Group List of Dictionary Data by Particular Key in Python
Group List of Dictionary Data by Particular Key in Python can be done using itertools.groupby() method.
Itertools.groupby()
This method calculates the keys for each element present in iterable. It returns key and iterable of grouped items.
Syntax: itertools.groupby(iterable, key_func)
Parameters:
- iterable: Iterable can be of any kind (list, tuple, dictionary).
- key_func: A function that calculates keys for each element present in iterable.
Return type: It returns consecutive keys and groups from the iterable. If the key function is not specified or is None, key defaults to an identity function and returns the element unchanged.
Let’s see the examples: Example 1: Suppose we have list of dictionary of employee and company.
INFO = [
{'employee': 'XYZ_1', 'company': 'ABC_1'},
{'employee': 'XYZ_2', 'company': 'ABC_2'},
{'employee': 'XYZ_3', 'company': 'ABC_3'},
{'employee': 'XYZ_4', 'company': 'ABC_3'},
{'employee': 'XYZ_5', 'company': 'ABC_2'},
{'employee': 'XYZ_6', 'company': 'ABC_3'},
{'employee': 'XYZ_7', 'company': 'ABC_1'},
{'employee': 'XYZ_8', 'company': 'ABC_2'},
{'employee': 'XYZ_9', 'company': 'ABC_1'}
]
Now we need to display all the data group by the ‘company’ key name.
Code:
Python3
from itertools import groupby
INFO = [
{ 'employee' : 'XYZ_1' , 'company' : 'ABC_1' },
{ 'employee' : 'XYZ_2' , 'company' : 'ABC_2' },
{ 'employee' : 'XYZ_3' , 'company' : 'ABC_3' },
{ 'employee' : 'XYZ_4' , 'company' : 'ABC_3' },
{ 'employee' : 'XYZ_5' , 'company' : 'ABC_2' },
{ 'employee' : 'XYZ_6' , 'company' : 'ABC_3' },
{ 'employee' : 'XYZ_7' , 'company' : 'ABC_1' },
{ 'employee' : 'XYZ_8' , 'company' : 'ABC_2' },
{ 'employee' : 'XYZ_9' , 'company' : 'ABC_1' }
]
def key_func(k):
return k[ 'company' ]
INFO = sorted (INFO, key = key_func)
for key, value in groupby(INFO, key_func):
print (key)
print ( list (value))
|
Output:
ABC_1 [{’employee’: ‘XYZ_1’, ‘company’: ‘ABC_1′}, {’employee’: ‘XYZ_7’, ‘company’: ‘ABC_1′}, {’employee’: ‘XYZ_9’, ‘company’: ‘ABC_1′}] ABC_2 [{’employee’: ‘XYZ_2’, ‘company’: ‘ABC_2′}, {’employee’: ‘XYZ_5’, ‘company’: ‘ABC_2′}, {’employee’: ‘XYZ_8’, ‘company’: ‘ABC_2′}] ABC_3 [{’employee’: ‘XYZ_3’, ‘company’: ‘ABC_3′}, {’employee’: ‘XYZ_4’, ‘company’: ‘ABC_3′}, {’employee’: ‘XYZ_6’, ‘company’: ‘ABC_3’}]
Example 2: Suppose we have list of dictionary of student grades and marks.
students = [
{'mark': '65','grade': 'C'},
{'mark': '86','grade': 'A'},
{'mark': '73','grade': 'B'},
{'mark': '49','grade': 'D'},
{'mark': '91','grade': 'A'},
{'mark': '79','grade': 'B'}
]
Now we need to display all the data group by the ‘grade’ key.
Code:
Python3
from itertools import groupby
from operator import itemgetter
students = [
{ 'mark' : '65' , 'grade' : 'C' },
{ 'mark' : '86' , 'grade' : 'A' },
{ 'mark' : '73' , 'grade' : 'B' },
{ 'mark' : '49' , 'grade' : 'D' },
{ 'mark' : '91' , 'grade' : 'A' },
{ 'mark' : '79' , 'grade' : 'B' }
]
students = sorted (students,
key = itemgetter( 'grade' ))
for key, value in groupby(students,
key = itemgetter( 'grade' )):
print (key)
for k in value:
print (k)
|
Output:
A
{'mark': '86', 'grade': 'A'}
{'mark': '91', 'grade': 'A'}
B
{'mark': '73', 'grade': 'B'}
{'mark': '79', 'grade': 'B'}
C
{'mark': '65', 'grade': 'C'}
D
{'mark': '49', 'grade': 'D'}
The time complexity of the provided code is O(n log n) for sorting the students list using the sorted() method, and O(n) for grouping the students by grade using the groupby() method.
The auxiliary space complexity of the code is O(n), as it creates a sorted copy of the students list and stores it in memory, and also creates a dictionary-like object to store the grouped data.
Last Updated :
20 Mar, 2023
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