包含字典列表的字典

Question

Newbie here, so apologies for a duplicate/silly question but I can't find the answer.

I am trying to store the contents of a list (of dictionaries) into a new dictionary.

a={}
b=[]
c={"Name":"Claude","Surname":"Verde","Age":"35","City":"Paris"}
b.append(c)
c={"Name":"Jean","Surname":"Claude","Age":"22","City":"Paris"}
b.append(c)
c={"Name":"Sam","Surname":"Smith","Age":"42","City":"London"}
b.append(c)
c={"Name":"James","Surname":"Jones","Age":"44","City":"London"}
b.append(c)
for i in range(len(b)):
    if b[i]['City'] == 'Paris':
        a["Paris"]=([b[i]])
    elif b[i]['City'] == 'London':
        a["London"]=([b[i]])
a

Result:

{'Paris': [{'Name': 'Jean',
   'Surname': 'Claude',
   'Age': '22',
   'City': 'Paris'}],
 'London': [{'Name': 'James',
   'Surname': 'Jones',
   'Age': '44',
   'City': 'London'}]}

However, I want a dictionary containing all residents of a city.

The code above only stores the second, not the first, resident in the list.

ie hoping for the below output.

How can I achieve this?

Any help greatly appreciated. Thanks

{'Paris': [{'Name': 'Claude',
   'Surname': 'Verde',
   'Age': '35',
   'City': 'Paris'},{'Name': 'Jean',
   'Surname': 'Claude',
   'Age': '22',
   'City': 'Paris'}],
 'London': [{'Name': 'Sam',
   'Surname': 'Smith,
   'Age': '42',
   'City': 'London'},{'Name': 'James',
   'Surname': 'Jones',
   'Age': '44',
   'City': 'London'}]}

Answer 1

When you initialize a, I would approach it like this:

a = {}

b=[]
c={"Name":"Claude","Surname":"Verde","Age":"35","City":"Paris"}
b.append(c)
c={"Name":"Jean","Surname":"Claude","Age":"22","City":"Paris"}
b.append(c)
c={"Name":"Sam","Surname":"Smith","Age":"42","City":"London"}
b.append(c)
c={"Name":"James","Surname":"Jones","Age":"44","City":"London"}
b.append(c)

for i in range(len(b)):
    if b[i]['City'] not in a.keys():
        a[b[i]['City']] = []

for i in range(len(b)):
    a[b[i]['City']].append(b[i]])

As a note: I'm pretty sure you can combine those for loops like this:

for i in range(len(b)):
    if b[i]['City'] not in a.keys():
        a[b[i]['City']] = []
    a[b[i]['City']].append(b[i]])

Answer 2

Use itertools.groupby

>>> lst = [
... {"Name":"Claude","Surname":"Verde","Age":"35","City":"Paris"},
... {"Name":"Jean","Surname":"Claude","Age":"22","City":"Paris"},
... {"Name":"Sam","Surname":"Smith","Age":"42","City":"London"},
... {"Name":"James","Surname":"Jones","Age":"44","City":"London"}]
>>>
>>> from itertools import groupby
>>> f = lambda d: d['City']
>>> res = {k:list(v) for k,v in groupby(sorted(lst, key=f), f)}
>>> pprint(res)
{'London': [{'Age': '42', 'City': 'London', 'Name': 'Sam', 'Surname': 'Smith'},
            {'Age': '44', 'City': 'London', 'Name': 'James', 'Surname': 'Jones'}],
 'Paris': [{'Age': '35', 'City': 'Paris', 'Name': 'Claude', 'Surname': 'Verde'},
           {'Age': '22', 'City': 'Paris', 'Name': 'Jean', 'Surname': 'Claude'}]}

Answer 3

Use itertools.groupby :

from itertools import groupby

b = [{'Name': 'Claude', 'Surname': 'Verde', 'Age': '35', 'City': 'Paris'}, 
     {'Name': 'Jean', 'Surname': 'Claude', 'Age': '22', 'City': 'Paris'}, 
     {'Name': 'Sam', 'Surname': 'Smith', 'Age': '42', 'City': 'London'}, 
     {'Name': 'James', 'Surname': 'Jones', 'Age': '44', 'City': 'London'}]

print({k: list(g) for k, g in groupby(b, key=lambda x: x['City'])})

# {'Paris': [{'Name': 'Claude', 'Surname': 'Verde', 'Age': '35', 'City': 'Paris'}, {'Name': 'Jean', 'Surname': 'Claude', 'Age': '22', 'City': 'Paris'}], 
#  'London': [{'Name': 'Sam', 'Surname': 'Smith', 'Age': '42', 'City': 'London'}, {'Name': 'James', 'Surname': 'Jones', 'Age': '44', 'City': 'London'}]}

Answer 4

You can use the get method that has a default argument:

a["Paris"]= a.get("Paris", []) + [b[i]]

The drawback is that it will create a new list every time you insert an element. However, it's fine if you don't have much data.

This gives, with a few improvements:

for person in b:
    if person['City'] in ('Paris', 'London'):
        a[person['City']] = a.get(person['City'], []) + [person]

If you don't want to recreate a new list:

for person in b:
    if person['City'] in ('Paris', 'London'):
        new_city = a.get(person['City'], [])
        new_city.append(person)
        a[person['City']] = new_city

Or you could also use a defaultdict instead of a dict .