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 5 hours ago
coldspeed
121k21121204
asked 9 hours ago
MatMat
595
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 5 hours ago
coldspeed
121k21121204
asked 9 hours ago
MatMat
595
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 5 hours ago
coldspeed
121k21121204
asked 9 hours ago
MatMat
595
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 5 hours ago
coldspeed
121k21121204
asked 9 hours ago
MatMat
595
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 5 hours ago
coldspeed
121k21121204
asked 9 hours ago
MatMat
595
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 5 hours ago
coldspeed
121k21121204
edited 5 hours ago
coldspeed
121k21121204
edited 5 hours ago
coldspeed
121k21121204
121k21121204
asked 9 hours ago
MatMat
595
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 9 hours ago
MatMat
595
asked 9 hours ago
MatMat
595
595
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 9 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 9 hours ago
Daniel MesejoDaniel Mesejo
14.7k11028
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 9 hours ago
schwobasegglschwobaseggl
36.8k32441
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 8 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 9 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 9 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 9 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 9 hours ago
coldspeedcoldspeed
121k21121204
edited 4 hours ago
answered 9 hours ago
coldspeedcoldspeed
121k21121204
answered 9 hours ago
coldspeedcoldspeed
121k21121204
answered 9 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 9 hours ago
Daniel MesejoDaniel Mesejo
14.7k11028
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 9 hours ago
Daniel MesejoDaniel Mesejo
14.7k11028
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 9 hours ago
Daniel MesejoDaniel Mesejo
14.7k11028
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 9 hours ago
Daniel MesejoDaniel Mesejo
14.7k11028
answered 9 hours ago
Daniel MesejoDaniel Mesejo
14.7k11028
answered 9 hours ago
Daniel MesejoDaniel Mesejo
14.7k11028
answered 9 hours ago
Daniel MesejoDaniel Mesejo
14.7k11028
14.7k11028
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 9 hours ago
schwobasegglschwobaseggl
36.8k32441
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 9 hours ago
schwobasegglschwobaseggl
36.8k32441
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 9 hours ago
schwobasegglschwobaseggl
36.8k32441
cycle is fine, but I shall add this modulo based approach:
{L1[i]: L2[i % len(L2)] for i in range(len(L1))]}
answered 9 hours ago
schwobasegglschwobaseggl
36.8k32441
answered 9 hours ago
schwobasegglschwobaseggl
36.8k32441
answered 9 hours ago
schwobasegglschwobaseggl
36.8k32441
answered 9 hours ago
schwobasegglschwobaseggl
36.8k32441
36.8k32441
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 8 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 8 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 8 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 8 hours ago
RoadRunnerRoadRunner
10.9k31340
edited 8 hours ago
answered 8 hours ago
RoadRunnerRoadRunner
10.9k31340
answered 8 hours ago
RoadRunnerRoadRunner
10.9k31340
answered 8 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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