Stack Overflow Asked by Anis Boudieb on December 11, 2021
I have three lists:
W = ['Anis', 'James', 'Arthur']
C = ['City1', 'City2', 'City3']
Cost = [53, 27, 13, 80, 47, 67, 53, 73, 47]
I want these results in this format:
{('Anis', 'City1'): 53}
{('Anis', 'City2'): 27}
{('Anis', 'City3'): 13}
{('James', 'City1'): 80}
{('James', 'City2'): 47}
{('James', 'City3'): 67}
{('Arthur', 'City1'): 53}
{('Arthur', 'City2'): 73}
{('Arthur', 'City3'): 47}
I have tried this
`for i in range(n1):
for j in range(n1):
for c in range(n2):
combinations, mt = multidict({(W[i], C[j]): Cost[c]})
print(mt)`
I got this instead:
{('Anis', 'City1'): 47}
{('Anis', 'City2'): 47}
{('Anis', 'City3'): 47}
{('James', 'City1'): 47}
{('James', 'City2'): 47}
{('James', 'City3'): 47}
{('Arthur', 'City1'): 47}
{('Arthur', 'City2'): 47}
{('Arthur', 'City3'): 47}
The cost values is not incrementing in the for loop (I am using gurobi optimizer library).
I have two ways in which you can do it. In the end you get your decired dicts but in a list containing the dicts.
name_place=[]
for name in W:
for city in C:
name_place.append((name, city))
final_list=[]
for i, value in enumerate(Cost):
final_list.append({name_place[i]:value})
print(final_list)
Alternatively for a little more compact code you can use listcomprehension:
print([{[(name,city) for name in W for city in C][i]: value} for i, value in enumerate(Cost)])
Answered by Alex S on December 11, 2021
Just another way, using an iterator:
cost = iter(Cost)
for w in W:
for c in C:
print({(w, c): next(cost)})
Output:
{('Anis', 'City1'): 53}
{('Anis', 'City2'): 27}
{('Anis', 'City3'): 13}
{('James', 'City1'): 80}
{('James', 'City2'): 47}
{('James', 'City3'): 67}
{('Arthur', 'City1'): 53}
{('Arthur', 'City2'): 73}
{('Arthur', 'City3'): 47}
Answered by Kelly Bundy on December 11, 2021
This can be done using a combination of itertools.product
(every combination of values) for W
and C
, and then zip
for pairs of corresponding values between the resulting tuples and your Cost
list:
import itertools
W = ['Anis', 'James', 'Arthur']
C = ['City1', 'City2', 'City3']
Cost = [53, 27, 13, 80, 47, 67, 53, 73, 47]
for k, v in zip(itertools.product(W, C), Cost):
d = {k: v}
print(d)
This gives:
{('Anis', 'City1'): 53}
{('Anis', 'City2'): 27}
{('Anis', 'City3'): 13}
{('James', 'City1'): 80}
{('James', 'City2'): 47}
{('James', 'City3'): 67}
{('Arthur', 'City1'): 53}
{('Arthur', 'City2'): 73}
{('Arthur', 'City3'): 47}
If in fact what you want is a single dictionary rather than a sequence of 1-element dictionaries, then modify this to:
d = {k: v for k, v in zip(itertools.product(W, C), Cost)}
print(d)
which gives:
{('Anis', 'City1'): 53, ('Anis', 'City2'): 27, ('Anis', 'City3'): 13, ('James', 'City1'): 80, ('James', 'City2'): 47, ('James', 'City3'): 67, ('Arthur', 'City1'): 53, ('Arthur', 'City2'): 73, ('Arthur', 'City3'): 47}
or more simply (thanks to user Heap Overflow for the suggestion):
d = dict(zip(itertools.product(W, C), Cost))
Answered by alani on December 11, 2021
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