select subarrays delimited by zeros in python

Question

Given a list like:

A = [18, 7, 0, 0, 0, 9, 12, 0, 0, 11, 2, 3, 3, 0, 0, 7, 8]

is there a simple way to create subarrays, with those elements that are separated by zeros (or at least by NaNs)? I mean, like:

A1 = [18, 7]
A2 = [9, 12]
A3 = [11, 2, 3, 3]
A4 = [7, 8]

I've written:

q=0
for i in range(0,len(A)):
    if A[i]-A[i-1] < 1:
        q=q+1

to retrieve the number of zeros-packets present in the list. But I need to populate subarrays, as long as I encounter them through the list... maybe something with the split function? Thank you in advance.

Of course there is a way, have you tried anything? Show us your code please. — Julien, Aug 01 '16 at 16:08
Possible duplicate : http://stackoverflow.com/questions/4322705/split-a-list-into-nested-lists-on-a-value — Yotam Salmon, Aug 01 '16 at 16:09

score 3 · Accepted Answer · answered Aug 01 '16 at 16:19

Well, itertools has the solution for you: groupby(list, filter).

If you want to group by zeroes, start with doing:

B = itertools.groupby(A, lambda x:x == 0)

The lambda expression "decides" which of the values should be a separator. You could separate by Nones using lambda x: x == None (for example). That will return you an iterable object. So, using a list comprehension, let's iterate through it (every iteration gives us a 2 values tuple):

C = [(i, list(j)) for i, j in B] # j is cast to a list because it's originally an object, not a list.

Output will be something like:

[(False, [18, 7]), (True, [0]), (True, [0]), (True, [0]), ... ]

Now, every list j that is the separator has a value True for i. So we can filter it:

C = [list(j) for i, j in B if not i]

Now, the result is a 2d list:

[[18, 7], [9, 12], [11, 2, 3, 3], [7, 8]]

So a one liner function:

def splitArr():
    return [list(j) for i, j in itertools.groupby(A, lambda x:x == 0) if not i]

@urgeo You're welcome! Hope you read everything and not just the final solution ;-) — Yotam Salmon, Aug 01 '16 at 16:59

score 1 · Answer 2 · answered Aug 01 '16 at 16:17

1

Try this:

import itertools as it
A = [18, 7, 0, 0, 0, 9, 12, 0, 0, 11, 2, 3, 3, 0, 0, 7, 8]

[list(v) for k, v in it.groupby(A, lambda x: not x) if not k]
=> [[18, 7], [9, 12], [11, 2, 3, 3], [7, 8]]

answered Aug 01 '16 at 16:17

Óscar López

232,561
37
312
386

score 0 · Answer 3 · answered Aug 01 '16 at 16:26

0

naive solution:

A = [18, 7, 0, 0, 0, 9, 12, 0, 0, 11, 2, 3, 3, 0, 0, 7, 8]
b = []
c = []
for i in A:
    if i == 0:
        if len(c):
            b.append(c)
        c = []
        continue
    c.append(i)
if len(c):
    b.append(c)

answered Aug 01 '16 at 16:26

Aviad Levy

750
4
13

select subarrays delimited by zeros in python

3 Answers3