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Home/ Questions/Q 889153
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Editorial Team
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Editorial Team
Asked: May 15, 20262026-05-15T13:31:22+00:00 2026-05-15T13:31:22+00:00

Current version: def chop(ar,size): p=len(ar)/size for i in xrange(p): yield ar[(i*size):((i+1)*size)] ar is type

  • 0

Current version:

def chop(ar,size):
    p=len(ar)/size
    for i in xrange(p):
        yield ar[(i*size):((i+1)*size)]

ar is type of list().

What i want is that chop() takes iterator and return iterator.

for i in chop(xrange(9),3):
    for j in i:
       print j,
    print

prints

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  1. Editorial Team
    Editorial Team
    2026-05-15T13:31:23+00:00Added an answer on May 15, 2026 at 1:31 pm

    There’s an implementation in the itertools documentation:

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