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?
Add a comment
|
5 Answers
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,)]
Comments
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))
3 Comments
inspectorG4dget
x if None not in x else tuple(y for y in x if y is not None). The x if None not in x is redundant here, as the else takes care of it. In the worst case, the else would return an empty tuple. Also, if there's a None in any tuple, the if would kill that tuple and pass it on to the else
Ry-
@inspectorG4dget: Thanks. But now it's not quite as fun :D
Thane Plummer
Works nicely in Python 2.7, but not in 3.5. For 3.5 use
itertools.zip_longest.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.
2 Comments
steveha
I don't remember what I was thinking in 2012 when I wrote this, but I think I probably just wanted to solve the problem. I have always felt that if there is a library function that does what you need, you almost never want to write your own instead of using it. Use the library function.
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):])
Comments
This answer is an extension of the top answer that allows for arbitrary inputs instead of only two.
import itertools
sentinel = object()
def zip_longest_no_fill(*args):
for i in itertools.zip_longest(*args, fillvalue=sentinel):
yield tuple(x for x in i if x is not sentinel)
