Python – Flatten Dictionary with List

Given a list and dictionary, flatten dictionary with keys and values at position of available element of key in list.

Input : test_list = [“Gfg”, “is”, “Best”, “For”, “Geeks”], subs_dict = {“Gfg” : 7} 
Output : [‘Gfg’, 7, ‘is’, ‘Best’, ‘For’, ‘Geeks’] 
Explanation : “Gfg” is replaced, followed by its value in dictionary. 

Input : test_list = [“Gfg”, “is”, “Best”, “For”, “Geeks”], subs_dict = {“gfg” : 7, “best” : 8} 
Output : [‘Gfg’, ‘is’, ‘Best’, ‘For’, ‘Geeks’] 
Explanation : No replacement. No matching values.

Method #1 : Using list comprehension + get()

The combination of above functionalities can be used to solve this problem. In this, we append all the key if present checking using get(), along with values in list.

Python3

# Python3 code to demonstrate working of
# Flatten Dictionary with List
# Using get() + list comprehension

# initializing list
test_list = ["Gfg", "is", "Best", "For", "Geeks"]

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

# initializing subs. Dictionary
subs_dict = {"Gfg" : 7, "Geeks" : 8}

# get() to perform presence checks and assign value
temp = object()
res = [ele for sub in ((ele, subs_dict.get(ele, temp))
       for ele in test_list) for ele in sub if ele != temp]

# printing result
print("The list after substitution : " + str(res))

Output

The original list : ['Gfg', 'is', 'Best', 'For', 'Geeks']
The list after substitution : ['Gfg', 7, 'is', 'Best', 'For', 'Geeks', 8]

Time Complexity: O(n), where n is the elements of dictionary
Auxiliary Space: O(n), where n is the size of dictionary

Method #2 : Using chain.from_iterable() + list comprehension 

This is yet another way in which this task can be performed. In this, we form key value pair list and append if present and if not retain the element. Next, the flattening of key-value lists is performed using chain.from_iterable().

Python3

# Python3 code to demonstrate working of
# Flatten Dictionary with List
# Using chain.from_iterable() + list comprehension
from itertools import chain

# initializing list
test_list = ["Gfg", "is", "Best", "For", "Geeks"]

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

# initializing subs. Dictionary
subs_dict = {"Gfg" : 7, "Geeks" : 8}

temp = ([ele, subs_dict[ele]] if ele in subs_dict
                  else [ele] for ele in test_list)

# chain.from_iterable() for flattening
res = list(chain.from_iterable(temp))

# printing result
print("The list after substitution : " + str(res))

Output

The original list : ['Gfg', 'is', 'Best', 'For', 'Geeks']
The list after substitution : ['Gfg', 7, 'is', 'Best', 'For', 'Geeks', 8]

Time Complexity: O(n*n) where n is the number of elements in the list “test_list”. 
Auxiliary Space: O(n) where n is the number of elements in the list “test_list”. 

Method #3: Using a for loop and append method for flattening the dictionary:

Step-by-step approach:

• Create an empty list called res.
• Iterate through each element ele in test_list.
• If ele is a key in subs_dict, then append ele and the corresponding value from subs_dict to res.
• Otherwise, append ele to res.
• Print the resulting list res.

Python3

# Python3 code to demonstrate working of
# Flatten Dictionary with List
# Using for loop and append()

# initializing list
test_list = ["Gfg", "is", "Best", "For", "Geeks"]

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

# initializing subs. Dictionary
subs_dict = {"Gfg" : 7, "Geeks" : 8}

res = []
for ele in test_list:
    if ele in subs_dict:
        res.append(ele)
        res.append(subs_dict[ele])
    else:
        res.append(ele)

# printing result
print("The list after substitution : " + str(res))

Output

The original list : ['Gfg', 'is', 'Best', 'For', 'Geeks']
The list after substitution : ['Gfg', 7, 'is', 'Best', 'For', 'Geeks', 8]

Time complexity: O(n) where n is the length of the input list.
Auxiliary space: O(n) where n is the length of the input list.