从Python中的字典列表创建列表字典

Question

I have a list of dictionaries as follows.

[{'a' : 1, 'b' : 2, 'c' : 2},
 {'a' : 2, 'b' : 3, 'c' : 3},
 {'a' : 3, 'b' : 5, 'c' : 6},
 {'a' : 4, 'b' : 7, 'c' : 8},
 {'a' : 1, 'b' : 8, 'c' : 9},
 {'a' : 2, 'b' : 0, 'c' : 0},
 {'a' : 5, 'b' : 1, 'c' : 3},
 {'a' : 7, 'b' : 4, 'c' : 5}]

I want to create a dictionary of lists from above list which should be as follows.

{1 : [{'a' : 1, 'b' : 2, 'c' : 2}, {'a' : 1, 'b' : 8, 'c' : 9}]
 2 : [{'a' : 2, 'b' : 3, 'c' : 3}, {'a' : 2, 'b' : 0, 'c' : 0}]
 3 : [{'a' : 3, 'b' : 5, 'c' : 6}]
 4 : [{'a' : 4, 'b' : 7, 'c' : 8}]
 5 : [{'a' : 5, 'b' : 1, 'c' : 3}]
 7 : [{'a' : 7, 'b' : 4, 'c' : 5}]

Basically I want to pick one of the keys in dictionary say 'a', and create new dictionary with the values of that key (1, 2, 3, 4, 5, 7) as keys for new dictionary to be created, and values for new dictionary should be list of all the dictionaries containing that value as value for key 'a'.

I know the simplest approach is iterating over the list and build the required dictionary. I am just curious is there another way of doing it.

Answer 1

A collections.defaultdict will be the most efficient:

from collections import defaultdict

l = [{'a': 1, 'b': 2, 'c': 2},
 {'a': 2, 'b': 3, 'c': 3},
 {'a': 3, 'b': 5, 'c': 6},
 {'a': 4, 'b': 7, 'c': 8},
 {'a': 1, 'b': 8, 'c': 9},
 {'a': 2, 'b': 0, 'c': 0},
 {'a': 5, 'b': 1, 'c': 3},
 {'a': 7, 'b': 4, 'c': 5}]

dct = defaultdict(list)

for d in l:
    dct[d["a"]].append(d)

from pprint import pprint as pp
pp(dict(dct))

Output:

{1: [{'a': 1, 'b': 2, 'c': 2}, {'a': 1, 'b': 8, 'c': 9}],
 2: [{'a': 2, 'b': 3, 'c': 3}, {'a': 2, 'b': 0, 'c': 0}],
 3: [{'a': 3, 'b': 5, 'c': 6}],
 4: [{'a': 4, 'b': 7, 'c': 8}],
 5: [{'a': 5, 'b': 1, 'c': 3}],
 7: [{'a': 7, 'b': 4, 'c': 5}]}

Answer 2

Normal dictionary with setdefault method can be used

Code:

data=[{'a' : 1, 'b' : 2, 'c' : 2},
{'a' : 2, 'b' : 3, 'c' : 3},
{'a' : 3, 'b' : 5, 'c' : 6},
{'a' : 4, 'b' : 7, 'c' : 8},
{'a' : 1, 'b' : 8, 'c' : 9},
{'a' : 2, 'b' : 0, 'c' : 0},
{'a' : 5, 'b' : 1, 'c' : 3},
{'a' : 7, 'b' : 4, 'c' : 5}]

dictionary_list={}
for row in data:
    dictionary_list.setdefault(row["a"],[]).append(row)
print dictionary_list

Output:

{1: [{'a': 1, 'c': 2, 'b': 2}, {'a': 1, 'c': 9, 'b': 8}],
2: [{'a': 2, 'c': 3, 'b': 3}, {'a': 2, 'c': 0, 'b': 0}],
3: [{'a': 3, 'c': 6, 'b': 5}],
4: [{'a': 4, 'c': 8, 'b': 7}],
5: [{'a': 5, 'c': 3, 'b': 1}],
7: [{'a': 7, 'c': 5, 'b': 4}]}

Answer 3

You can do it in following way

mylist = [
   {'a' : 1, 'b' : 2, 'c' : 2},
   {'a' : 2, 'b' : 3, 'c' : 3},
   {'a' : 3, 'b' : 5, 'c' : 6},
   {'a' : 4, 'b' : 7, 'c' : 8},
   {'a' : 1, 'b' : 8, 'c' : 9},
   {'a' : 2, 'b' : 0, 'c' : 0},
   {'a' : 5, 'b' : 1, 'c' : 3},
   {'a' : 7, 'b' : 4, 'c' : 5}
]

def get_dict(mylist, required_key):
    result_dict = {}

    for mydict in mylist:
        result_dict.setdefault(mydict[required_key], [])
        result_dict[mydict[required_key]].append(mydict)

    return result_dict

result_dict = get_dict(mylist, required_key = 'a')
print(result_dict)