How to find the cumulative sum of numbers in a lis

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].

I tried the following:

for i in time_interval:
    t1 = time_interval[0]
    t2 = time_interval[1] + t1
    t3 = time_interval[2] + t2
    print(t1, t2, t3)

4 10 22
4 10 22
4 10 22

Answer 1

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 2

First, you want a running list of subsequences:

subseqs = (seq[:i] for i in range(1, len(seq)+1))

Then you just call sum on each subsequence:

sums = [sum(subseq) for subseq in subseqs]

(This isn't the most efficient way to do it, because you're adding all of the prefixes repeatedly. But that probably won't matter for most use cases, and it's easier to understand if you don't have to think of the running totals.)

If you're using Python 3.2 or newer, you can use itertools.accumulate to do it for you:

sums = itertools.accumulate(seq)

And if you're using 3.1 or earlier, you can just copy the "equivalent to" source straight out of the docs (except for changing next(it) to it.next() for 2.5 and earlier).

Answer 3

A pure python oneliner for cumulative sum:

cumsum = lambda X: X[:1] + cumsum([X[0]+X[1]] + X[2:]) if X[1:] else X

This is a recursive version inspired by recursive cumulative sums. Some explanations:

1. The first term X[:1] is a list containing the previous element and is almost the same as [X[0]] (which would complain for empty lists).
2. The recursive cumsum call in the second term processes the current element [1] and remaining list whose length will be reduced by one.
3. if X[1:] is shorter for if len(X)>1.

Test:

cumsum([4,6,12])
#[4, 10, 22]

cumsum([])
#[]

And simular for cumulative product:

cumprod = lambda X: X[:1] + cumprod([X[0]*X[1]] + X[2:]) if X[1:] else X

Test:

cumprod([4,6,12])
#[4, 24, 288]

Answer 4

In Python 2 you can define your own generator function like this:

def accumu(lis):
    total = 0
    for x in lis:
        total += x
        yield total

In [4]: list(accumu([4,6,12]))
Out[4]: [4, 10, 22]

And in Python 3.2+ you can use itertools.accumulate():

In [1]: lis = [4,6,12]

In [2]: from itertools import accumulate

In [3]: list(accumulate(lis))
Out[3]: [4, 10, 22]

Answer 5

Behold:

a = [4, 6, 12]
reduce(lambda c, x: c + [c[-1] + x], a, [0])[1:]

Will output (as expected):

[4, 10, 22]