45

I have a list of tuples (always pairs) like this:

[(0, 1), (2, 3), (5, 7), (2, 1)]

I'd like to find the sum of the first items in each pair, i.e.:

0 + 2 + 5 + 2

How can I do this in Python? At the moment I'm iterating through the list: 

sum = 0
for pair in list_of_pairs:
   sum += pair[0]

I have a feeling there must be a more Pythonic way.

gsamaras
  • 71,951
  • 46
  • 188
  • 305
Ben
  • 66,838
  • 37
  • 84
  • 108

7 Answers7

80

In modern versions of Python I'd suggest what SilentGhost posted (repeating here for clarity):

sum(i for i, j in list_of_pairs)

In an earlier version of this answer I had suggested this, which was necessary because SilentGhost's version didn't work in the version of Python (2.3) that was current at the time:

sum([pair[0] for pair in list_of_pairs])

Now that version of Python is beyond obsolete, and SilentGhost's code works in all currently-maintained versions of Python, so there's no longer any reason to recommend the version I had originally posted.

David Z
  • 128,184
  • 27
  • 255
  • 279
47
sum(i for i, j in list_of_pairs)

will do too.

SilentGhost
  • 307,395
  • 66
  • 306
  • 293
  • I thought so at first, but when I tried that in the quickest Python I could access it raised a syntax error :-( Turns out I was testing on Python 2.3, though... +1 anyway – David Z Mar 12 '09 at 10:47
  • +1: prefer this -- tuples have a fixed size and you usually know what the size is. – S.Lott Mar 12 '09 at 10:56
  • 1
    I like this approach, too. But Davids solution works with n-tuples, too, which might be preferable, depending on the actual problem. –  Mar 12 '09 at 11:15
  • 3
    I should probably note that this code is about 15% faster than the one from accepted answer. – SilentGhost May 23 '09 at 19:28
  • `i` and `j` should be replaced with `first` and `second` ;) – Arnab Datta Aug 15 '12 at 22:58
  • `i` and `j` are fine. If I had to type `first` and `second` for _every_ sequence iteration, I'd have bleeding stumps for fingers within a man-month. – Cecil Curry Nov 30 '16 at 06:01
24

I recommend:

sum(i for i, _ in list_of_pairs)

Note:

Using the variable _(or __ to avoid confliction with the alias of gettext) instead of j has at least two benefits:

  1. _(which stands for placeholder) has better readability
  2. pylint won't complain: "Unused variable 'j'"
Hui Zheng
  • 10,084
  • 2
  • 35
  • 40
6

If you have a very large list or a generator that produces a large number of pairs you might want to use a generator based approach. For fun I use itemgetter() and imap(), too. A simple generator based approach might be enough, though.

import operator
import itertools

idx0 = operator.itemgetter(0)
list_of_pairs = [(0, 1), (2, 3), (5, 7), (2, 1)]
sum(itertools.imap(idx0, list_of_pairs))

Note that itertools.imap() is available in Python >= 2.3. So you can use a generator based approach there, too.

borlafu
  • 80
  • 1
  • 6
4

Obscure (but fun) answer:

>>> sum(zip(*list_of_pairs)[0])
9

Or when zip's are iterables only this should work:

>>> sum(zip(*list_of_pairs).__next__())
9
SilentGhost
  • 307,395
  • 66
  • 306
  • 293
Ali Afshar
  • 40,967
  • 12
  • 95
  • 109
0

Below is sample code, you can also specify the list range.

def test_lst_sum():
    lst = [1, 3, 5]
    print sum(lst)  # 9
    print sum(lst[1:])  # 8

    print sum(lst[5:])  # 0  out of range so return 0
    print sum(lst[5:-1])  # 0

    print sum(lst[1: -1])  # 3

    lst_tp = [('33', 1), ('88', 2), ('22', 3), ('44', 4)]
    print sum(x[1] for x in lst_tp[1:])  # 9
Jayhello
  • 5,931
  • 3
  • 49
  • 56
-2
s,p=0,0
for i in l:
  s=s+i[0]
  p=p+i[1]
print(tuple(s,p))
InSync
  • 4,851
  • 4
  • 8
  • 30
Aman
  • 5
  • 1