How to find the cumulative sum of numbers in a list?

Question

time_interval = [4, 6, 12]

I want to sum up the numbers like [4, 4+6, 4+6+12] in order to get the list t = [4, 10, 22].

Answer 1

Assignment expressions from PEP 572 (new in Python 3.8) offer yet another way to solve this:

time_interval = [4, 6, 12]

total_time = 0
cum_time = [total_time := total_time + t for t in time_interval]

Answer 2

Somewhat hacky, but seems to work:

def cumulative_sum(l):
  y = [0]
  def inc(n):
    y[0] += n
    return y[0]
  return [inc(x) for x in l]

I did think that the inner function would be able to modify the y declared in the outer lexical scope, but that didn't work, so we play some nasty hacks with structure modification instead. It is probably more elegant to use a generator.

Answer 3

lst = [4,6,12]

[sum(lst[:i+1]) for i in xrange(len(lst))]

If you are looking for a more efficient solution (bigger lists?) a generator could be a good call (or just use numpy if you really care about perf).

def gen(lst):
    acu = 0
    for num in lst:
        yield num + acu
        acu += num

print list(gen([4, 6, 12]))

Answer 4

Try this: accumulate function, along with operator add performs the running addition.

import itertools  
import operator  
result = itertools.accumulate([1,2,3,4,5], operator.add)  
list(result)

Answer 5

values = [4, 6, 12]
total  = 0
sums   = []

for v in values:
  total = total + v
  sums.append(total)

print 'Values: ', values
print 'Sums:   ', sums

Running this code gives

Values: [4, 6, 12]
Sums:   [4, 10, 22]

Answer 6

If You want a pythonic way without numpy working in 2.7 this would be my way of doing it

l = [1,2,3,4]
_d={-1:0}
cumsum=[_d.setdefault(idx, _d[idx-1]+item) for idx,item in enumerate(l)]

now let's try it and test it against all other implementations

import timeit, sys
L=list(range(10000))
if sys.version_info >= (3, 0):
    reduce = functools.reduce
    xrange = range


def sum1(l):
    cumsum=[]
    total = 0
    for v in l:
        total += v
        cumsum.append(total)
    return cumsum


def sum2(l):
    import numpy as np
    return list(np.cumsum(l))

def sum3(l):
    return [sum(l[:i+1]) for i in xrange(len(l))]

def sum4(l):
    return reduce(lambda c, x: c + [c[-1] + x], l, [0])[1:]

def this_implementation(l):
    _d={-1:0}
    return [_d.setdefault(idx, _d[idx-1]+item) for idx,item in enumerate(l)]


# sanity check
sum1(L)==sum2(L)==sum3(L)==sum4(L)==this_implementation(L)
>>> True    

# PERFORMANCE TEST
timeit.timeit('sum1(L)','from __main__ import sum1,sum2,sum3,sum4,this_implementation,L', number=100)/100.
>>> 0.001018061637878418

timeit.timeit('sum2(L)','from __main__ import sum1,sum2,sum3,sum4,this_implementation,L', number=100)/100.
>>> 0.000829620361328125

timeit.timeit('sum3(L)','from __main__ import sum1,sum2,sum3,sum4,this_implementation,L', number=100)/100.
>>> 0.4606760001182556 

timeit.timeit('sum4(L)','from __main__ import sum1,sum2,sum3,sum4,this_implementation,L', number=100)/100.
>>> 0.18932826995849608

timeit.timeit('this_implementation(L)','from __main__ import sum1,sum2,sum3,sum4,this_implementation,L', number=100)/100.
>>> 0.002348129749298096