Woo hoo! Another coincidence. I just happened to be playing with this problem today: You have a large list - xs - of N numbers. It's necessary to compute slice sums sum(xs[i:j]) for a great many slices, 0 <= i <= j <= N. For concreteness, say xs is a time series representing a toll booth receipt's by hour across years. "Management" may ask for all sorts of sums - by 24-hour period, by week, by month, by year, by season, ... A little thought showed that sum(xs[i:j]) = sum(xs[:j]) - sum(xs[:i]), so if we precomputed just the prefix sums, the sum across an arbitrary slice could be computed thereafter in constant time. Hard to beat that ;-) But computing the prefix sums is exactly what accumulate() does! With one twist: while we have N numbers, there are N+1 slice indices. So accumulate(xs) doesn't quite work. It needs to also have a 0 inserted as the first prefix sum (the empty prefix sum(xs[:0]). Which is exactly what a this_is_the_initial_value=0 argument would do for us. As is, using the chain trick: class SliceSummer: def __init__(self, xs): from itertools import accumulate, chain self.N = N = len(xs) if not N: raise ValueError("need a non-empty sequence") self.prefixsum = list(accumulate(chain([0], xs))) assert len(self.prefixsum) == N+1 def slicesum(self, i, j): N = self.N if not 0 <= i <= j <= N: raise ValueError(f"need 0 <= {i} <= {j} <= {N}") return self.prefixsum[j] - self.prefixsum[i] def test(N): from random import randrange xs = [randrange(-10, 11) for _ in range(N)] ntried = 0 ss = SliceSummer(xs) NP1 = N + 1 for i in range(NP1): for j in range(i, NP1): ntried += 1 assert ss.slicesum(i, j) == sum(xs[i:j]) assert ntried == N * NP1 // 2 + NP1, ntried