Iterate an iterator by chunks (of n) in Python? -

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this question has answer here:

can think of nice way (maybe itertools) split iterator chunks of given size?

therefore l=[1,2,3,4,5,6,7] chunks(l,3) becomes iterator [1,2,3], [4,5,6], [7]

i can think of small program not nice way maybe itertools.

the grouper() recipe itertools documentation's recipes comes close want:

def grouper(n, iterable, fillvalue=none):     "grouper(3, 'abcdefg', 'x') --> abc def gxx"     args = [iter(iterable)] * n     return izip_longest(fillvalue=fillvalue, *args)

it fill last chunk fill value, though.

a less general solution works on sequences handle last chunk desired is

[my_list[i:i + chunk_size] in range(0, len(my_list), chunk_size)]

finally, solution works on general iterators behaves desired is

def grouper(n, iterable):     = iter(iterable)     while true:        chunk = tuple(itertools.islice(it, n))        if not chunk:            return        yield chunk

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