List comprehension with an accumulator

Question

What is the best way to replicate this simple function using a list comprehension (or another compact approach)?

import numpy as np

sum=0
array=[]
for i in np.random.rand(100):
   sum+=i
   array.append(sum)

Answer 1

In Python 3, you'd use itertools.accumulate():

from itertools import accumulate

array = list(accumulate(rand(100)))

Accumulate yields the running result of adding up the values of the input iterable, starting with the first value:

>>> from itertools import accumulate
>>> list(accumulate(range(10)))
[0, 1, 3, 6, 10, 15, 21, 28, 36, 45]

You can pass in a different operation as a second argument; this should be a callable that takes the accumulated result and the next value, returning the new accumulated result. The operator module is very helpful in providing standard mathematical operators for this kind of work; you could use it to produce a running multiplication result for example:

>>> import operator
>>> list(accumulate(range(1, 10), operator.mul))
[1, 2, 6, 24, 120, 720, 5040, 40320, 362880]

The functionality is easy enough to backport to older versions (Python 2, or Python 3.0 or 3.1):

# Python 3.1 or before

import operator

def accumulate(iterable, func=operator.add):
    'Return running totals'
    # accumulate([1,2,3,4,5]) --> 1 3 6 10 15
    # accumulate([1,2,3,4,5], operator.mul) --> 1 2 6 24 120
    it = iter(iterable)
    total = next(it)
    yield total
    for element in it:
        total = func(total, element)
        yield total