Zipping unequal lists in python in to a list which does not drop any element from longer list being zipped

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Question :

Zipping unequal lists in python in to a list which does not drop any element from longer list being zipped

I have two lists

a = [1,2,3]
b = [9,10]

I want to combine (zip) these two lists into one list c such that

c = [(1,9), (2,10), (3, )]

Is there any function in standard library in Python to do this?

Asked By: Dilawar


Answer #1:

Normally, you use itertools.zip_longest for this:

>>> import itertools
>>> a = [1, 2, 3]
>>> b = [9, 10]
>>> for i in itertools.zip_longest(a, b): print(i)
(1, 9)
(2, 10)
(3, None)

But zip_longest pads the shorter iterable with Nones (or whatever value you pass as the fillvalue= parameter). If that’s not what you want then you can use a comprehension to filter out the Nones:

for i in (tuple(p for p in pair if p is not None) 
          for pair in itertools.zip_longest(a, b)):

but note that if either of the iterables has None values this will filter them out too. If you don’t want that, define your own object for fillvalue= and filter that instead of None:

_marker = object()

def zip_longest_no_fill(a, b):
    for i in itertools.zip_longest(a, b, fillvalue=_marker):
        yield tuple(x for x in i if x is not _marker)

list(zip_longest_no_fill(a, b))  # [(1, 9), (2, 10), (3,)]

On Python 2, use itertools.izip_longest instead.

Answered By: inspectorG4dget

Answer #2:

Another way is map:

a = [1, 2, 3]
b = [9, 10]
c = map(None, a, b)

Although that will too contain (3, None) instead of (3,). To do that, here’s a fun line:

c = (tuple(y for y in x if y is not None) for x in map(None, a, b))
Answered By: Ry-

Answer #3:

It’s not too hard to just write the explicit Python to do the desired operation:

def izip_short(a, b):
    ia = iter(a)
    ib = iter(b)
    for x in ia:
            y = next(ib)
            yield (x, y)
        except StopIteration:
            yield (x,)
    for x in ia:
        yield (x,)
    for y in ib:
        yield (None, y)

a = [1, 2, 3]
b = [9, 10]
list(izip_short(a, b))
list(izip_short(b, a))

I wasn’t sure how you would want to handle the b sequence being longer than the a sequence, so I just stuff in a None for the first value in the tuple in that case.

Get an explicit iterator for each sequence. Run the a iterator as a for loop, while manually using next(ib) to get the next value from the b sequence. If we get a StopIteration on the b sequence, we break the loop and then for x in ia: gets the rest of the a sequence; after that for y in ib: will do nothing because that iterator is already exhausted. Alternatively, if the first for x in ia: loop exhausts the a iterator, the second for x in ia: does nothing but there could be values left in the b sequence and the for y in ib: loop collects them.

Answered By: steveha

Answer #4:

Single line:

c = zip(a, b) + [(x,) for x in a[len(b):]] + [(x,) for x in b[len(a):]]

If you want to reuse this:

def mergeUsNicely(a, b):
    def tupleMe(val):
        return (val,)
    return zip(a, b) + map(tupleMe, a[len(b):]) + map(tupleMe, b[len(a):])
Answered By: SeeknInspYre

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