group elements in an array and sum

Question

I have the following array

a = [0,0,1,0,0,0,1,1,0,0,0,0,1,1,1,0,0,0,0,0,1]

I want to group every 3rd element and sum all of the elements within each group. So I can get a new array with a new size showing this sum

b = [1,0,2,0,3,0,1]

Any suggestions?

Is it guaranteed that `len(a) % 3 == 0` (the length is a multiple of three)? — intboolstring, Jul 30 '16 at 00:57
Related: [How do you split a list into evenly sized chunks in Python?](http://stackoverflow.com/q/312443/190597). — unutbu, Jul 30 '16 at 00:59

score 3 · Answer 1 · answered Jul 30 '16 at 01:00

Simply, most pythonicly would be the following

b = [sum(a[i:i+3]) for i in range(0, len(a), 3)]

where your input array is a.

>>> a = [0,0,1,0,0,0,1,1,0,0,0,0,1,1,1,0,0,0,0,0,1]
>>> b = [sum(a[i:i+3]) for i in range(0, len(a), 3)]
>>> b
[1, 0, 2, 0, 3, 0, 1]

DurgaDatta · Answer 2 · 2016-07-30T01:18:55.250

3

You can split in chunk and sum:

step = 3
[sum(a[i:i+step]) for i in range(0, len(a),step)]
[1, 0, 2, 0, 3, 0, 1]

If the length is not the multiple of step, last chunk might be smaller.

edited Jul 30 '16 at 01:18

answered Jul 30 '16 at 01:02

DurgaDatta

score 2 · Answer 3 · answered Jul 30 '16 at 00:59

2

Maybe something like this:

a = [0,0,1,0,0,0,1,1,0,0,0,0,1,1,1,0,0,0,0,0,1]
b = []

for i in range(0,len(a),3):
    b.append(sum(a[i:i+3]))
print b

Output:

[1, 0, 2, 0, 3, 0, 1]

answered Jul 30 '16 at 00:59

Psidom · Accepted Answer · 2016-07-30T01:32:54.333

1

Another option using groupby from itertools:

from itertools import groupby
[sum(v for _, v in g) for _, g in groupby(enumerate(a), key = lambda x: x[0]/3)]
# [1, 0, 2, 0, 3, 0, 1]

Or another way to use zip:

[sum(v) for v in zip(a[::3], a[1::3], a[2::3])]
# [1, 0, 2, 0, 3, 0, 1]

edited Jul 30 '16 at 01:32

answered Jul 30 '16 at 01:17

Psidom

4 Answers4