In this article, we will update the values of a list of dictionaries.
Method 1: Using append() function
The append function is used to insert a new value in the list of dictionaries, we will use pop() function along with this to eliminate the duplicate data.
Syntax:
- dictionary[row][‘key’].append(‘value’)
- dictionary[row][‘key’].pop(position)
Where:
- dictionary is the input list of dictionaries
- row is the row that we want to update
- value is the new value to be updated
- key is the column to be updated
- position is the place where the old value is there
Python program to create a list of dictionaries
Python3
# create a list of dictionaries# with student datadata = [ {'name': 'sravan', 'subjects': ['java', 'python']}, {'name': 'bobby', 'subjects': ['c/cpp', 'java']}, {'name': 'ojsawi', 'subjects': ['iot', 'cloud']}, {'name': 'rohith', 'subjects': ['php', 'os']}, {'name': 'gnanesh', 'subjects': ['html', 'sql']}]# display first studentprint(data[0])# display all studentdata |
Output:
{'name': 'sravan', 'subjects': ['java', 'python']}
[{'name': 'sravan', 'subjects': ['java', 'python']},
{'name': 'bobby', 'subjects': ['c/cpp', 'java']},
{'name': 'ojsawi', 'subjects': ['iot', 'cloud']},
{'name': 'rohith', 'subjects': ['php', 'os']},
{'name': 'gnanesh', 'subjects': ['html', 'sql']}]
Time Complexity: O(1)
Auxiliary Space: O(1)
Update values in the above list of dictionaries
Python3
# update first student python subject# to htmldata[0]['subjects'].append('html')data[0]['subjects'].pop(1)# update third student java subject# to dbmsdata[2]['subjects'].append('dbms')data[2]['subjects'].pop(1)# update fourth student php subject# to php-mysqldata[3]['subjects'].append('php-mysql')data[3]['subjects'].pop(0)# display updated listdata |
Output:
[{'name': 'sravan', 'subjects': ['java', 'html']},
{'name': 'bobby', 'subjects': ['c/cpp', 'java']},
{'name': 'ojsawi', 'subjects': ['iot', 'dbms']},
{'name': 'rohith', 'subjects': ['os', 'php-mysql']},
{'name': 'gnanesh', 'subjects': ['html', 'sql']}]
Time complexity: O(1)
Auxiliary space: O(1)
Method 2: Using insert() function
This function is used to insert updated data based on an index.
Syntax:
- dictionary[row][‘key’].insert(index,’value’)
- dictionary[row][‘key’].pop(position)
Where,
- dictionary is the input list of dictionaries
- row is the row that we want to update
- value is the new value to be updated
- index is the position to be updated
- key is the column to be updated
- position is the place where the old value is there
Example: Python program to update the values in a list of dictionaries
Python3
# update first student python subject# to htmldata[0]['subjects'].insert(0, 'html')data[0]['subjects'].pop(1)# update third student java subject# to dbmsdata[2]['subjects'].insert(0, 'dbms')data[2]['subjects'].pop(1)# update fourth student php subject# to php-mysqldata[3]['subjects'].insert(1, 'php-mysql')data[3]['subjects'].pop(0)# display updated listdata |
Output:
[{'name': 'sravan', 'subjects': ['html', 'python']},
{'name': 'bobby', 'subjects': ['c/cpp', 'java']},
{'name': 'ojsawi', 'subjects': ['dbms', 'cloud']},
{'name': 'rohith', 'subjects': ['php-mysql', 'os']},
{'name': 'gnanesh', 'subjects': ['html', 'sql']}]
The time complexity of updating the subjects of each student is O(1), since we are only inserting and popping elements from a list which takes constant time.
The space complexity of the program is also O(1), since we are not using any additional space other than the given list of dictionaries data.
