Python Dict Comprehension to Create and Update Dictionary
Question:
I have a list of dictionaries (data) and want to convert it into dictionary (x) as below.
I am using following ‘for loop’ to achieve.
data = [{'Dept': '0123', 'Name': 'Tom'},
{'Dept': '0123', 'Name': 'Cheryl'},
{'Dept': '0123', 'Name': 'Raj'},
{'Dept': '0999', 'Name': 'Tina'}]
x = {}
for i in data:
if i['Dept'] in x:
x[i['Dept']].append(i['Name'])
else:
x[i['Dept']] = [i['Name']]
Output:
x -> {'0999': ['Tina'], '0123': ['Tom', 'Cheryl', 'Raj']}
Is it possible to implement the above logic in dict comprehension or any other more pythonic way?
Answers:
It seems way too complicated to be allowed into any code that matters even the least bit, but just for fun, here you go:
{
dept: [item['Name'] for item in data if item['Dept'] == dept]
for dept in {item['Dept'] for item in data}
}
The dict comprehension, even though not impossible, might not be the best choice. May I suggest using a defaultdict
:
from collections import defaultdict
dic = defaultdict(list)
for i in data:
dic[i['Dept']].append(i['Name'])
I have a list of dictionaries (data) and want to convert it into dictionary (x) as below.
I am using following ‘for loop’ to achieve.
data = [{'Dept': '0123', 'Name': 'Tom'},
{'Dept': '0123', 'Name': 'Cheryl'},
{'Dept': '0123', 'Name': 'Raj'},
{'Dept': '0999', 'Name': 'Tina'}]
x = {}
for i in data:
if i['Dept'] in x:
x[i['Dept']].append(i['Name'])
else:
x[i['Dept']] = [i['Name']]
Output:
x -> {'0999': ['Tina'], '0123': ['Tom', 'Cheryl', 'Raj']}
Is it possible to implement the above logic in dict comprehension or any other more pythonic way?
It seems way too complicated to be allowed into any code that matters even the least bit, but just for fun, here you go:
{
dept: [item['Name'] for item in data if item['Dept'] == dept]
for dept in {item['Dept'] for item in data}
}
The dict comprehension, even though not impossible, might not be the best choice. May I suggest using a defaultdict
:
from collections import defaultdict
dic = defaultdict(list)
for i in data:
dic[i['Dept']].append(i['Name'])