相当于python中的Haskell scanl

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.

我想知道在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。

Something that does this:

事情是这样的:

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).

问题不在于如何实现它，我已经有两个实现，如下所示(但是，如果您有一个更优雅的实现，请在这里展示)。

First implementation:

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))

Gives:

给:

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

Second implementation:

第二个实现:

 # 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:

给:

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

Thank you :)

谢谢你:)

3 个解决方案

#1

You can use this, if its more elegant:

你可以用这个，如果它更优雅:

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

Use it like:

使用它:

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:

Python 3。出现x itertools。积累(iterable,func = operator.add)。实现如下。实施可能会给你一些建议:

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

#2

I had a similar need. This version uses the python list comprehension

我也有类似的需要。这个版本使用python列表理解

def scanl(data):
    '''
    returns list of successive reduced values from the list (see haskell foldl)
    '''
    return [0] + [sum(data[:(k+1)]) for (k,v) in enumerate(data)]


>>> scanl(range(1,11))

gives:

给:

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

#3

As usual, the Python ecosystem is also overflowing with solutions:

与往常一样，Python生态系统也充满了解决方案:

Toolz has an accumulate capable of taking a user-supplied function as an argument. I tested it with lambda expressions.

工具z有一个累积能力，可以将用户提供的函数作为参数。我用lambda表达式测试了它。

https://github.com/pytoolz/toolz/blob/master/toolz/itertoolz.py

https://pypi.python.org/pypi/toolz

as does more_itertools

more_itertools一样

http://more-itertools.readthedocs.io/en/stable/api.html

I did not test the version from more-itertools, but it also can take a user-supplied function.

我没有从more itertools中测试这个版本，但是它也可以使用用户提供的函数。

#1