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


Asked By – Oliver Zheng

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 look like if it was written with more verbose code?

Now we will see solution for issue: How does zip(*[iter(s)]*n) work in Python?


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)

This question is answered By – Ignacio Vazquez-Abrams

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