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.

It wasn't me but it's traditional to say something like "I have this code" (and show it) and it doesn't quite work. You haven't shown what you've tried. — doctorlove
– doctorlove, Commented Jul 22, 2014 at 10:41
well, i can just go through the array with a loop and mark the start and end indices, I thought people who know python would understand it easily — Shan
– Shan, Commented Jul 22, 2014 at 10:48
If pure Python is an option then you can use itertools.groupby for this. — Ashwini Chaudhary
– Ashwini Chaudhary, Commented Jul 22, 2014 at 11:17
For the record, that syntax doesn't construct an array but a list. — DSM
– DSM, Commented Jul 22, 2014 at 11:58

Warren Weckesser · Accepted Answer · 2014-07-22 16:40:20Z

55

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]])

edited Jul 22, 2014 at 16:40

answered Jul 22, 2014 at 16:00

Warren Weckesser

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5 Comments

toine Over a year ago

any way to do that with pandas ?

MikeL Over a year ago

Nice solution! Is there a reason for using np.int8 data type in iszero? I think if we simply use booleans we can also avoid np.abs() and simply set absdiff = np.diff(iszero). Am I missing something?

Jason Over a year ago

This is a clever solution. I modified it to count the number of ties in the Mann-Kendall trend test stackoverflow.com/a/68442829/2005415

Joren Over a year ago

Oneliner: np.ediff1d(np.r_[0, a == 0, 0]).nonzero()[0].reshape(-1, 2)

Rainald62 Mar 1 at 11:12

More readable with diff since numpy 1.16: np.diff(booleanArray, prepend=False, append=False).nonzero()[0].reshape(-1, 2).

cs95 · Accepted Answer · 2017-08-17 05:57:30Z

1

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)]

edited Aug 17, 2017 at 5:57

cs95

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answered Jul 22, 2014 at 11:53

rajeshv90

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1 Comment

Ashwini Chaudhary Over a year ago

This will return [19, 19], I think OP expects just [19]. And instead of creating an unnecessary list b, try to use enumerate(a).

Bestasttung · Accepted Answer · 2014-07-22 12:01:27Z

-1

Here is a custom function, not sure the most efficient but works :

def getZeroIndexes(li):
  begin = 0
  end = 0
  indexes = []
  zero = False
  for ind,elt in enumerate(li):
    if not elt and not zero:
      begin = ind
      zero = True
    if not elt and zero:
      end = ind
    if elt and zero:
      zero = False
      if begin == end:
        indexes.append(begin)
      else:
        indexes.append((begin, end))

  return indexes

answered Jul 22, 2014 at 12:01

Bestasttung

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