# Python – Flatten Dictionary with List

• Last Updated : 01 Aug, 2020

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 listtest_list = ["Gfg", "is", "Best", "For", "Geeks"]  # printing original listprint("The original list : " + str(test_list))  # initializing subs. Dictionarysubs_dict = {"Gfg" : 7, "Geeks" : 8}  # get() to perform presence checks and assign valuetemp = 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]

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 listtest_list = ["Gfg", "is", "Best", "For", "Geeks"]  # printing original listprint("The original list : " + str(test_list))  # initializing subs. Dictionarysubs_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]

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