18

是否有 Pythonic 等价于 Ruby 的#each_cons

在 Ruby 中,您可以这样做:

array = [1,2,3,4]
array.each_cons(2).to_a
=> [[1,2],[2,3],[3,4]]
4

8 回答 8

21

我认为没有,我查看了内置模块itertools,这是我期望的。您可以简单地创建一个:

def each_cons(xs, n):
    return [xs[i:i+n] for i in range(len(xs)-n+1)]
于 2011-05-04T04:17:30.737 回答
11

对于此类事情,itertools您应该查看的模块是:

from itertools import tee, izip

def pairwise(iterable):
    "s -> (s0,s1), (s1,s2), (s2, s3), ..."
    a, b = tee(iterable)
    next(b, None)
    return izip(a, b)

然后:

>>> list(pairwise([1, 2, 3, 4]))
[(1, 2), (2, 3), (3, 4)]

对于更通用的解决方案,请考虑以下内容:

def split_subsequences(iterable, length=2, overlap=0):
    it = iter(iterable)
    results = list(itertools.islice(it, length))
    while len(results) == length:
        yield results
        results = results[length - overlap:]
        results.extend(itertools.islice(it, length - overlap))
    if results:
        yield results

这允许任意长度的子序列和任意重叠。用法:

>> list(split_subsequences([1, 2, 3, 4], length=2))
[[1, 2], [3, 4]]
>> list(split_subsequences([1, 2, 3, 4], length=2, overlap=1))
[[1, 2], [2, 3], [3, 4], [4]]
于 2011-05-04T04:12:12.570 回答
6

我的列表解决方案(Python2):

import itertools
def each_cons(xs, n):
    return itertools.izip(*(xs[i:] for i in xrange(n)))

编辑:itertools.izip不再使用 Python 3 ,因此您使用 plain zip

def each_cons(xs, n):
    return zip(*(xs[i:] for i in range(n)))
于 2012-10-14T06:44:08.913 回答
5

快速单线:

a = [1, 2, 3, 4]

out = [a[i:i + 2] for i in range(len(a) - 1)]
于 2011-05-04T04:16:06.710 回答
5

Python 肯定可以做到这一点。如果你不想那么急切,使用itertool 的islice 和izip。此外,重要的是要记住普通切片将创建一个副本,因此如果内存使用很重要,您还应该考虑 itertool 等价物。

each_cons = lambda l: zip(l[:-1], l[1:])

于 2011-05-04T04:24:27.187 回答
5

更新:没关系我在下面的回答,只需使用toolz.itertoolz.sliding_window()- 它会做正确的事情。


each_cons对于在序列/生成器长度不足时保留 Ruby 行为的真正惰性实现:

import itertools
def each_cons(sequence, n):
    return itertools.izip(*(itertools.islice(g, i, None)
                          for i, g in
                          enumerate(itertools.tee(sequence, n))))

例子:

>>> print(list(each_cons(xrange(5), 2)))
[(0, 1), (1, 2), (2, 3), (3, 4)]
>>> print(list(each_cons(xrange(5), 5)))
[(0, 1, 2, 3, 4)]
>>> print(list(each_cons(xrange(5), 6)))
[]
>>> print(list(each_cons((a for a in xrange(5)), 2)))
[(0, 1), (1, 2), (2, 3), (3, 4)]

请注意,在 izip 的参数上使用的元组解包应用于大小为的元组nitertools.tee(xs, n)即“窗口大小”),而不是我们想要迭代的序列。

于 2014-01-12T14:08:46.910 回答
2

接近@Blender 的解决方案,但有一个修复:

a = [1, 2, 3, 4]
n = 2
out = [a[i:i + n] for i in range(len(a) - n + 1)]
# => [[1, 2], [2, 3], [3, 4]]

或者

a = [1, 2, 3, 4]
n = 3
out = [a[i:i + n] for i in range(len(a) - n + 1)]
# => [[1, 2, 3], [2, 3, 4]]
于 2019-04-11T13:14:47.327 回答
1

Elias 代码的 Python 3 版本

from itertools import islice, tee

def each_cons(sequence, n):
    return zip(
        *(
            islice(g, i, None)
            for i, g in
            enumerate(tee(sequence, n))
        )
    )

$ ipython

...    

In [2]: a_list = [1, 2, 3, 4, 5]

In [3]: list(each_cons(a_list, 2))
Out[3]: [(1, 2), (2, 3), (3, 4), (4, 5)]

In [4]: list(each_cons(a_list, 3))
Out[4]: [(1, 2, 3), (2, 3, 4), (3, 4, 5)]

In [5]: list(each_cons(a_list, 5))
Out[5]: [(1, 2, 3, 4, 5)]

In [6]: list(each_cons(a_list, 6))
Out[6]: []
于 2014-11-25T17:56:55.857 回答