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Python – Common keys in list and dictionary

  • Last Updated : 14 Oct, 2020

Given a dictionary and list, extract all the keys and list which are common.

Input : test_dict = {“Gfg”: 3, “is” : 5, “best” : 9, “for” : 0, “geeks” : 3}, test_list = [“Gfg”, “best”, “CS”]
Output : [‘Gfg’, ‘best’]
Explanation : Gfg and best are present in both dictionary and List.

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Input : test_dict = {“Gfg”: 3, “is” : 5, “best” : 9, “for” : 0, “geeks” : 3}, test_list = [“Gfg”, “good”, “CS”]
Output : [‘Gfg’]
Explanation : Gfg is present in both dictionary and List.



Method #1 : Using list comprehension

This is one of the ways in which this task can be performed. In this, we iterate for all the dictionary and list values, if find a match, they are added to result.

Python3




# Python3 code to demonstrate working of 
# Common keys in list and dictionary
# Using list comprehension
  
# initializing dictionary
test_dict = {"Gfg": 3, "is" : 5, "best" : 9, "for" : 0, "geeks" : 3}
  
# printing original dictionary
print("The original dictionary is : " + str(test_dict))
  
# initializing test_list 
test_list = ["Gfg", "best", "geeks"]
  
# using in operator to check for match 
res = [ele for ele in test_dict if ele in test_list]
  
# printing result 
print("The required result : " + str(res)) 
Output
The original dictionary is : {'Gfg': 3, 'is': 5, 'best': 9, 'for': 0, 'geeks': 3}
The required result : ['Gfg', 'best', 'geeks']

Method #2 : Using set() + intersection()

This is yet another way in which this task can be performed. In this, we convert both the containers, list and dictionary keys to set() and then intersect to find required match.

Python3




# Python3 code to demonstrate working of 
# Common keys in list and dictionary
# Using set() + intersection()
  
# initializing dictionary
test_dict = {"Gfg": 3, "is" : 5, "best" : 9, "for" : 0, "geeks" : 3}
  
# printing original dictionary
print("The original dictionary is : " + str(test_dict))
  
# initializing test_list 
test_list = ["Gfg", "best", "geeks"]
  
# intersection() used to get Common elements 
res = set(test_list).intersection(set(test_dict))
  
# printing result 
print("The required result : " + str(list(res))) 
Output
The original dictionary is : {'Gfg': 3, 'is': 5, 'best': 9, 'for': 0, 'geeks': 3}
The required result : ['best', 'geeks', 'Gfg']



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