Elegant pythonic cumsum

Question

What would be an elegant and pythonic way to implement cumsum?
Alternatively - if there\'a already a built-in way to do it, that would be even better of course...

Answer 1

in place:

a=[1,2,3,4,5]
def cumsum(a):
    for i in range(1,len(a)):
        a[i]+=a[i-1]

cumsum(a)
print a
"[1, 3, 6, 10, 15]"

Answer 2

I have updated @GrantJ 's answer and have timed everything with Python 3 in Jupyter. I have added in the following simple accumulate example:

def cumsum_acc(iterable):
    return accumulate(iterable)

The timings show that this approach is the fastest:

reduce(lambda acc, itm: (iadd(acc[0], [acc[1] + itm]), acc[1] + itm), values_ten, ([], 0))[0]
Average: 4.18553415873987e-06
reduce(lambda acc, itm: (iadd(acc[0], [acc[1] + itm]), acc[1] + itm), values_mil, ([], 0))[0]
Average: 0.4559302114011018
reduce(lambda acc, itm: iadd(acc, [acc[-1] + itm]), values_ten, [0])[1:]
Average: 3.3726942356950078e-06
reduce(lambda acc, itm: iadd(acc, [acc[-1] + itm]), values_mil, [0])[1:]
Average: 0.4217967894903154
cumsum_rking(values_ten)
Average: 2.2734952410773416e-06
cumsum_rking(values_mil)
Average: 0.24106813411868303
list(cumsum_larsmans(values_ten))
Average: 1.5819200433296032e-06
list(cumsum_larsmans(values_mil))
Average: 0.17061943569953542
list(cumsum_acc(values_ten))
Average: 9.456979988264607e-07
list(cumsum_acc(values_mil))
Average: 0.11057746014014924

The code behind it (by GrantJ, modified for Python 3):

from timeit import timeit
from itertools import accumulate
from functools import reduce

def bmark(prog, setup, number):
    duration = timeit(prog, setup=setup, number=number)
    print(prog)
    print('Average:', duration / number)

values_ten = list(range(10))
values_mil = list(range(1000000))

from operator import iadd

bmark('reduce(lambda acc, itm: (iadd(acc[0], [acc[1] + itm]), acc[1] + itm), \
values_ten, ([], 0))[0]',
      setup='from __main__ import iadd, reduce, values_ten', number=1000000)
bmark('reduce(lambda acc, itm: (iadd(acc[0], [acc[1] + itm]), acc[1] + itm), \
values_mil, ([], 0))[0]',
      setup='from __main__ import iadd, reduce, values_mil', number=10)

bmark('reduce(lambda acc, itm: iadd(acc, [acc[-1] + itm]), \
values_ten, [0])[1:]',
      setup='from __main__ import iadd, reduce, values_ten', number=1000000)
bmark('reduce(lambda acc, itm: iadd(acc, [acc[-1] + itm]), \
values_mil, [0])[1:]',
      setup='from __main__ import iadd, reduce, values_mil', number=10)

def cumsum_rking(iterable):
    values = list(iterable)
    for pos in range(1, len(values)):
        values[pos] += values[pos - 1]
    return values

bmark('cumsum_rking(values_ten)',
      setup='from __main__ import cumsum_rking, values_ten', number=1000000)
bmark('cumsum_rking(values_mil)',
      setup='from __main__ import cumsum_rking, values_mil', number=10)

def cumsum_larsmans(iterable):
    total = 0
    for value in iterable:
        total += value
        yield total

bmark('list(cumsum_larsmans(values_ten))',
      setup='from __main__ import cumsum_larsmans, values_ten', number=1000000)
bmark('list(cumsum_larsmans(values_mil))',
      setup='from __main__ import cumsum_larsmans, values_mil', number=10)

def cumsum_acc(iterable):
    return accumulate(iterable)

bmark('list(cumsum_acc(values_ten))',
      setup='from __main__ import cumsum_acc, accumulate, values_ten', number=1000000)
bmark('list(cumsum_acc(values_mil))',
      setup='from __main__ import cumsum_acc, accumulate, values_mil', number=10)

Answer 3

Starting Python 3.8, and the introduction of assignment expressions (PEP 572) (:= operator), we can use and increment a variable within a list comprehension:

total = 0
[total := total + x for x in [1, 2, 3 ,4, 5]]
# [1, 3, 6, 10, 15]

This:

• Initializes a variable total to 0 which symbolizes the cumulative sum
• For each item, this both:
  • increments total with the current looped item (total := total + x) via an assignment expression
  • and at the same time, maps x to the new value of total

Answer 4

for loops are pythonic

def cumsum(vec):
    r = [vec[0]]
    for val in vec[1:]:
        r.append(r[-1] + val)
    return r

Answer 5

def cumsum(a):
     return map(lambda x: sum( a[0:x+1] ), range( 0, len(a) ))

cumsum([1,2,3])

> [1, 3, 6]

Answer 6

a=[1,2,3,4,5]

def cumsum(a):
    a=iter(a)
    cc=[next(a)]
    for i in a:
        cc.append(cc[-1]+i)
    return cc

print cumsum(a)
"[1, 3, 6, 10, 15]"