What is the 'pythonic' equivalent to the 'fold' function from functional programming?

Question

What is the most idiomatic way to achieve something like the following, in Haskell:

foldl (+) 0 [1,2,3,4,5]
--> 15

Or its equivalent in

Answer 1

In Python 3, the reduce has been removed: Release notes. Nevertheless you can use the functools module

import operator, functools
def product(xs):
    return functools.reduce(operator.mul, xs, 1)

On the other hand, the documentation expresses preference towards for-loop instead of reduce, hence:

def product(xs):
    result = 1
    for i in xs:
        result *= i
    return result