How to zip two lists of lists in Python?

Question

I have two lists of lists that have equivalent numbers of items. The two lists look like this:

L1 = [[1, 2], [3, 4], [5, 6]]  L2 =[[a, b], [c, d], [e, f]] 

I am looking to create one list that looks like this:

Lmerge = [[1, 2, a, b], [3, 4, c, d], [5, 6, e, f]] 

I was attempting to use zip() something like this:

for list1, list2 in zip(*L1, *L2):     Lmerge = [list1, list2] 

What is the best way to combine two lists of lists? Thanks in advance.

Answer 1

>>> map(list.__add__, L1, L2) [[1, 2, 'a', 'b'], [3, 4, 'c', 'd'], [5, 6, 'e', 'f']] 

Answer 2

>>> L1 = [[1, 2], [3, 4], [5, 6]] >>> L2 =[["a", "b"], ["c", "d"], ["e", "f"]] >>> [x + y for x,y in zip(L1,L2)] [[1, 2, 'a', 'b'], [3, 4, 'c', 'd'], [5, 6, 'e', 'f']] 

Or,

>>> [sum(x,[]) for x in zip(L1,L2)] [[1, 2, 'a', 'b'], [3, 4, 'c', 'd'], [5, 6, 'e', 'f']] 

or,

>>> import itertools >>> [list(itertools.chain(*x)) for x in zip(L1,L2)] [[1, 2, 'a', 'b'], [3, 4, 'c', 'd'], [5, 6, 'e', 'f']] 

We can also do it without zip():

>>> [L1[i] + L2[i] for i in xrange(min(len(L1), len(L2)))]   [[1, 2, 'a', 'b'], [3, 4, 'c', 'd'], [5, 6, 'e', 'f']]  >>> [x + L2[i] for i, x in enumerate(L1)]  # assuming len(L1) == len(l2) [[1, 2, 'a', 'b'], [3, 4, 'c', 'd'], [5, 6, 'e', 'f']]  >>> # same as above, but deals with different lengths >>> Lx, Ly = ((L2,L1), (L1,L2))[len(L1)<=len(L2)] # shortcut for if/else >>> [x + Ly[i] for i, x in enumerate(Lx)] 

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Some benchmarks

Here are some benchmarks for the answers provided so far.

It looks like the most popular answer ([x + y for x,y in zip(L1,L2)]) is pretty much on par with @hammar's map solution. On the other hand, the alternative solutions I've given have proven to be rubbish!

However, the fastest solutions (for now) seems to be the ones that uses list comprehension without zip().

[me@home]$ SETUP="L1=[[x,x+1] for x in xrange(10000)];L2=[[x+2,x+3] for x in xrange(10000)]"  [me@home]$ # this raises IndexError if len(L1) > len(L2) [me@home]$ python -m timeit "$SETUP" "[x + L2[i] for i, x in enumerate(L1)]" 100 loops, best of 3: 10.6 msec per loop  [me@home]$ # same as above, but deals with length inconsistencies [me@home]$ python -m timeit "$SETUP" "Lx,Ly=((L2,L1),(L1,L2))[len(L1)<=len(L2)];[x + Ly[i] for i, x in enumerate(Lx)]" 100 loops, best of 3: 10.6 msec per loop  [me@home]$ # almost as fast as above, but easier to read [me@home]$ python -m timeit "$SETUP" "[L1[i] + L2[i] for i in xrange(min(len(L1),len(L2)))]" 100 loops, best of 3: 10.8 msec per loop  [me@home]$ python -m timeit "$SETUP" "L3=[x + y for x,y in zip(L1,L2)]" 100 loops, best of 3: 13.4 msec per loop  [me@home]$ python -m timeit "$SETUP" "L3=map(list.__add__, L1, L2)"  100 loops, best of 3: 13.5 msec per loop  [me@home]$ python -m timeit "$SETUP" "L3=[sum(x,[]) for x in zip(L1,L2)]" 100 loops, best of 3: 18.1 msec per loop  [me@home]$ python -m timeit "$SETUP;import itertools" "L3=[list(itertools.chain(*x)) for x in zip(L1,L2)]" 10 loops, best of 3: 32.9 msec per loop 

@Zac's suggestion is really quick, but then we're comparing apples and oranges here since it does a list extension in-place on L1 instead of creating a third list. So, if L1 is not longer needed, this is a great solution.

[me@home]$ python -m timeit "$SETUP" "for index, x in enumerate(L1): x.extend(L2[index])" 100 loops, best of 3: 9.46 msec per loop 

However, if L1 has to be kept intact, then performance would be sub par once you include the deepcopy.

[me@home]$ python -m timeit "$SETUP;from copy import deepcopy" "L3=deepcopy(L1) > for index, x in enumerate(L1): x.extend(L2[index])" 10 loops, best of 3: 116 msec per loop