CODEWITHSUNDEEP
pythonlistdictionary
LiamNeesonFan
LiamNeesonFan

Reputation: 2813

How to Combine Each of the Elements of Two Lists in Python?

I have a huge group of lists within lists that I want to combine: It looks sort of like this: 

[[1,2,3,4,5], [6,7,8,9,0], [2,5,7,9,4], [4,7,8,43,6]...]

up to about 20 of these lists in a list. I now want to combine the first list and second list to look like this: 

[[1,6], [2,7], [3,8], [4,9], [5,0]]

And then I want to do it again with the 1st and 3rd, all the way until the end. And then do it again starting with the second list to 3rd,4th...last line (but not the first one because that was already done with 1st to 2nd list). How can I write code that will do this?

Here is what I have so far:

xcols = the column with all the lists like I showed above

def MakeLists(xcols):
    multilist = []
    for i in xcols:
        for j in xcols[index(i):]:
            currentlist = map(list.__add__, i, j)
            multilist.append(currentlist)

Gives me an error when I run it though, probably at the map part because I don't know how to first convert each element into a list and then map them. Any help would be great. Thanks!

Upvotes: 8

Views: 24800

Answers (5)

stranac
stranac

Reputation: 28226

How about something like this:

>>> import itertools
>>> foo = [[1, 2, 3], [4, 5, 6], [7, 8, 8]]
>>> for p in itertools.permutations(foo, 2):
...     print zip(*p)
... 
[(1, 4), (2, 5), (3, 6)]
[(1, 7), (2, 8), (3, 8)]
[(4, 1), (5, 2), (6, 3)]
[(4, 7), (5, 8), (6, 8)]
[(7, 1), (8, 2), (8, 3)]
[(7, 4), (8, 5), (8, 6)]

Edit: In case you only want to zip a list with those after it, as people in comments are explaining:

>>> import itertools
>>> for p in itertools.combinations(foo, 2):
...     print zip(*p)
... 
[(1, 4), (2, 5), (3, 6)]
[(1, 7), (2, 8), (3, 8)]
[(4, 7), (5, 8), (6, 8)]

Upvotes: 7

Jan Vorcak
Jan Vorcak

Reputation: 19989

a=[[1,2,3,4,5],[6,7,8,9,0],[2,5,7,9,4],[4,7,8,43,6]]

i=0

for l in a[i:]:
    for inner in a[i+1:]:
        print [list(b) for b in zip(l, inner)]
    i += 1

prints

[[1, 6], [2, 7], [3, 8], [4, 9], [5, 0]]
[[1, 2], [2, 5], [3, 7], [4, 9], [5, 4]]
[[1, 4], [2, 7], [3, 8], [4, 43], [5, 6]]
[[6, 2], [7, 5], [8, 7], [9, 9], [0, 4]]
[[6, 4], [7, 7], [8, 8], [9, 43], [0, 6]]
[[2, 4], [5, 7], [7, 8], [9, 43], [4, 6]]

Upvotes: 2

Tadeck
Tadeck

Reputation: 137310

The shortest solution is (lsts is a list of the lists you have):

[zip(lsts[i],lsts[j]) for i in xrange(len(lsts)) for j in xrange(i,len(lsts)) if i!=j]

It will do exactly what you said. Try this out.

Is it what you expected?

Upvotes: 1

Ceasar
Ceasar

Reputation: 23073

def foo(li):
  for element in li[1:]:
    for pair in zip(li[0], element): 
      yield pair

>>> from test import foo
>>> bar = [[1, 2, 3, 5, 6], [6, 7, 8, 9, 10], [11, 12, 13, 14, 15]]
>>> foo(bar)
<generator object foo at 0x10592df50>
>>> [e for e in foo(bar)]
[(1, 6), (2, 7), (3, 8), (5, 9), (6, 10), (1, 11), (2, 12), (3, 13), (5, 14), (6, 15)]

Upvotes: 2

jordanm
jordanm

Reputation: 34924

You will only be able to accomplish your intended result if you have an even number of lists. This code will produce the intended result. There may be something more "pythonic", however.

foo = [[1,2,3,4,5],[6,7,8,9,0],[2,5,7,9,4],[4,7,8,43,6]]
newlist = []

for i in xrange(len(foo)):
    if i % 2 == 0:
        list1 = foo[i]
        list2 = foo[i + 1]
        for n in xrange(len(list1)):
            newlist.append([list1[n],list2[n]])

print newlist

Result:

[[1, 6], [2, 7], [3, 8], [4, 9], [5, 0], [2, 4], [5, 7], [7, 8], [9, 43], [4, 6]]

Upvotes: 2

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