Zipping lists of unequal size

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?

Answers:

Thank you for visiting the Q&A section on Magenaut. Please note that all the answers may not help you solve the issue immediately. So please treat them as advisements. If you found the post helpful (or not), leave a comment & I’ll get back to you as soon as possible.

Method 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)):
...     print(i)
... 
(1, 9)
(2, 10)
(3,)

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:

sentinel = object()

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

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

Method 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))

Method 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:
        try:
            y = next(ib)
            yield (x, y)
        except StopIteration:
            yield (x,)
            break
    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.

Method 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):])

All methods was sourced from stackoverflow.com or stackexchange.com, is licensed under cc by-sa 2.5, cc by-sa 3.0 and cc by-sa 4.0