how to interleaving lists

Question

I have two lists that could be not equal in lengths and I want to be able to interleave them. I want to be able to append the extra values in the longer list at the end of my interleaved list.I have this:

def interleave(xs,ys):
a=xs
b=ys
minlength=[len(a),len(b)]
extralist= list()
interleave= list()
for i in range((minval(minlength))):
    pair=a[i],b[i]
    interleave.append(pair)
    flat=flatten(interleave)
    c=a+b
if len(b)>len(a):
    remainder=len(b)-len(a)
    print(remainder)
    for j in range(remainder,-1,-1):
        extra=b[-j]
        extralist.append(extra)
        print(extralist)
if len(a)>len(b):
    remainder=len(a)-len(b)
    print(remainder)
    for j in range(remainder,-1,-1):
        extra=a[-j]
        extralist.append(extra)
final=flat+extralist
return final

but if I test it:

>>> interleave([1,2,3], ["hi", "bye",True, False, 33])
[1, 'hi', 2, 'bye', 3, True]
>>>

The False and 33 don't appear. What is it that Im doing wrong?

EDIT: I found the solution to it. I indexed backwards :)

@Marcin I want to be able to do it using for loops and if else statements. — user300, Nov 10 '13 at 15:57
@Marcin I have a zip function I defined that zips them but it doesn't zip the remaining elements. — user300, Nov 10 '13 at 15:59

score 3 · Answer 1 · 2013-11-10T19:28:31.103

3

Here is a solution using itertools.zip_longest and a list comprehension:

>>> from itertools import zip_longest
>>> a = [1, 2, 3]
>>> b = ["hi", "bye", True, False, 33]
>>> [y for x in zip_longest(a, b) for y in x]
[1, 'hi', 2, 'bye', 3, True, None, False, None, 33]
>>> [y for x in zip_longest(a, b) for y in x if y is not None]
[1, 'hi', 2, 'bye', 3, True, False, 33]
>>>

Edit:

If you don't want to use itertools, I guess you could do:

>>> a = [1, 2, 3]
>>> b = ["hi", "bye", True, False, 33]
>>> items = sorted((a,b), key=len)
>>> out = []
>>> for x,y in enumerate(items[1]):
...     out.append(y)
...     try:
...         out.append(items[0][x])
...     except IndexError:
...         out.extend(items[1][x+1:])
...         break
...
>>> out
['hi', 1, 'bye', 2, True, 3, False, 33]
>>>

But I must say this method isn't as efficient as my first solution.

Edit 2:

Changing enumerate to zip reduces the efficiency again. However, if you must, then you can do this:

>>> a = [1, 2, 3]
>>> b = ["hi", "bye", True, False, 33]
>>> out = []
>>> items = sorted((a,b), key=len)
>>> for x,y in zip(range(len(items[1])), items[1]):
...     out.append(y)
...     try:
...         out.append(items[0][x])
...     except IndexError:
...         out.extend(items[1][x+1:])
...         break
...
>>> out
['hi', 1, 'bye', 2, True, 3, False, 33]
>>>

edited Nov 10 '13 at 19:28

answered Nov 10 '13 at 15:56

Is there a way to do it in the simplest form? – user300 Nov 10 '13 at 15:58
2

@user2829177 - What do you mean by "simplest form"? My solution is small and efficient. Plus, `itertools` is a standard library, so it will always be there. – Nov 10 '13 at 16:06
I don't want to use any functions that I don't know yet – user300 Nov 10 '13 at 16:12
@user2829177 - Well, see my edit. Maybe that is what you want. – Nov 10 '13 at 16:20
1

Heh. Everyone's reinventing the same kinds of solutions that we came up with one of the [last times](http://stackoverflow.com/questions/19883826/interleaving-2-lists-of-unequal-lengths) the OP asked. – DSM Nov 10 '13 at 16:28
@iCodez can i use zip instead of enumerate? – user300 Nov 10 '13 at 16:46
@user2829177 - I don't recommend doing this because it reduces efficiency. However, there _is_ a way if you really want to. See my edit. Also, I made a slight change to the `except` blocks to improve efficiency. – Nov 10 '13 at 19:29

score 0 · Answer 2 · answered Nov 10 '13 at 16:23

0

Another option

from itertools import chain
x = [1, 2, 3]
y = iter(["hi", "bye",True, False, 33])
list(chain.from_iterable(zip(x, y))) + list(y)
[1, 'hi', 2, 'bye', 3, True, False, 33]

answered Nov 10 '13 at 16:23

iruvar

22,736
7
53
82

how to interleaving lists

2 Answers2