How to zip two differently sized lists?

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Question :

How to zip two differently sized lists?

I want to zip two list with different length

for example

A = [1,2,3,4,5,6,7,8,9]
B = ["A","B","C"]

and I expect this

[(1, 'A'), (2, 'B'), (3, 'C'), (4, 'A'), (5, 'B'), (6, 'C'), (7, 'A'), (8, 'B'), (9, 'C')]

But the build-in zip won’t repeat to pair with the list with larger size .
Does there exist any build-in way can achieve this?

here is my code

idx = 0
zip_list = []
for value in larger:
    idx += 1
    if idx == len(smaller):
        idx = 0

Answer #1:

You can use itertools.cycle:

Make an iterator returning elements from the iterable and saving a copy of each. When the iterable is exhausted, return elements from the saved copy. Repeats indefinitely.


A = [1,2,3,4,5,6,7,8,9]
B = ["A","B","C"]

from itertools import cycle
zip_list = zip(A, cycle(B)) if len(A) > len(B) else zip(cycle(A), B)
Answered By: sloth

Answer #2:

Try this.

A = [1,2,3,4,5,6,7,8,9]
B = ["A","B","C"]
Z = []
for i, a in enumerate(A):
    Z.append((a, B[i % len(B)]))

Just make sure that the larger list is in A.

Answered By: Eser Ayg├╝n

Answer #3:

You can use itertools.cycle:

from itertools import cycle

my_list = [1, 2, 3, 5, 5, 9]
another_list = ['Yes', 'No']

cyc = cycle(another_list)

print([[i, next(cyc)] for i in my_list])
# [[1, 'Yes'], [2, 'No'], [3, 'Yes'], [5, 'No'], [5, 'Yes'], [9, 'No']]
Answered By: Henry Yik

Answer #4:

Do you know that the second list is shorter?

import itertools
list(zip(my_list, itertools.cycle(another_list)))

This will actually give you a list of tuples rather than a list of lists. I hope that’s okay.

Answered By: Frank Yellin

Answer #5:

Solution for an arbitrary number of iterables, and you don’t know which one is longest (also allowing a default for any empty iterables):

from itertools import cycle, zip_longest

def zip_cycle(*iterables, empty_default=None):
    cycles = [cycle(i) for i in iterables]
    for _ in zip_longest(*iterables):
        yield tuple(next(i, empty_default) for i in cycles)

for i in zip_cycle(range(2), range(5), ['a', 'b', 'c'], []):


(0, 0, 'a', None)
(1, 1, 'b', None)
(0, 2, 'c', None)
(1, 3, 'a', None)
(0, 4, 'b', None)
Answered By: Dane White

Answer #6:

You can use the modulo % operator in a loop that counts up

my_list=[1, 2, 3, 5, 5, 9]

new_list = []
for cur in range(len(my_list)):
    new_list.append([my_list[cur], another_list[cur % 2]])
# [[1, 'Yes'], [2, 'No'], [3, 'Yes'], [5, 'No'], [5, 'Yes'], [9, 'No']]

2 can be replaced with len(another_list)

Answered By: Dexus

Answer #7:

A very simple approach is to multiply the short list so it’s longer:

my_list = [1, 2, 3, 5, 5, 9]
another_list = ['Yes', 'No']

zip(my_list, another_list*3))
#[(1, 'Yes'), (2, 'No'), (3, 'Yes'), (5, 'No'), (5, 'Yes'), (9, 'No')]

Note here that the multiplier doesn’t need to be carefully calculated since zip only goes out to the length of the shortest list (and the point of the multiplier is to make sure the shortest list is my_list). That is, the result would be the same if 100 were used instead of 3.

Answered By: tom10

Answer #8:

symmetric, no conditionals one liner

[*zip(A*(len(B)//len(A) + 1), B*(len(A)//len(B) + 1))]

which strictly answers ‘How to zip two differently sized lists?’

needs a patch for equal sized lists to be general:

[*(zip(A, B) if len(A) == len(B)
         else zip(A*(len(B)//len(A) + 1),
                  B*(len(A)//len(B) + 1)))]
Answered By: f5r5e5d

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