Pythonic way to combine (interleave, interlace, intertwine) two lists in an alternating fashion?
Question:
I have two lists, the first of which is guaranteed to contain exactly one more item than the second. I would like to know the most Pythonic way to create a new list whose even-index values come from the first list and whose odd-index values come from the second list.
# example inputs
list1 = ['f', 'o', 'o']
list2 = ['hello', 'world']
# desired output
['f', 'hello', 'o', 'world', 'o']
This works, but isn’t pretty:
list3 = []
while True:
try:
list3.append(list1.pop(0))
list3.append(list2.pop(0))
except IndexError:
break
How else can this be achieved? What’s the most Pythonic approach?
If you need to handle lists of mismatched length (e.g. the second list is longer, or the first has more than one element more than the second), some solutions here will work while others will require adjustment. For more specific answers, see How to interleave two lists of different length? to leave the excess elements at the end, or How to elegantly interleave two lists of uneven length? to try to intersperse elements evenly, or Insert element in Python list after every nth element for the case where a specific number of elements should come before each "added" element.
Answers:
There’s a recipe for this in the itertools
documentation (note: for Python 3):
from itertools import cycle, islice
def roundrobin(*iterables):
"roundrobin('ABC', 'D', 'EF') --> A D E B F C"
# Recipe credited to George Sakkis
num_active = len(iterables)
nexts = cycle(iter(it).__next__ for it in iterables)
while num_active:
try:
for next in nexts:
yield next()
except StopIteration:
# Remove the iterator we just exhausted from the cycle.
num_active -= 1
nexts = cycle(islice(nexts, num_active))
In Python 2, this should do what you want:
>>> iters = [iter(list1), iter(list2)]
>>> print list(it.next() for it in itertools.cycle(iters))
['f', 'hello', 'o', 'world', 'o']
Here’s one way to do it by slicing:
>>> list1 = ['f', 'o', 'o']
>>> list2 = ['hello', 'world']
>>> result = [None]*(len(list1)+len(list2))
>>> result[::2] = list1
>>> result[1::2] = list2
>>> result
['f', 'hello', 'o', 'world', 'o']
I’d do the simple:
chain.from_iterable( izip( list1, list2 ) )
It’ll come up with an iterator without creating any additional storage needs.
I’m too old to be down with list comprehensions, so:
import operator
list3 = reduce(operator.add, zip(list1, list2))
Here’s a one liner that does it:
list3 = [ item for pair in zip(list1, list2 + [0]) for item in pair][:-1]
Here’s a one liner using list comprehensions, w/o other libraries:
list3 = [sub[i] for i in range(len(list2)) for sub in [list1, list2]] + [list1[-1]]
Here is another approach, if you allow alteration of your initial list1 by side effect:
[list1.insert((i+1)*2-1, list2[i]) for i in range(len(list2))]
My take:
a = "hlowrd"
b = "el ol"
def func(xs, ys):
ys = iter(ys)
for x in xs:
yield x
yield ys.next()
print [x for x in func(a, b)]
Stops on the shortest:
def interlace(*iters, next = next) -> collections.Iterable:
"""
interlace(i1, i2, ..., in) -> (
i1-0, i2-0, ..., in-0,
i1-1, i2-1, ..., in-1,
.
.
.
i1-n, i2-n, ..., in-n,
)
"""
return map(next, cycle([iter(x) for x in iters]))
Sure, resolving the next/__next__ method may be faster.
def combine(list1, list2):
lst = []
len1 = len(list1)
len2 = len(list2)
for index in range( max(len1, len2) ):
if index+1 <= len1:
lst += [list1[index]]
if index+1 <= len2:
lst += [list2[index]]
return lst
Without itertools and assuming l1 is 1 item longer than l2:
>>> sum(zip(l1, l2+[0]), ())[:-1]
('f', 'hello', 'o', 'world', 'o')
In python 2, using itertools and assuming that lists don’t contain None:
>>> filter(None, sum(itertools.izip_longest(l1, l2), ()))
('f', 'hello', 'o', 'world', 'o')
This one is based on Carlos Valiente’s contribution above
with an option to alternate groups of multiple items and make sure that all items are present in the output :
A=["a","b","c","d"]
B=[1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16]
def cyclemix(xs, ys, n=1):
for p in range(0,int((len(ys)+len(xs))/n)):
for g in range(0,min(len(ys),n)):
yield ys[0]
ys.append(ys.pop(0))
for g in range(0,min(len(xs),n)):
yield xs[0]
xs.append(xs.pop(0))
print [x for x in cyclemix(A, B, 3)]
This will interlace lists A and B by groups of 3 values each:
['a', 'b', 'c', 1, 2, 3, 'd', 'a', 'b', 4, 5, 6, 'c', 'd', 'a', 7, 8, 9, 'b', 'c', 'd', 10, 11, 12, 'a', 'b', 'c', 13, 14, 15]
I know the questions asks about two lists with one having one item more than the other, but I figured I would put this for others who may find this question.
Here is Duncan’s solution adapted to work with two lists of different sizes.
list1 = ['f', 'o', 'o', 'b', 'a', 'r']
list2 = ['hello', 'world']
num = min(len(list1), len(list2))
result = [None]*(num*2)
result[::2] = list1[:num]
result[1::2] = list2[:num]
result.extend(list1[num:])
result.extend(list2[num:])
result
This outputs:
['f', 'hello', 'o', 'world', 'o', 'b', 'a', 'r']
This is nasty but works no matter the size of the lists:
list3 = [
element for element in
list(itertools.chain.from_iterable([
val for val in itertools.izip_longest(list1, list2)
]))
if element != None
]
import itertools
print([x for x in itertools.chain.from_iterable(itertools.zip_longest(list1,list2)) if x])
I think this is the most pythonic way of doing it.
Multiple one-liners inspired by answers to another question:
import itertools
list(itertools.chain.from_iterable(itertools.izip_longest(list1, list2, fillvalue=object)))[:-1]
[i for l in itertools.izip_longest(list1, list2, fillvalue=object) for i in l if i is not object]
[item for sublist in map(None, list1, list2) for item in sublist][:-1]
Might be a bit late buy yet another python one-liner. This works when the two lists have equal or unequal size. One thing worth nothing is it will modify a and b. If it’s an issue, you need to use other solutions.
a = ['f', 'o', 'o']
b = ['hello', 'world']
sum([[a.pop(0), b.pop(0)] for i in range(min(len(a), len(b)))],[])+a+b
['f', 'hello', 'o', 'world', 'o']
How about numpy? It works with strings as well:
import numpy as np
np.array([[a,b] for a,b in zip([1,2,3],[2,3,4,5,6])]).ravel()
Result:
array([1, 2, 2, 3, 3, 4])
An alternative in a functional & immutable way (Python 3):
from itertools import zip_longest
from functools import reduce
reduce(lambda lst, zipped: [*lst, *zipped] if zipped[1] != None else [*lst, zipped[0]], zip_longest(list1, list2),[])
from itertools import chain
list(chain(*zip('abc', 'def'))) # Note: this only works for lists of equal length
['a', 'd', 'b', 'e', 'c', 'f']
If both lists have equal length, you can do:
[x for y in zip(list1, list2) for x in y]
As the first list has one more element, you can add it post hoc:
[x for y in zip(list1, list2) for x in y] + [list1[-1]]
Edit: To illustrate what is happening in that first list comprehension, this is how you would spell it out as a nested for loop:
result = []
for y in zip(list1, list2): # y is is a 2-tuple, containining one element from each list
for x in y: # iterate over the 2-tuple
result.append(x) # append each element individually
itertools.zip_longest
returns an iterator of tuple pairs with any missing elements in one list replaced with fillvalue=None
(passing fillvalue=object
lets you use None
as a value). If you flatten these pairs, then filter fillvalue
in a list comprehension, this gives:
>>> from itertools import zip_longest
>>> def merge(a, b):
... return [
... x for y in zip_longest(a, b, fillvalue=object)
... for x in y if x is not object
... ]
...
>>> merge("abc", "defgh")
['a', 'd', 'b', 'e', 'c', 'f', 'g', 'h']
>>> merge([0, 1, 2], [4])
[0, 4, 1, 2]
>>> merge([0, 1, 2], [4, 5, 6, 7, 8])
[0, 4, 1, 5, 2, 6, 7, 8]
Generalized to arbitrary iterables:
>>> def merge(*its):
... return [
... x for y in zip_longest(*its, fillvalue=object)
... for x in y if x is not object
... ]
...
>>> merge("abc", "lmn1234", "xyz9", [None])
['a', 'l', 'x', None, 'b', 'm', 'y', 'c', 'n', 'z', '1', '9', '2', '3', '4']
>>> merge(*["abc", "x"]) # unpack an iterable
['a', 'x', 'b', 'c']
Finally, you may want to return a generator rather than a list comprehension:
>>> def merge(*its):
... return (
... x for y in zip_longest(*its, fillvalue=object)
... for x in y if x is not object
... )
...
>>> merge([1], [], [2, 3, 4])
<generator object merge.<locals>.<genexpr> at 0x000001996B466740>
>>> next(merge([1], [], [2, 3, 4]))
1
>>> list(merge([1], [], [2, 3, 4]))
[1, 2, 3, 4]
If you’re OK with other packages, you can try more_itertools.roundrobin
:
>>> list(roundrobin('ABC', 'D', 'EF'))
['A', 'D', 'E', 'B', 'F', 'C']
using for loop also we can achive this easily:
list1 = ['f', 'o', 'o']
list2 = ['hello', 'world']
list3 = []
for i in range(len(list1)):
#print(list3)
list3.append(list1[i])
if i < len(list2):
list3.append(list2[i])
print(list3)
output :
['f', 'hello', 'o', 'world', 'o']
Further by using list comprehension this can be reduced. But for understanding this loop can be used.
Obviously late to the party, but here’s a concise one for equal-length lists:
output = [e for sub in zip(list1,list2) for e in sub]
It generalizes for an arbitrary number of equal-length lists, too:
output = [e for sub in zip(list1,list2,list3) for e in sub]
etc.
My approach looks as follows:
from itertools import chain, zip_longest
def intersperse(*iterators):
# A random object not occurring in the iterators
filler = object()
r = (x for x in chain.from_iterable(zip_longest(*iterators, fillvalue=filler)) if x is not filler)
return r
list1 = ['f', 'o', 'o']
list2 = ['hello', 'world']
print(list(intersperse(list1, list2)))
It works for an arbitrary number of iterators and yields an iterator, so I applied list()
in the print line.
def alternate_elements(small_list, big_list):
mew = []
count = 0
for i in range(len(small_list)):
mew.append(small_list[i])
mew.append(big_list[i])
count +=1
return mew+big_list[count:]
if len(l2)>len(l1):
res = alternate_elements(l1,l2)
else:
res = alternate_elements(l2,l1)
print(res)
Here we swap lists based on size and perform, can someone provide better solution with time complexity O(len(l1)+len(l2))
I have two lists, the first of which is guaranteed to contain exactly one more item than the second. I would like to know the most Pythonic way to create a new list whose even-index values come from the first list and whose odd-index values come from the second list.
# example inputs
list1 = ['f', 'o', 'o']
list2 = ['hello', 'world']
# desired output
['f', 'hello', 'o', 'world', 'o']
This works, but isn’t pretty:
list3 = []
while True:
try:
list3.append(list1.pop(0))
list3.append(list2.pop(0))
except IndexError:
break
How else can this be achieved? What’s the most Pythonic approach?
If you need to handle lists of mismatched length (e.g. the second list is longer, or the first has more than one element more than the second), some solutions here will work while others will require adjustment. For more specific answers, see How to interleave two lists of different length? to leave the excess elements at the end, or How to elegantly interleave two lists of uneven length? to try to intersperse elements evenly, or Insert element in Python list after every nth element for the case where a specific number of elements should come before each "added" element.
There’s a recipe for this in the itertools
documentation (note: for Python 3):
from itertools import cycle, islice
def roundrobin(*iterables):
"roundrobin('ABC', 'D', 'EF') --> A D E B F C"
# Recipe credited to George Sakkis
num_active = len(iterables)
nexts = cycle(iter(it).__next__ for it in iterables)
while num_active:
try:
for next in nexts:
yield next()
except StopIteration:
# Remove the iterator we just exhausted from the cycle.
num_active -= 1
nexts = cycle(islice(nexts, num_active))
In Python 2, this should do what you want:
>>> iters = [iter(list1), iter(list2)]
>>> print list(it.next() for it in itertools.cycle(iters))
['f', 'hello', 'o', 'world', 'o']
Here’s one way to do it by slicing:
>>> list1 = ['f', 'o', 'o']
>>> list2 = ['hello', 'world']
>>> result = [None]*(len(list1)+len(list2))
>>> result[::2] = list1
>>> result[1::2] = list2
>>> result
['f', 'hello', 'o', 'world', 'o']
I’d do the simple:
chain.from_iterable( izip( list1, list2 ) )
It’ll come up with an iterator without creating any additional storage needs.
I’m too old to be down with list comprehensions, so:
import operator
list3 = reduce(operator.add, zip(list1, list2))
Here’s a one liner that does it:
list3 = [ item for pair in zip(list1, list2 + [0]) for item in pair][:-1]
Here’s a one liner using list comprehensions, w/o other libraries:
list3 = [sub[i] for i in range(len(list2)) for sub in [list1, list2]] + [list1[-1]]
Here is another approach, if you allow alteration of your initial list1 by side effect:
[list1.insert((i+1)*2-1, list2[i]) for i in range(len(list2))]
My take:
a = "hlowrd"
b = "el ol"
def func(xs, ys):
ys = iter(ys)
for x in xs:
yield x
yield ys.next()
print [x for x in func(a, b)]
Stops on the shortest:
def interlace(*iters, next = next) -> collections.Iterable:
"""
interlace(i1, i2, ..., in) -> (
i1-0, i2-0, ..., in-0,
i1-1, i2-1, ..., in-1,
.
.
.
i1-n, i2-n, ..., in-n,
)
"""
return map(next, cycle([iter(x) for x in iters]))
Sure, resolving the next/__next__ method may be faster.
def combine(list1, list2):
lst = []
len1 = len(list1)
len2 = len(list2)
for index in range( max(len1, len2) ):
if index+1 <= len1:
lst += [list1[index]]
if index+1 <= len2:
lst += [list2[index]]
return lst
Without itertools and assuming l1 is 1 item longer than l2:
>>> sum(zip(l1, l2+[0]), ())[:-1]
('f', 'hello', 'o', 'world', 'o')
In python 2, using itertools and assuming that lists don’t contain None:
>>> filter(None, sum(itertools.izip_longest(l1, l2), ()))
('f', 'hello', 'o', 'world', 'o')
This one is based on Carlos Valiente’s contribution above
with an option to alternate groups of multiple items and make sure that all items are present in the output :
A=["a","b","c","d"]
B=[1,2,3,4,5,6,7,8,9,10,11,12,13,14,15,16]
def cyclemix(xs, ys, n=1):
for p in range(0,int((len(ys)+len(xs))/n)):
for g in range(0,min(len(ys),n)):
yield ys[0]
ys.append(ys.pop(0))
for g in range(0,min(len(xs),n)):
yield xs[0]
xs.append(xs.pop(0))
print [x for x in cyclemix(A, B, 3)]
This will interlace lists A and B by groups of 3 values each:
['a', 'b', 'c', 1, 2, 3, 'd', 'a', 'b', 4, 5, 6, 'c', 'd', 'a', 7, 8, 9, 'b', 'c', 'd', 10, 11, 12, 'a', 'b', 'c', 13, 14, 15]
I know the questions asks about two lists with one having one item more than the other, but I figured I would put this for others who may find this question.
Here is Duncan’s solution adapted to work with two lists of different sizes.
list1 = ['f', 'o', 'o', 'b', 'a', 'r']
list2 = ['hello', 'world']
num = min(len(list1), len(list2))
result = [None]*(num*2)
result[::2] = list1[:num]
result[1::2] = list2[:num]
result.extend(list1[num:])
result.extend(list2[num:])
result
This outputs:
['f', 'hello', 'o', 'world', 'o', 'b', 'a', 'r']
This is nasty but works no matter the size of the lists:
list3 = [
element for element in
list(itertools.chain.from_iterable([
val for val in itertools.izip_longest(list1, list2)
]))
if element != None
]
import itertools
print([x for x in itertools.chain.from_iterable(itertools.zip_longest(list1,list2)) if x])
I think this is the most pythonic way of doing it.
Multiple one-liners inspired by answers to another question:
import itertools
list(itertools.chain.from_iterable(itertools.izip_longest(list1, list2, fillvalue=object)))[:-1]
[i for l in itertools.izip_longest(list1, list2, fillvalue=object) for i in l if i is not object]
[item for sublist in map(None, list1, list2) for item in sublist][:-1]
Might be a bit late buy yet another python one-liner. This works when the two lists have equal or unequal size. One thing worth nothing is it will modify a and b. If it’s an issue, you need to use other solutions.
a = ['f', 'o', 'o']
b = ['hello', 'world']
sum([[a.pop(0), b.pop(0)] for i in range(min(len(a), len(b)))],[])+a+b
['f', 'hello', 'o', 'world', 'o']
How about numpy? It works with strings as well:
import numpy as np
np.array([[a,b] for a,b in zip([1,2,3],[2,3,4,5,6])]).ravel()
Result:
array([1, 2, 2, 3, 3, 4])
An alternative in a functional & immutable way (Python 3):
from itertools import zip_longest
from functools import reduce
reduce(lambda lst, zipped: [*lst, *zipped] if zipped[1] != None else [*lst, zipped[0]], zip_longest(list1, list2),[])
from itertools import chain
list(chain(*zip('abc', 'def'))) # Note: this only works for lists of equal length
['a', 'd', 'b', 'e', 'c', 'f']
If both lists have equal length, you can do:
[x for y in zip(list1, list2) for x in y]
As the first list has one more element, you can add it post hoc:
[x for y in zip(list1, list2) for x in y] + [list1[-1]]
Edit: To illustrate what is happening in that first list comprehension, this is how you would spell it out as a nested for loop:
result = []
for y in zip(list1, list2): # y is is a 2-tuple, containining one element from each list
for x in y: # iterate over the 2-tuple
result.append(x) # append each element individually
itertools.zip_longest
returns an iterator of tuple pairs with any missing elements in one list replaced with fillvalue=None
(passing fillvalue=object
lets you use None
as a value). If you flatten these pairs, then filter fillvalue
in a list comprehension, this gives:
>>> from itertools import zip_longest
>>> def merge(a, b):
... return [
... x for y in zip_longest(a, b, fillvalue=object)
... for x in y if x is not object
... ]
...
>>> merge("abc", "defgh")
['a', 'd', 'b', 'e', 'c', 'f', 'g', 'h']
>>> merge([0, 1, 2], [4])
[0, 4, 1, 2]
>>> merge([0, 1, 2], [4, 5, 6, 7, 8])
[0, 4, 1, 5, 2, 6, 7, 8]
Generalized to arbitrary iterables:
>>> def merge(*its):
... return [
... x for y in zip_longest(*its, fillvalue=object)
... for x in y if x is not object
... ]
...
>>> merge("abc", "lmn1234", "xyz9", [None])
['a', 'l', 'x', None, 'b', 'm', 'y', 'c', 'n', 'z', '1', '9', '2', '3', '4']
>>> merge(*["abc", "x"]) # unpack an iterable
['a', 'x', 'b', 'c']
Finally, you may want to return a generator rather than a list comprehension:
>>> def merge(*its):
... return (
... x for y in zip_longest(*its, fillvalue=object)
... for x in y if x is not object
... )
...
>>> merge([1], [], [2, 3, 4])
<generator object merge.<locals>.<genexpr> at 0x000001996B466740>
>>> next(merge([1], [], [2, 3, 4]))
1
>>> list(merge([1], [], [2, 3, 4]))
[1, 2, 3, 4]
If you’re OK with other packages, you can try more_itertools.roundrobin
:
>>> list(roundrobin('ABC', 'D', 'EF'))
['A', 'D', 'E', 'B', 'F', 'C']
using for loop also we can achive this easily:
list1 = ['f', 'o', 'o']
list2 = ['hello', 'world']
list3 = []
for i in range(len(list1)):
#print(list3)
list3.append(list1[i])
if i < len(list2):
list3.append(list2[i])
print(list3)
output :
['f', 'hello', 'o', 'world', 'o']
Further by using list comprehension this can be reduced. But for understanding this loop can be used.
Obviously late to the party, but here’s a concise one for equal-length lists:
output = [e for sub in zip(list1,list2) for e in sub]
It generalizes for an arbitrary number of equal-length lists, too:
output = [e for sub in zip(list1,list2,list3) for e in sub]
etc.
My approach looks as follows:
from itertools import chain, zip_longest
def intersperse(*iterators):
# A random object not occurring in the iterators
filler = object()
r = (x for x in chain.from_iterable(zip_longest(*iterators, fillvalue=filler)) if x is not filler)
return r
list1 = ['f', 'o', 'o']
list2 = ['hello', 'world']
print(list(intersperse(list1, list2)))
It works for an arbitrary number of iterators and yields an iterator, so I applied list()
in the print line.
def alternate_elements(small_list, big_list):
mew = []
count = 0
for i in range(len(small_list)):
mew.append(small_list[i])
mew.append(big_list[i])
count +=1
return mew+big_list[count:]
if len(l2)>len(l1):
res = alternate_elements(l1,l2)
else:
res = alternate_elements(l2,l1)
print(res)
Here we swap lists based on size and perform, can someone provide better solution with time complexity O(len(l1)+len(l2))