Skip to content
Related Articles

Related Articles

Improve Article

Python – Flatten Dictionary with List

  • Last Updated : 01 Aug, 2020
Geek Week

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]

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]

 Attention geek! Strengthen your foundations with the Python Programming Foundation Course and learn the basics.  

To begin with, your interview preparations Enhance your Data Structures concepts with the Python DS Course. And to begin with your Machine Learning Journey, join the Machine Learning – Basic Level Course




My Personal Notes arrow_drop_up
Recommended Articles
Page :