How does zip(*[iter(s)]*n) work in Python?
iterator, python
Solution
`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(list(zip(x, x, x)))
Problem
``` s = [1,2,3,4,5,6,7,8,9] n = 3 list(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? This is a technique used for splitting a list into chunks of equal size - see that question for a general overview of the problem.