Elegant pythonic cumsum

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

# Using list comprehention
cumsum = [sum(a[:i+1]) for i in range(len(a))]           # [1, 3, 6, 10, 15]

# Using map()
cumsum = map(lambda i:  sum(a[:i+1]), range(len(a)))     # [1, 3, 6, 10, 15]

My idea was to use reduce in a functional manner:

from operator import iadd
reduce(lambda acc, itm: iadd(acc, [acc[-1] + itm]), [1, 2, 3, 4, 5], [0])[1:]
>>> [1, 3, 6, 10, 15]

iadd from the operator module has the unique property of doing an in-place addition and returning the destination as a result.

If that [1:] copy makes you cringe, you could likewise do:

from operator import iadd
reduce(lambda acc, itm: (iadd(acc[0], [acc[1] + itm]), acc[1] + itm),
       [1, 2, 3, 4, 5], ([], 0))[0]
>>> [1, 3, 6, 10, 15]

But I discovered that locally the first example is much faster and IMO generators are more pythonic than functional programming like 'reduce':

reduce(lambda acc, itm: (iadd(acc[0], [acc[1] + itm]), acc[1] + itm), values_ten, ([], 0))[0]
Average: 6.4593828736e-06
reduce(lambda acc, itm: (iadd(acc[0], [acc[1] + itm]), acc[1] + itm), values_mil, ([], 0))[0]
Average: 0.727404361961
reduce(lambda acc, itm: iadd(acc, [acc[-1] + itm]), values_ten, [0])[1:]
Average: 5.16271911336e-06
reduce(lambda acc, itm: iadd(acc, [acc[-1] + itm]), values_mil, [0])[1:]
Average: 0.524223491301
cumsum_rking(values_ten)
Average: 1.9828751369e-06
cumsum_rking(values_mil)
Average: 0.234241141632
list(cumsum_larsmans(values_ten))
Average: 2.02786211569e-06
list(cumsum_larsmans(values_mil))
Average: 0.201473119335

Here's the benchmark script, YMMV:

from timeit import timeit

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

values_ten = list(xrange(10))
values_mil = list(xrange(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, 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, values_mil', number=10)

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

def cumsum_rking(iterable):
    values = list(iterable)
    for pos in xrange(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)

And here's my Python version string:

Python 2.7 (r27:82525, Jul  4 2010, 09:01:59) [MSC v.1500 32 bit (Intel)] on win32

It's available in Numpy:

>>> import numpy as np
>>> np.cumsum([1,2,3,4,5])
array([ 1,  3,  6, 10, 15])

Or use itertools.accumulate since Python 3.2:

>>> from itertools import accumulate
>>> list(accumulate([1,2,3,4,5]))
[ 1,  3,  6, 10, 15]

If Numpy is not an option, a generator loop would be the most elegant solution I can think of:

def cumsum(it):
    total = 0
    for x in it:
        total += x
        yield total

Ex.

>>> list(cumsum([1,2,3,4,5]))
>>> [1, 3, 6, 10, 15]