What is Python – Update values of a list of dictionaries?

In this article, we will learn Python – Update values of a list of dictionaries,This free Python tutorial for complete beginners will help you learn Python from scratch.

Python – Update values of a list of dictionaries

Python - Add values to Dictionary of List

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 data
data = [
    {'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 student
print(data[0])
 
# display all student
data

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 html
data[0]['subjects'].append('html')
data[0]['subjects'].pop(1)
 
# update third student java subject
# to dbms
data[2]['subjects'].append('dbms')
data[2]['subjects'].pop(1)
 
# update fourth student php subject
# to php-mysql
data[3]['subjects'].append('php-mysql')
data[3]['subjects'].pop(0)
 
# display updated list
data

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 html
data[0]['subjects'].insert(0, 'html')
data[0]['subjects'].pop(1)
 
# update third student java subject
# to dbms
data[2]['subjects'].insert(0, 'dbms')
data[2]['subjects'].pop(1)
 
# update fourth student php subject
# to php-mysql
data[3]['subjects'].insert(1, 'php-mysql')
data[3]['subjects'].pop(0)
 
# display updated list
data

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.

Tags: