How to change the major order of lists in python?

Question

I have a list of N lists, each sub-list containing M elements.

How do I create a list of M lists with N elements each, containing the n-th elements from the initial lists?

Let's say I have this

myList = [[1, 2, 3],[4, 5, 6]]

I want to have

[[1, 4],[2, 5],[3, 6]]

Performance is also important. I have sublists of millions of elements (though, one dimension will always be small: like 1.000.000 x 8 )

score 3 · Accepted Answer · answered Mar 18 '13 at 09:56

3

This will give tuples, but it's trivial to extend this to contain lists:

zip(*myList)

i.e.

[list(i) for i in zip(*myList)]

answered Mar 18 '13 at 09:56

filmor

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score 3 · Answer 2 · edited May 23 '17 at 12:28

>>> lst = [[1, 2, 3], [4, 5, 6]]
>>> zip(*lst)
[(1, 4), (2, 5), (3, 6)]

zip returns a list in Python 2, but an iterator in Python 3, which you have to convert to a list if you need to (list(zip(lst)))

See also: Matrix Transpose in Python, Transpose/Unzip Function (inverse of zip)?

What zip(*lst) does is unpacks the elements of lst using the * operator into separate arguments of the zip function.

We know what happens when we put two lists in:

>>> zip([1, 2, 3], [4, 5, 6])
[(1, 4), (2, 5), (3, 6)]

So if we have a list containing those two lists, the * will unpack the list to form two separate arguments - the equivalent of the above call.

>>> zip(*[[1, 2, 3], [4, 5, 6]])
[(1, 4), (2, 5), (3, 6)]

score 3 · Answer 3 · answered Mar 18 '13 at 09:56

3

Use

myList = [[1, 2, 3],[4, 5, 6]] 
zip(*myList )

answered Mar 18 '13 at 09:56

Arty

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score 2 · Answer 4 · edited Mar 19 '13 at 02:34

2

Maybe with the help of itertools.izip()

itertools.izip(*myList)

To create lists from this iterator do the following:

map(list, itertools.izip(*myList))

edited Mar 19 '13 at 02:34

jamylak

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answered Mar 18 '13 at 09:56

Cédric Julien

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dmg · Answer 5 · 2013-03-18T10:18:36.757

0

Since you mentioned efficiency, I tried these with PyPy and somewhere around 100x1000 they start to be better than zip (on CPython however they are worse):

myList = [[1, 2, 3],[4, 5, 6]]
newList = [[] for _ in xrange(len(myList[0]))]
    for innerList in myList:
        for elIdx in xrange(len(innerList)):
            newList[elIdx].append(innerList[elIdx])

print newList

And my first attempt (which is worse), but still better then zip on PyPy:

newList = [[None for i in xrange(len(myList))] for j in xrange(len(myList[0]))]
for i in xrange(len(myList)):
    for j in xrange(len(myList[i])):
        newList[j][i] = myList[i][j]

print newList

edited Mar 18 '13 at 10:18

answered Mar 18 '13 at 10:00

dmg

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This is not more perfomant than using builtins in Python 2. For the test data `[range(1000000) for i in xrange(8)]` this code executes in 3 seconds on my system while my code takes 1.8 seconds. If you don't build lists but stick with tuples instead it is done in 0.26 seconds. – filmor Mar 18 '13 at 10:08
@filmor You are right, I was running it with PyPy and interestingly enough somewhere around 100x1000 these turn out to be faster than `zip` (tuple or list). I could post some numbers if you want. – dmg Mar 18 '13 at 10:24
This is indeed interesting, seems like PyPy is optimising better than the C compiler in this case ;) – filmor Mar 19 '13 at 08:38

How to change the major order of lists in python?

5 Answers5