CODEWITHSUNDEEP
pythonpython-3.xdictionary
Rahul Sharma
Rahul Sharma

Reputation: 2495

Replace dictionary keys from values of another dictionary

I have three dictionaries:

packed_items = {0: [0, 3],
                2: [1], 
                1: [2]}
trucks_dict = {0: [9.5, 5.5, 5.5],
               1: [13.0, 5.5, 7.0],
               2: [16.0, 6.0, 7.0]}
items_dict = {0: [4.6, 4.3, 4.3],
              1: [4.6, 4.3, 4.3],
              2: [6.0, 5.6, 9.0],
              3: [8.75, 5.6, 6.6]}

packed_items consists of trucks as keys and values as list of items. I want to change my packed_dict such that it gives me output in this format

packed_dict = {[9.5, 5.5, 5.5]:[[4.6, 4.3, 4.3],[8.75, 5.6, 6.6]]
               [16.0, 6.0, 7.0]:[[4.6, 4.3, 4.3]]
               [13.0, 5.5, 7.0]:[[6.0, 5.6, 9.0]]}

Basically I want to replace my keys in packed_items with the values in trucks_dict, and values in packed_items with values in items_dict.

Upvotes: 1

Views: 1133

Answers (3)

blhsing
blhsing

Reputation: 106455

You can use a dict comprehension to map the lists in trucks_dict to items in items_dict. The lists have to be converted to tuples so that they can be hashable as keys:

{tuple(trucks_dict[k]): [items_dict[i] for i in l] for k, l in packed_items.items()}

This returns:

{(9.5, 5.5, 5.5): [[4.6, 4.3, 4.3], [8.75, 5.6, 6.6]],
 (13.0, 5.5, 7.0): [[6.0, 5.6, 9.0]],
 (16.0, 6.0, 7.0): [[4.6, 4.3, 4.3]]}

Upvotes: 0

Austin
Austin

Reputation: 26039

You cannot have lists as dictionary keys because they are unhashable.

Because you asked for string keys, you can do:

from collections import defaultdict

packed_items = {0: [0, 3],
                2: [1], 
                1: [2]}
trucks_dict = {0: [9.5, 5.5, 5.5],
               1: [13.0, 5.5, 7.0],
               2: [16.0, 6.0, 7.0]}
items_dict = {0: [4.6, 4.3, 4.3],
              1: [4.6, 4.3, 4.3],
              2: [6.0, 5.6, 9.0],
              3: [8.75, 5.6, 6.6]}

d = defaultdict(list)

for k1, v1 in trucks_dict.items():
    for k2, v2 in items_dict.items():
        if k1 == k2 % 3:
            d[str(v1)].append(v2)

print(d)
# {'[9.5, 5.5, 5.5]': [[4.6, 4.3, 4.3], [8.75, 5.6, 6.6]], '[16.0, 6.0, 7.0]': [[4.6, 4.3, 4.3]], '[13.0, 5.5, 7.0]': [[6.0, 5.6, 9.0]]}

Upvotes: 1

Stephen Rauch
Stephen Rauch

Reputation: 49784

By converting your list keys to tuples, you can do that with something like:

Code:

result = {}
for k, v in packed_items.items():
    for i in v:
        result.setdefault(tuple(trucks_dict[k]), []).append(items_dict[i])

Test Code:

packed_items = {0: [0, 3],
                2: [1],
                1: [2]}
trucks_dict = {0: [9.5, 5.5, 5.5],
               1: [13.0, 5.5, 7.0],
               2: [16.0, 6.0, 7.0]}
items_dict = {0: [4.6, 4.3, 4.3],
              1: [4.6, 4.3, 4.3],
              2: [6.0, 5.6, 9.0],
              3: [8.75, 5.6, 6.6]}

result = {}
for k, v in packed_items.items():
    for i in v:
        result.setdefault(tuple(trucks_dict[k]), []).append(items_dict[i])
print(result)

Results:

{(9.5, 5.5, 5.5): [[4.6, 4.3, 4.3], [8.75, 5.6, 6.6]], 
 (16.0, 6.0, 7.0): [[4.6, 4.3, 4.3]], 
 (13.0, 5.5, 7.0): [[6.0, 5.6, 9.0]]
}

Upvotes: 1

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