Transpose list of lists Transpose list of lists python python

# Transpose list of lists

``map(list, zip(*l))--> [[1, 4, 7], [2, 5, 8], [3, 6, 9]]``

For python 3.x users can use

``list(map(list, zip(*l))) # short circuits at shortest nested list if table is jaggedlist(map(list, itertools.zip_longest(*l, fillvalue=None))) # discards no data if jagged and fills short nested lists with None``

Explanation:

There are two things we need to know to understand what's going on:

1. The signature of zip: `zip(*iterables)` This means `zip` expects an arbitrary number of arguments each of which must be iterable. E.g. `zip([1, 2], [3, 4], [5, 6])`.
2. Unpacked argument lists: Given a sequence of arguments `args`, `f(*args)` will call `f` such that each element in `args` is a separate positional argument of `f`.
3. `itertools.zip_longest` does not discard any data if the number of elements of the nested lists are not the same (homogenous), and instead fills in the shorter nested lists then zips them up.

Coming back to the input from the question `l = [[1, 2, 3], [4, 5, 6], [7, 8, 9]]`, `zip(*l)` would be equivalent to `zip([1, 2, 3], [4, 5, 6], [7, 8, 9])`. The rest is just making sure the result is a list of lists instead of a list of tuples.

One way to do it is with NumPy transpose. For a list, a:

``>>> import numpy as np>>> np.array(a).T.tolist()[[1, 4, 7], [2, 5, 8], [3, 6, 9]]``

Or another one without zip:

``>>> map(list,map(None,*a))[[1, 4, 7], [2, 5, 8], [3, 6, 9]]``

Equivalently to Jena's solution:

``>>> l=[[1,2,3],[4,5,6],[7,8,9]]>>> [list(i) for i in zip(*l)]... [[1, 4, 7], [2, 5, 8], [3, 6, 9]]``