字典列表中的字典

Question

I have a list of dicts.

dictList = [
  {'name': 'some name'},
  {'name': 'some other name'},
  {'age': 'some age'},
  {'last_name': 'some last name'}
]

In that list of dicts each dict has one key and one value for that key, as shown above.

I need to create a dict that has the keys from all the dicts and each value for every key is a set with item values from the list of dicts. In the example, it'd be something like this:

expected_dict = {
    'name': ['some name', 'some other name'],
    'age': ['some age'],
    'last_name': ['some last name']
}

How can I do this in Python?

Answer 1

collections.defaultdict is one way:

from collections import defaultdict

d = defaultdict(list)

dictList = [
  {'name': 'some name'},
  {'name': 'some other name'},
  {'age': 'some age'},
  {'last_name': 'some last name'}
]

for i in dictList:
    for k, v in i.items():
        d[k].append(v)

# defaultdict(list,
#             {'age': ['some age'],
#              'last_name': ['some last name'],
#              'name': ['some name', 'some other name']})

Answer 2

You can use the builtin setdefault() function.

dictList = [
  {'name': 'some name'},
  {'name': 'some other name'},
  {'age': 'some age'},
  {'last_name': 'some last name'}
]

expected_dict = {}

for dictionary in dictList:
    for key, val in dictionary.items():
        expected_dict.setdefault(key, []).append(val)

print(expected_dict)

Output:

{
    'name': ['some name', 'some other name'], 
    'age': ['some age'], 
    'last_name': ['some last name']
}

Note: Using collections.defaultdict (as shown in this answer ) is simpler and faster than using dict.setdefault() .

From the documentation :

Working of collections.defaultdict :

When each key is encountered for the first time, it is not already in the mapping; so an entry is automatically created using the default_factory function which returns an empty list. The list.append() operation then attaches the value to the new list. When keys are encountered again, the look-up proceeds normally (returning the list for that key) and the list.append() operation adds another value to the list. This technique is simpler and faster than an equivalent technique using dict.setdefault() .

Answer 3

bigD = {}
for element in dictList:
    for key in element:
        if key in bigD:
            bigD[key].append(element[key])
        else:
            bigD[key] = element[key]

Answer 4

You can use itertools.groupby :

import import itertools
dictList = [
    {'name': 'some name'},
    {'name': 'some other name'},
    {'age': 'some age'},
    {'last_name': 'some last name'}
]
new_list = {a:[c for [[_, c]] in b] for a, b in itertools.groupby(map(lambda x:x.items(), dictList), key=lambda x:x[0][0])}

Output:

{'age': ['some age'], 'last_name': ['some last name'], 'name': ['some name', 'some other name']}

Answer 5

Very simply.

dict = {} 
list = [1,2,3]
dict['numbs'] = list
print(dict)

Output : {'numbs': [1, 2, 3]}