問題描述

我正在嘗試用 Python 編寫 Haskell 函數(shù)splitEvery".這是它的定義:

I'm trying to write the Haskell function 'splitEvery' in Python. Here is it's definition:

splitEvery :: Int -> [e] -> [[e]]
    @'splitEvery' n@ splits a list into length-n pieces.  The last
    piece will be shorter if  does not evenly divide the length of
    the list.

它的基本版本可以正常工作，但我想要一個可以與生成器表達(dá)式、列表和迭代器一起使用的版本.而且，如果有一個生成器作為輸入，它應(yīng)該返回一個生成器作為輸出！

The basic version of this works fine, but I want a version that works with generator expressions, lists, and iterators. And, if there is a generator as an input it should return a generator as an output!

# should not enter infinite loop with generators or lists
splitEvery(itertools.count(), 10)
splitEvery(range(1000), 10)

# last piece must be shorter if n does not evenly divide
assert splitEvery(5, range(9)) == [[0, 1, 2, 3, 4], [5, 6, 7, 8]]

# should give same correct results with generators
tmp = itertools.islice(itertools.count(), 10)
assert list(splitEvery(5, tmp)) == [[0, 1, 2, 3, 4], [5, 6, 7, 8]]

當(dāng)前實(shí)施

這是我目前擁有的代碼，但它不適用于簡單的列表.

Current Implementation

Here is the code I currently have but it doesn't work with a simple list.

def splitEvery_1(n, iterable):
    res = list(itertools.islice(iterable, n))
    while len(res) != 0:
        yield res
        res = list(itertools.islice(iterable, n))

這個不適用于生成器表達(dá)式(感謝 jellybean 修復(fù)它):

This one doesn't work with a generator expression (thanks to jellybean for fixing it):

def splitEvery_2(n, iterable): 
    return [iterable[i:i+n] for i in range(0, len(iterable), n)]

必須有一段簡單的代碼來進(jìn)行拆分.我知道我可以擁有不同的功能，但它似乎應(yīng)該很容易做到.我可能陷入了一個不重要的問題，但這真的讓我很煩惱.

There has to be a simple piece of code that does the splitting. I know I could just have different functions but it seems like it should be and easy thing to do. I'm probably getting stuck on an unimportant problem but it's really bugging me.

類似于 http://docs.python.org 的 grouper/library/itertools.html#itertools.groupby 但我不希望它填充額外的值.

It is similar to grouper from http://docs.python.org/library/itertools.html#itertools.groupby but I don't want it to fill extra values.

def grouper(n, iterable, fillvalue=None):
    "grouper(3, 'ABCDEFG', 'x') --> ABC DEF Gxx"
    args = [iter(iterable)] * n
    return izip_longest(fillvalue=fillvalue, *args)

它確實(shí)提到了一種截?cái)嘧詈笠粋€值的方法.這也不是我想要的.

It does mention a method that truncates the last value. This isn't what I want either.

保證迭代的從左到右的評估順序.這使得使用 izip(*[iter(s)]*n) 將數(shù)據(jù)系列聚類為 n 長度組成為可能.

The left-to-right evaluation order of the iterables is guaranteed. This makes possible an idiom for clustering a data series into n-length groups using izip(*[iter(s)]*n).

list(izip(*[iter(range(9))]*5)) == [[0, 1, 2, 3, 4]]
# should be [[0, 1, 2, 3, 4], [5, 6, 7, 8]]

推薦答案

from itertools import islice

def split_every(n, iterable):
    i = iter(iterable)
    piece = list(islice(i, n))
    while piece:
        yield piece
        piece = list(islice(i, n))

一些測試:

>>> list(split_every(5, range(9)))
[[0, 1, 2, 3, 4], [5, 6, 7, 8]]

>>> list(split_every(3, (x**2 for x in range(20))))
[[0, 1, 4], [9, 16, 25], [36, 49, 64], [81, 100, 121], [144, 169, 196], [225, 256, 289], [324, 361]]

>>> [''.join(s) for s in split_every(6, 'Hello world')]
['Hello ', 'world']

>>> list(split_every(100, []))
[]

這篇關(guān)于在python中每隔n個項(xiàng)目拆分一個生成器/可迭代(splitEvery)的文章就介紹到這了，希望我們推薦的答案對大家有所幫助，也希望大家多多支持html5模板網(wǎng)！