在python中等效的Haskell scanl

Question

我想知道python中是否有内置函数用于等效的Haskell scanl,reduce相当于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,如果你有一个更优雅的实现,请随时在这里显示).

首次实施:

 # 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))

得到:

[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))

得到:

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

谢谢 :)

Answer 1

你可以使用它,如果它更优雅:

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

使用它像:

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

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

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

Answer 2

开始Python 3.8，并引入赋值表达式（PEP 572）（:=运算符），它提供了命名表达式结果的可能性，我们可以使用列表理解来复制左扫描操作：

acc = 0
scanned = [acc := acc + x for x in [1, 2, 3, 4, 5]]
# scanned = [1, 3, 6, 10, 15]

或者以通用方式，给定一个列表、一个归约函数和一个初始化的累加器：

items = [1, 2, 3, 4, 5]
f = lambda acc, x: acc + x
accumulator = 0

我们可以items从左侧扫描并使用以下方法减少它们f：

scanned = [accumulator := f(accumulator, x) for x in items]
# scanned = [1, 3, 6, 10, 15]