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 Ruby:
[1,2,3,4,5].inject(0) {|m,x| m + x}
#> 15
Obviously, Python provides the reduce function, which is an implementation of fold, exactly as above, however, I was told that the 'pythonic' way of programming was to avoid lambda terms and higher-order functions, preferring list-comprehensions where possible. Therefore, is there a preferred way of folding a list, or list-like structure in Python that isn't the reduce function, or is reduce the idiomatic way of achieving this?
Starting Python 3.8, and the introduction of assignment expressions (PEP 572) (:= operator), which gives the possibility to name the result of an expression, we can use a list comprehension to replicate what other languages call fold/foldleft/reduce operations:
Given a list, a reducing function and an accumulator:
items = [1, 2, 3, 4, 5]
f = lambda acc, x: acc * x
accumulator = 1
we can fold items with f in order to obtain the resulting accumulation:
[accumulator := f(accumulator, x) for x in items]
# accumulator = 120
or in a condensed formed:
acc = 1; [acc := acc * x for x in [1, 2, 3, 4, 5]]
# acc = 120
Note that this is actually also a "scanleft" operation as the result of the list comprehension represents the state of the accumulation at each step:
acc = 1
scanned = [acc := acc * x for x in [1, 2, 3, 4, 5]]
# scanned = [1, 2, 6, 24, 120]
# acc = 120