相当于python中的Haskell scanl [英] Equivalent of Haskell scanl in python

问题描述

我想知道 python 中是否有一个内置函数用于等效的 Haskell scanl，因为 reduce 相当于 foldl>.

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.

这样做的东西:

Prelude> scanl (+) 0 [1 ..10]
[0,1,3,6,10,15,21,28,36,45,55]

问题不在于如何实现它，我已经有 2 个实现，如下所示(但是，如果您有更优雅的实现，请随时在此处展示).

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).

第一个实现:

 # 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))

给出:

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

第二次实施:

 # 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))

给出:

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

谢谢:)

推荐答案

你可以使用这个，如果它更优雅:

You can use this, if its more elegant:

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

像这样使用:

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

Python 3.x 具有 itertools.accumulate(iterable,func= operator.add).它的实现如下.实施可能会给你一些想法:

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

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