Create a dictionary by zipping together two lists of uneven length
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'}
python list dictionary zip
edited 11 hours ago
coldspeed
121k21121204
asked 15 hours ago
MatMat
795
New contributor
Mat is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
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add a comment |
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'}
python list dictionary zip
edited 11 hours ago
coldspeed
121k21121204
asked 15 hours ago
MatMat
795
New contributor
Mat is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
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'}
python list dictionary zip
edited 11 hours ago
coldspeed
121k21121204
asked 15 hours ago
MatMat
795
New contributor
Mat is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
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'}
python list dictionary zip
python list dictionary zip
edited 11 hours ago
coldspeed
121k21121204
asked 15 hours ago
MatMat
795
New contributor
Mat is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 11 hours ago
coldspeed
121k21121204
asked 15 hours ago
MatMat
795
New contributor
Mat is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
edited 11 hours ago
coldspeed
121k21121204
edited 11 hours ago
coldspeed
121k21121204
edited 11 hours ago
coldspeed
121k21121204
121k21121204
asked 15 hours ago
MatMat
795
New contributor
Mat is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
asked 15 hours ago
MatMat
795
asked 15 hours ago
MatMat
795
795
New contributor
Mat is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
New contributor
Mat is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
Mat is a new contributor to this site. Take care in asking for clarification, commenting, and answering.
Check out our Code of Conduct.
add a comment |
add a comment |
4 Answers
4
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oldest
votes
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'}
answered 15 hours ago
coldspeedcoldspeed
121k21121204
add a comment |
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)
answered 15 hours ago
Daniel MesejoDaniel Mesejo
14.8k21028
add a comment |
cycle is fine, but I shall add this modulo based approach:
{L1[i]: L2[i % len(L2)] for i in range(len(L1))]}
answered 15 hours ago
schwobasegglschwobaseggl
36.8k32442
add a comment |
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.
answered 15 hours ago
RoadRunnerRoadRunner
10.9k31340
add a comment |
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4 Answers
4
active
oldest
votes
4 Answers
4
active
oldest
votes
active
oldest
votes
active
oldest
votes
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'}
answered 15 hours ago
coldspeedcoldspeed
121k21121204
add a comment |
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'}
answered 15 hours ago
coldspeedcoldspeed
121k21121204
add a comment |
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'}
answered 15 hours ago
coldspeedcoldspeed
121k21121204
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'}
answered 15 hours ago
coldspeedcoldspeed
121k21121204
edited 10 hours ago
answered 15 hours ago
coldspeedcoldspeed
121k21121204
answered 15 hours ago
coldspeedcoldspeed
121k21121204
answered 15 hours ago
coldspeedcoldspeed
121k21121204
121k21121204
add a comment |
add a comment |
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)
answered 15 hours ago
Daniel MesejoDaniel Mesejo
14.8k21028
add a comment |
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)
answered 15 hours ago
Daniel MesejoDaniel Mesejo
14.8k21028
add a comment |
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)
answered 15 hours ago
Daniel MesejoDaniel Mesejo
14.8k21028
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)
answered 15 hours ago
Daniel MesejoDaniel Mesejo
14.8k21028
answered 15 hours ago
Daniel MesejoDaniel Mesejo
14.8k21028
answered 15 hours ago
Daniel MesejoDaniel Mesejo
14.8k21028
answered 15 hours ago
Daniel MesejoDaniel Mesejo
14.8k21028
14.8k21028
add a comment |
add a comment |
cycle is fine, but I shall add this modulo based approach:
{L1[i]: L2[i % len(L2)] for i in range(len(L1))]}
answered 15 hours ago
schwobasegglschwobaseggl
36.8k32442
add a comment |
cycle is fine, but I shall add this modulo based approach:
{L1[i]: L2[i % len(L2)] for i in range(len(L1))]}
answered 15 hours ago
schwobasegglschwobaseggl
36.8k32442
add a comment |
cycle is fine, but I shall add this modulo based approach:
{L1[i]: L2[i % len(L2)] for i in range(len(L1))]}
answered 15 hours ago
schwobasegglschwobaseggl
36.8k32442
cycle is fine, but I shall add this modulo based approach:
{L1[i]: L2[i % len(L2)] for i in range(len(L1))]}
answered 15 hours ago
schwobasegglschwobaseggl
36.8k32442
answered 15 hours ago
schwobasegglschwobaseggl
36.8k32442
answered 15 hours ago
schwobasegglschwobaseggl
36.8k32442
answered 15 hours ago
schwobasegglschwobaseggl
36.8k32442
36.8k32442
add a comment |
add a comment |
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.
answered 15 hours ago
RoadRunnerRoadRunner
10.9k31340
add a comment |
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.
answered 15 hours ago
RoadRunnerRoadRunner
10.9k31340
add a comment |
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.
answered 15 hours ago
RoadRunnerRoadRunner
10.9k31340
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.
answered 15 hours ago
RoadRunnerRoadRunner
10.9k31340
edited 14 hours ago
answered 15 hours ago
RoadRunnerRoadRunner
10.9k31340
answered 15 hours ago
RoadRunnerRoadRunner
10.9k31340
answered 15 hours ago
RoadRunnerRoadRunner
10.9k31340
10.9k31340
add a comment |
add a comment |
Mat is a new contributor. Be nice, and check out our Code of Conduct.
Mat is a new contributor. Be nice, and check out our Code of Conduct.
Mat is a new contributor. Be nice, and check out our Code of Conduct.
Mat is a new contributor. Be nice, and check out our Code of Conduct.
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