Python Running Sum in List

Question

Given the following list:

a=[1,2,3]

I'd like to generate a new list where each number is the sum of it and the values before it, like this:

result = [1,3,6]

Logic:

1 has no preceding value, so it stays the same.

3 is from the first value (1) added to the value of the second number in the list (2)

6 is from the sum of 1 and 2 from the first two elements, plus the third value of 3.

Thanks in advance!

score 17 · Answer 1 · edited Apr 09 '20 at 00:27

17

Python 3 has itertools.accumulate for exactly this purpose:

>>> from itertools import accumulate
>>> a=[1,2,3]
>>> list(accumulate(a))
[1, 3, 6]

edited Apr 09 '20 at 00:27

personal_cloud

3,943
3
28
38

answered Dec 30 '16 at 05:42

niemmi

17,113
7
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score 8 · Accepted Answer · answered Dec 30 '16 at 05:48

8

If you'd like a numpy solution

from numpy import cumsum
result = list(cumsum(a))

answered Dec 30 '16 at 05:48

gsmafra

2,434
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I wounder if it is that much faster then the built in Python version. – SudoKid Dec 30 '16 at 05:53

score 3 · Answer 3 · answered Dec 30 '16 at 05:45

3

How about an ordinary loop?

a = [1,2,3]
result = []
s = 0
for item in a:
    s += item
    result.append(s)

print(result)

answered Dec 30 '16 at 05:45

TigerhawkT3

48,464
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score 2 · Answer 4 · answered Dec 30 '16 at 05:44

2

Python has a function for this.

import itertools

result = list(itertools.accumlate([1, 2, 3]))

Python itertools solve some problems really well you should take some time and read over them.

https://docs.python.org/3/library/itertools.html

answered Dec 30 '16 at 05:44

SudoKid

439
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Shobeir · Answer 5 · 2016-12-31T06:35:00.573

2

Avinash Raj's code doesn't work correctly.

a = [1,2,3]
b = [sum(a[:(i+1)]) for i, j in enumerate(a)]
print(b)

Edited based on @Avinash Raj

edited Dec 31 '16 at 06:35

answered Dec 30 '16 at 05:45

Shobeir

127
7

yep, much simpler one `[sum(a[:i+1]) for i,j in enumerate(a)]` – Avinash Raj Dec 30 '16 at 05:47
While this has the advantage of being declarative and thus nicely readable, it notably has a worse time complexity (O(n^2)) compared to the simple iterative solution with a mutable cumulation variable (which has O(n)). – Mario Boley Aug 15 '20 at 21:58

score 2 · Answer 6 · answered Dec 30 '16 at 05:49

2

try this..

def running_sum(a):
  tot = 0
  for item in a:
    tot += item
    yield tot

a = [1,2,3,4]
print list(running_sum(a))

answered Dec 30 '16 at 05:49

Shivkumar kondi

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Dan Lenski · Answer 7 · 2016-12-30T16:42:42.030

There are about a hundred different ways to do this kind of cumulative sum. Depending on what you actually want to use the result for, a less obvious or less general-purpose solution might be more time- or memory-efficient—although the simple solution below is O(1) in terms of memory and O(N) in time.

The most straightforward procedural approach in virtually every imperative programming language goes something like this:

csum=0
result=[]
for val in a:
    csum += val
    result.append(csum)

The Python standard library also includes a function to do just this: itertools.accumulate.

import itertools
result = list(itertools.accumulate(a))

Python Running Sum in List

7 Answers7