Open In App

Python – Append items at beginning of dictionary

Last Updated : 05 May, 2023
Improve
Improve
Like Article
Like
Save
Share
Report

Given a dictionary, append another dictionary at beginning of it.

Input : test_dict = {“Gfg” : 5, “is” : 3, “best” : 10}, updict = {“pre1” : 4} Output : {‘pre1’: 4, ‘Gfg’: 5, ‘is’: 3, ‘best’: 10} Explanation : New dictionary updated at front of dictionary. Input : test_dict = {“Gfg” : 5}, updict = {“pre1” : 4} Output : {‘pre1’: 4, ‘Gfg’: 5} Explanation : New dictionary updated at front of dictionary, “pre1” : 4.

Method #1 : Using update()

This is one of the ways in which this task can be performed. In this, we use update function to update old dictionary after the new one so that new dictionary is appended at beginning.

Python3




# Python3 code to demonstrate working of
# Append items at beginning of dictionary
# Using update()
 
# initializing dictionary
test_dict = {"Gfg" : 5, "is" : 3, "best" : 10}
 
# printing original dictionary
print("The original dictionary is : " + str(test_dict))
 
# initializing update dictionary
updict = {"pre1" : 4, "pre2" : 8}
 
# update() on new dictionary to get desired order
updict.update(test_dict)
 
# printing result
print("The required dictionary : " + str(updict))


Output

The original dictionary is : {'Gfg': 5, 'is': 3, 'best': 10}
The required dictionary : {'pre1': 4, 'pre2': 8, 'Gfg': 5, 'is': 3, 'best': 10}

Method #2 : Using ** operator

This is yet another way in which this task can be performed. In this, we perform packing and unpacking of items into custom made dictionary using ** operator.

Python3




# Python3 code to demonstrate working of
# Append items at beginning of dictionary
# Using ** operator
 
# initializing dictionary
test_dict = {"Gfg" : 5, "is" : 3, "best" : 10}
 
# printing original dictionary
print("The original dictionary is : " + str(test_dict))
 
# initializing update dictionary
updict = {"pre1" : 4, "pre2" : 8}
 
# ** operator for packing and unpacking items in order
res = {**updict, **test_dict}
 
# printing result
print("The required dictionary : " + str(res))


Output

The original dictionary is : {'Gfg': 5, 'is': 3, 'best': 10}
The required dictionary : {'pre1': 4, 'pre2': 8, 'Gfg': 5, 'is': 3, 'best': 10}

Method #3 : Using for loop

Approach

  1. Initiate a for loop to traverse keys of test_dict
  2. Add keys and values of test_dict to updict using assignment operator =
  3. Display updict

Python3




# Python3 code to demonstrate working of
# Append items at beginning of dictionary
 
 
# initializing dictionary
test_dict = {"Gfg" : 5, "is" : 3, "best" : 10}
 
# printing original dictionary
print("The original dictionary is : " + str(test_dict))
 
# initializing update dictionary
updict = {"pre1" : 4, "pre2" : 8}
for i in list(test_dict.keys()):
    updict[i]=test_dict[i]
 
 
# printing result
print("The required dictionary : " + str(updict))


Output

The original dictionary is : {'Gfg': 5, 'is': 3, 'best': 10}
The required dictionary : {'pre1': 4, 'pre2': 8, 'Gfg': 5, 'is': 3, 'best': 10}

Time Complexity : O(N) N – length of test_dict

Auxiliary Space : O(N) N – length of updated dictionary up_dict



Like Article
Suggest improvement
Previous
Next
Share your thoughts in the comments

Similar Reads