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
# import a groupby() method # from itertools module from itertools import groupby # dictionary 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' } ] # define a function for key def key_func(k): return k[ 'company' ] # sort INFO data by 'company' key. 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
# import required methods from itertools import groupby from operator import itemgetter # dictionary 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' } ] # Sort students data by grade key. students = sorted (students, key = itemgetter( 'grade' )) # Display data grouped by 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.