問題描述
你能想出一個很好的方法(也許使用 itertools)來將迭代器分成給定大小的塊嗎?
Can you think of a nice way (maybe with itertools) to split an iterator into chunks of given size?
因此 l=[1,2,3,4,5,6,7] 與 chunks(l,3) 成為迭代器 [1,2,3], [4,5,6], [7]
Therefore l=[1,2,3,4,5,6,7] with chunks(l,3) becomes an iterator [1,2,3], [4,5,6], [7]
我可以想到一個小程序來做到這一點，但可能不是使用 itertools 的好方法.
I can think of a small program to do that but not a nice way with maybe itertools.
推薦答案
itertools 文檔的 grouper() 配方.python.org/3/library/itertools.html#itertools-recipes" rel="nofollow noreferrer">recipes 接近你想要的:
The grouper() recipe from the itertools documentation's recipes comes close to what you want:
def grouper(n, iterable, fillvalue=None):
"grouper(3, 'ABCDEFG', 'x') --> ABC DEF Gxx"
args = [iter(iterable)] * n
return izip_longest(fillvalue=fillvalue, *args)
不過，它會用一個填充值填充最后一個塊.
It will fill up the last chunk with a fill value, though.
一種不太通用的解決方案，它只適用于序列，但可以根據(jù)需要處理最后一個塊
A less general solution that only works on sequences but does handle the last chunk as desired is
[my_list[i:i + chunk_size] for i in range(0, len(my_list), chunk_size)]
最后，一個適用于通用迭代器并按需要運行的解決方案是
Finally, a solution that works on general iterators and behaves as desired is
def grouper(n, iterable):
it = iter(iterable)
while True:
chunk = tuple(itertools.islice(it, n))
if not chunk:
return
yield chunk
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