[英]Equivalent of Haskell scanl in python
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python中是否有内置函数用于等效的Haskell
scanl
,因为reduce
相当于foldl
。
Something that does this: 这样做的东西:
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). 问题不是如何实现它,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)。
First implementation: 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))
Gives: 得到:
[0, 1, 3, 6, 10, 15, 21, 28, 36, 45, 55]
Second implementation: 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))
Gives: 得到:
[0, 1, 3, 6, 10, 15, 21, 28, 36, 45, 55]
Thank you :) 谢谢 :)
You can use this, if its more elegant: 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
Use it like: 使用它像:
import operator
list(scanl(operator.add, 0, range(1,11)))
Python 3.x has itertools.accumulate(iterable, func= operator.add)
. Python 3.x有
itertools.accumulate(iterable, func= operator.add)
。 It is implemented as below. 它的实现如下。 The implementation might give you ideas:
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
I had a similar need. 我有类似的需求。 This version uses the python list comprehension
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: 得到:
[0, 1, 3, 6, 10, 15, 21, 28, 36, 45, 55]
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 a scan left operation: 启动
Python 3.8
,并引入赋值表达式(PEP 572) ( :=
运算符),它可以命名表达式的结果,我们可以使用列表推导来复制扫描左操作:
acc = 0
scanned = [acc := acc + x for x in [1, 2, 3, 4, 5]]
# scanned = [1, 3, 6, 10, 15]
Or in a generic way, given a list, a reducing function and an initialized accumulator: 或者以通用方式,给定列表,缩减函数和初始化累加器:
items = [1, 2, 3, 4, 5]
f = lambda acc, x: acc + x
accumulator = 0
we can scan items
from the left and reduce them with f
: 我们可以从左边扫描
items
并用f
减少它们:
scanned = [accumulator := f(accumulator, x) for x in items]
# scanned = [1, 3, 6, 10, 15]
As usual, the Python ecosystem is also overflowing with solutions: 像往常一样,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. Toolz有一个能够将用户提供的函数作为参数的累加。 I tested it with lambda expressions.
我用lambda表达式测试了它。
https://github.com/pytoolz/toolz/blob/master/toolz/itertoolz.py https://github.com/pytoolz/toolz/blob/master/toolz/itertoolz.py
https://pypi.python.org/pypi/toolz https://pypi.python.org/pypi/toolz
as does more_itertools 和more_itertools一样
http://more-itertools.readthedocs.io/en/stable/api.html 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. 我没有测试更多的itertools版本,但它也可以采用用户提供的功能。
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