Equivalent of Haskell scanl in python

Question

I would like to know if there is a built in function in python for the equivalent Haskell

scanl

, as

reduce

is the equivalent of

foldl

.

Something that does this:

Prelude> scanl (+) 0 [1 ..10]

[0,1,3,6,10,15,21,28,36,45,55]

The question is not about how to implement it, I already have 2 implementations, shown below (however, if you have a more elegant one please feel free to show it here).

First implementation:

 # Inefficient, uses reduce multiple times

 def scanl(f, base, l):

   ls = [l[0:i] for i in range(1, len(l) + 1)]

   return [base] + [reduce(f, x, base) for x in ls]



  print scanl(operator.add, 0, range(1, 11))

Gives:

[0, 1, 3, 6, 10, 15, 21, 28, 36, 45, 55]

Second implementation:

 # Efficient, using an accumulator

 def scanl2(f, base, l):

   res = [base]

   acc = base

   for x in l:

     acc = f(acc, x)

     res += [acc]

   return res



 print scanl2(operator.add, 0, range(1, 11))

Gives:

[0, 1, 3, 6, 10, 15, 21, 28, 36, 45, 55]

Thank you :)

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Answer 1

You can use this, if its more elegant:

def scanl(f, base, l):
    for x in l:
        base = f(base, x)
        yield base

Use it like:

import operator
list(scanl(operator.add, 0, range(1,11)))

Python 3.x has itertools.accumulate(iterable, func= operator.add). It is implemented as below. The implementation might give you ideas:

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

Equivalent of Haskell scanl in python

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