How does zip(*[iter(s)]*n) work in Python?

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渐次进展 2020-11-22 05:55
s = [1,2,3,4,5,6,7,8,9]
n = 3

zip(*[iter(s)]*n) # returns [(1,2,3),(4,5,6),(7,8,9)]

How does zip(*[iter(s)]*n) work? What would it l

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  •  一生所求
    2020-11-22 06:13

    iter() is an iterator over a sequence. [x] * n produces a list containing n quantity of x, i.e. a list of length n, where each element is x. *arg unpacks a sequence into arguments for a function call. Therefore you're passing the same iterator 3 times to zip(), and it pulls an item from the iterator each time.

    x = iter([1,2,3,4,5,6,7,8,9])
    print zip(x, x, x)
    

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