 0, 0, 0, 0, 4, 5, 6, 0, 0, 0, 0, 9, 8, 7, 0, 10, 11]

In [237]: runs = zero_runs(a)

In [238]: runs
Out[238]: 
array([[ 3,  9],
       [12, 16],
       [19, 20]])

With this format, it is simple to get the number of zeros in each run:

In [239]: runs[:,1] - runs[:,0]
Out[239]: array([6, 4, 1])

It's always a good idea to check the edge cases:

In [240]: zero_runs([0,1,2])
Out[240]: array([[0, 1]])

In [241]: zero_runs([1,2,0])
Out[241]: array([[2, 3]])

In [242]: zero_runs([1,2,3])
Out[242]: array([], shape=(0, 2), dtype=int64)

In [243]: zero_runs([0,0,0])
Out[243]: array([[0, 3]])

Answer 2

You can use itertools to achieve your expected result.

from itertools import groupby
a= [1, 2, 3, 0, 0, 0, 0, 0, 0, 4, 5, 6, 0, 0, 0, 0, 9, 8, 7,0,10,11]
b = range(len(a))
for group in groupby(iter(b), lambda x: a[x]):
    if group[0]==0:
        lis=list(group[1])
        print [min(lis),max(lis)]