Python – Render Initials as Dictionary Key

Given List of Strings, convert to dictionary with Key as initial value of values. Won’t work in cases having words with similar initials. 

Input : test_list = [“geeksforgeeks”, “is”, “best”] 
Output : {‘g’: ‘geeksforgeeks’, ‘i’: ‘is’, ‘b’: ‘best’} 
Explanation : Keys constructed from initial character.

Input : test_list = [“geeksforgeeks”, “best”] 
Output : {‘g’: ‘geeksforgeeks’, ‘b’: ‘best’} 
Explanation : Keys constructed from initial character. 

Method #1 : Using loop 

In this, we create each dictionary by getting initial element using string element access and render value as list element.

Python3

# Python3 code to demonstrate working of
# Render Initials as Dictionary Key
# Using loop

# initializing list
test_list = ["geeksforgeeks", "is", "best"]

# printing original list
print("The original list is : " + str(test_list))

res = dict()
for ele in test_list:

    # assigning initials as key
    res[ele[0]] = ele

# printing result
print("Constructed Dictionary : " + str(res))

Output

The original list is : ['geeksforgeeks', 'is', 'best']
Constructed Dictionary : {'g': 'geeksforgeeks', 'i': 'is', 'b': 'best'}

Time Complexity: O(n), where n is the number of elements in the input list. 

Auxiliary Space: O(n)

Method #2 : Using dictionary comprehension 

In this, we create dictionary using shorthand method similar to above method, to provide one liner alternative to actual problem.

Python3

# Python3 code to demonstrate working of
# Render Initials as Dictionary Key
# Using dictionary comprehension

# initializing list
test_list = ["geeksforgeeks", "is", "best"]

# printing original list
print("The original list is : " + str(test_list))

# constructing dictionary
res = {ele[0] : ele for ele in test_list}

# printing result
print("Constructed Dictionary : " + str(res))

Output

The original list is : ['geeksforgeeks', 'is', 'best']
Constructed Dictionary : {'g': 'geeksforgeeks', 'i': 'is', 'b': 'best'}

Method #3: Using the reduce function from the functools module.

Step-by-step approach:

1. Define a function construct_dict(dict_so_far, current_word) that takes two arguments: dict_so_far, which is a dictionary that accumulates the key-value pairs as we process each element in test_list, and current_word, which is the current word being processed.
2. Inside the construct_dict function, we add a new key-value pair to the dictionary dict_so_far using current_word[0] as the key and current_word as the value. Then, we return the updated dictionary.
3. Call the reduce function from the functools module, passing construct_dict as the function to apply to each element in test_list, test_list as the list to apply the function to, and an empty dictionary {} as the initial value of the accumulator.
4. The reduce function iterates over the list test_list, calling construct_dict with the current element and the current value of the accumulator (which is initially an empty dictionary), and updates the accumulator with the key-value pair returned by construct_dict.
5. After all elements in test_list have been processed, reduce returns the final value of the accumulator, which is a dictionary where the keys are the first letters of the words in test_list and the values are the corresponding words.
6. Finally, we print the constructed dictionary using the print() function.

Python3

from functools import reduce

# initializing list
test_list = ["geeksforgeeks", "is", "best"]

# printing original list
print("The original list is : " + str(test_list))

# Define a function to construct the dictionary
def construct_dict(dict_so_far, current_word):
    dict_so_far[current_word[0]] = current_word
    return dict_so_far

# Call reduce with the function and list as arguments
res = reduce(construct_dict, test_list, {})

# printing result
print("Constructed Dictionary : " + str(res))

Output

The original list is : ['geeksforgeeks', 'is', 'best']
Constructed Dictionary : {'g': 'geeksforgeeks', 'i': 'is', 'b': 'best'}

Time complexity: O(n), where n is the length of the input list, since we only need to iterate over the list once.
Auxiliary space: O(n), since we need to create a dictionary with n key-value pairs.

Method #4: Using loop and defaultdict with tuple as default value:

Algorithm:

1. Import the defaultdict module from the collections library.
2. Initialize an empty defaultdict with the str function.
3. Iterate through each element of the input list using a for loop.
4. Extract the first character of each element and use it as a key to store the corresponding element in the defaultdict.
5. The defaultdict will automatically create a key with an empty string value if the key does not already exist.
6. Return the resulting defaultdict.

Python3

from collections import defaultdict

test_list = ["geeksforgeeks", "is", "best"]
# printing original list
print("The original list is : " + str(test_list))
res = defaultdict(str)

for ele in test_list:
    res[ele[0]] = ele

print("Constructed Dictionary : " + str(dict(res)))
#This code is contributed by Jyothi Pinjala

Output

The original list is : ['geeksforgeeks', 'is', 'best']
Constructed Dictionary : {'g': 'geeksforgeeks', 'i': 'is', 'b': 'best'}

Time Complexity: O(n), where n is the length of the input list. This is because we only iterate through the list once.

Auxiliary Space: O(n), where n is the length of the input list. This is because we create a dictionary with n key-value pairs. However, the space complexity can be slightly higher due to the overhead of using a defaultdict instead of a regular dictionary.