Create a dictionary by zipping together two lists of uneven length

Question

I have two lists different lengths, L1 and L2. L1 is longer than L2. I would like to get a dictionary with members of L1 as keys and members of L2 as values.

As soon as all the members of L2 are used up. I would like to start over and begin again with L2[0].

L1 = ['A', 'B', 'C', 'D', 'E']    
L2 = ['1', '2', '3']    
D = dict(zip(L1, L2))    
print(D)

As expected, the output is this:

{'A': '1', 'B': '2', 'C': '3'}

What I would like to achieve is the following:

{'A': '1', 'B': '2', 'C': '3', 'D': '1', 'E': '2'}

Curious as to the reason for the close votes. What about this question is too broad? If it is because the OP hasn't offered any solution, then I can understand OP may not have the first clue where to begin. Based on their explanation of the question, it does not seem like they could have googled for "cycle", "circular", or other similar keywords that would have led to a solution. — cs95, Jan 09 '19 at 09:14

cs95 · Accepted Answer · 2019-01-08T21:20:28.240

37

Use itertools.cycle to cycle around to the beginning of L2:

from itertools import cycle
dict(zip(L1, cycle(L2)))
# {'A': '1', 'B': '2', 'C': '3', 'D': '1', 'E': '2'}

In your case, concatenating L2 with itself also works.

# dict(zip(L1, L2 * 2))
dict(zip(L1, L2 + L2))
# {'A': '1', 'B': '2', 'C': '3', 'D': '1', 'E': '2'}

edited Jan 08 '19 at 21:20

answered Jan 08 '19 at 16:22

cs95

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score 25 · Answer 2 · answered Jan 08 '19 at 16:22

Use itertools.cycle:

from itertools import cycle

L1 = ['A', 'B', 'C', 'D', 'E']
L2 = ['1', '2', '3']

result = dict(zip(L1, cycle(L2)))

print(result)

Output

{'E': '2', 'B': '2', 'A': '1', 'D': '1', 'C': '3'}

As an alternative you could use enumerate and index L2 modulo the length of L2:

result = {v: L2[i % len(L2)] for i, v in enumerate(L1)}
print(result)

user2390182 · Answer 3 · 2021-10-11T13:36:29.267

16

cycle is fine, but I shall add this modulo based approach:

{x: L2[i % len(L2)] for i, x in enumerate(L1)]}

edited Oct 11 '21 at 13:36

answered Jan 08 '19 at 16:29

user2390182

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RoadRunner · Answer 4 · 2019-01-08T17:05:30.290

You can also use a collections.deque() to create an circular FIFO queue:

from collections import deque

L1 = ['A', 'B', 'C', 'D', 'E']    
L2 = deque(['1', '2', '3'])

result = {}
for letter in L1:
    number = L2.popleft()
    result[letter] = number
    L2.append(number)

print(result)
# {'A': '1', 'B': '2', 'C': '3', 'D': '1', 'E': '2'}

Which pops the left most item currently in L2 and appends it to the end once the number is added to the dictionary.

Note: Both collections.deque.popleft() and collections.deque.append() are O(1) operations, so the above is still O(N), since you need to traverse all the elements in L1.

score 7 · Answer 5 · answered Jan 09 '19 at 09:05

7

Other option without dependencies with good old for loop:

D = {}
for i, e in enumerate(L1):
  D[e] = L2[i%len(L2)]

D #=> {'A': '1', 'B': '2', 'C': '3', 'D': '1', 'E': '2'}

Or just:

{ e: L2[i%len(L2)] for i, e in enumerate(L1) }
#=> {'A': '1', 'B': '2', 'C': '3', 'D': '1', 'E': '2'}

answered Jan 09 '19 at 09:05

iGian

11,023
3
21
36

Isn't that already here? https://stackoverflow.com/a/54095907/4909087 and also in another answer. – cs95 Jan 09 '19 at 09:53

Create a dictionary by zipping together two lists of uneven length

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