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!

Answer 1

If you'd like a numpy solution

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

Answer 2

Python 3 has itertools.accumulate for exactly this purpose:

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

Answer 3

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

Answer 4

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

Answer 5

How about an ordinary loop?

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

print(result)

Answer 6

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))

Answer 7

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))