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?
zip(*[iter(s)]*n)
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.
iter()
[x] * n
n
x
*arg
zip()
x = iter([1,2,3,4,5,6,7,8,9]) print zip(x, x, x)
1.4m articles
1.4m replys
5 comments
57.0k users