Python – Extract Key’s Value, if Key Present in List and Dictionary

Given a list, dictionary, and a Key K, print the value of K from the dictionary if the key is present in both, the list and the dictionary.

Input : test_list = ["Gfg", "is", "Good", "for", "Geeks"], 
                test_dict = {"Gfg" : 5, "Best" : 6}, K = "Gfg" 
Output : 5 
Explanation : "Gfg" is present in list and has value 5 in dictionary. 

Input : test_list = ["Good", "for", "Geeks"], 
                test_dict = {"Gfg" : 5, "Best" : 6}, K = "Gfg" 
Output : None 
Explanation : "Gfg" not present in List.

Method #1 : Using all() + generator expression

The combination of the above functions offers one of the ways in which this problem can be solved. In this, we use all() to check for occurrence in both dictionary and list. If the result is true value is extracted from to result.

Python3

# Python3 code to demonstrate working of
# Extract Key's Value, if Key Present in List and Dictionary
# Using all() + list comprehension

# initializing list
test_list = ["Gfg", "is", "Good", "for", "Geeks"]

# initializing Dictionary
test_dict = {"Gfg" : 2, "is" : 4, "Best" : 6}

# initializing K
K = "Gfg"

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

# using all() to check for occurrence in list and dict
# encapsulating list and dictionary keys in list
res = None
if all(K in sub for sub in [test_dict, test_list]):
    res = test_dict[K]

# printing result
print("Extracted Value : " + str(res))

Output

The original list : ['Gfg', 'is', 'Good', 'for', 'Geeks']
The original Dictionary : {'Gfg': 2, 'is': 4, 'Best': 6}
Extracted Value : 2

Time complexity: O(n), where n is the total number of elements in the test_list. This is because, in the worst case, each element of the list needs to be checked for its presence in both the list and the dictionary, resulting in a time complexity of O(n).
Auxiliary space: O(1), as it only requires a constant amount of additional memory to store the result and temporary variables.

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

This is another way to check for the key’s presence in both containers. In this, we compute the intersection of all values of list and dict keys and test for the Key’s occurrence in that.

Python3

# Python3 code to demonstrate working of
# Extract Key's Value, if Key Present in List and Dictionary
# Using set() + intersection()

# initializing list
test_list = ["Gfg", "is", "Good", "for", "Geeks"]

# initializing Dictionary
test_dict = {"Gfg" : 2, "is" : 4, "Best" : 6}

# initializing K
K = "Gfg"

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

# conversion of lists to set and intersection with keys
# using intersection
res = None
if K in set(test_list).intersection(test_dict):
    res = test_dict[K]

# printing result
print("Extracted Value : " + str(res))

Output

The original list : ['Gfg', 'is', 'Good', 'for', 'Geeks']
The original Dictionary : {'Gfg': 2, 'is': 4, 'Best': 6}
Extracted Value : 2

Time complexity: O(n), where n is the length of the list.
Auxiliary space: O(m), where m is the number of unique keys in the dictionary that match the elements in the list.

Method #3: Using in operator

Python3

# Python3 code to demonstrate working of
# Extract Key's Value, if Key Present in List and Dictionary

# initializing list
test_list = ["Gfg", "is", "Good", "for", "Geeks"]

# initializing Dictionary
test_dict = {"Gfg" : 2, "is" : 4, "Best" : 6}

# initializing K
K = "Gfg"

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

if K in test_dict.keys() and K in test_list:
    res=test_dict[K]

# printing result
print("Extracted Value : " + str(res))

Output

The original list : ['Gfg', 'is', 'Good', 'for', 'Geeks']
The original Dictionary : {'Gfg': 2, 'is': 4, 'Best': 6}
Extracted Value : 2

Time Complexity: O(n)
Auxiliary Space: O(1)

Method 4: Using operator.countOf() method

Python3

# Python3 code to demonstrate working of
# Extract Key's Value, if Key Present in List and Dictionary
import operator as op
# initializing list
test_list = ["Gfg", "is", "Good", "for", "Geeks"]

# initializing Dictionary
test_dict = {"Gfg": 2, "is": 4, "Best": 6}

# initializing K
K = "Gfg"

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

if op.countOf(test_dict.keys(), K) > 0 and op.countOf(test_list, K) > 0:
    res = test_dict[K]

# printing result
print("Extracted Value : " + str(res))

Output

The original list : ['Gfg', 'is', 'Good', 'for', 'Geeks']
The original Dictionary : {'Gfg': 2, 'is': 4, 'Best': 6}
Extracted Value : 2

Time Complexity: O(N)
Auxiliary Space: O(1)

Method 5: Using any() + dictionary.get() method

Python3

# Python3 code to demonstrate working of
# Extract Key's Value, if Key Present in List and Dictionary
# Using any() + dictionary.get() method

# initializing list
test_list = ["Gfg", "is", "Good", "for", "Geeks"]

# initializing Dictionary
test_dict = {"Gfg": 2, "is": 4, "Best": 6}

# initializing K
K = "Gfg"

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

# using any() to check for occurrence in list and dict
# accessing value of key using dictionary.get() method
res = None
if K in test_list and K in test_dict:
    res = test_dict.get(K)

# printing result
print("Extracted Value : " + str(res))

Output

The original list : ['Gfg', 'is', 'Good', 'for', 'Geeks']
The original Dictionary : {'Gfg': 2, 'is': 4, 'Best': 6}
Extracted Value : 2

Time Complexity: O(N)
Auxiliary Space: O(1)

Method #6: Using try-except block to handle KeyErrors

Use a try-except block to handle the KeyError that may occur if K is not present in test_dict. First, check if K is present in both test_list and test_dict.keys(). If yes, extract the value of K from test_dict. If no, set the result to None. If a KeyError occurs, handle it by setting the result to None. Finally, print the result.

Python3

test_list = ["Gfg", "is", "Good", "for", "Geeks"]
test_dict = {"Gfg" : 2, "is" : 4, "Best" : 6}
K = "Gfg"

try:
    # check if K is present in both test_list and test_dict keys
    if K in test_list and K in test_dict.keys():
        # if yes, extract value of K from test_dict
        res = test_dict[K]
    else:
        # if no, set result to None
        res = None
except KeyError:
    # handle KeyError exception
    res = None

# print result
print("Extracted Value : " + str(res))

Output

Extracted Value : 2

Time complexity: O(n), where n is the length of the test_list. This is because we check if K is present in both test_list and test_dict.keys(), which takes O(n) time in the worst case.
Auxiliary space: O(1), because we use a constant amount of extra space to store the variables test_list, test_dict, K, and res, and we don’t create any additional data structures in the code.

Method #7: Using try-except block with .get() method

This method involves using a try-except block to check if the key exists in the dictionary and retrieve its value using the .get() method.

Follow the below steps to implement the above idea:

• Initialize the list, dictionary, and key, same as in the original code.
• Use a try-except block to handle the KeyError exception that occurs if the key is not found in the dictionary.
• Inside the try block, retrieve the value of the key using the .get() method on the dictionary. If the key is not found, .get() will return None by default.
• If the key is found in both the list and dictionary, return the value of the key. If not, return None.

Below is the implementation of the above approach:

Python3

# Python3 code to demonstrate working of
# Extract Key's Value, if Key Present in List and Dictionary
# Using try-except block with .get() method

# initializing list
test_list = ["Gfg", "is", "Good", "for", "Geeks"]

# initializing Dictionary
test_dict = {"Gfg" : 2, "is" : 4, "Best" : 6}

# initializing K
K = "Gfg"

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

# using try-except block with .get() method
res = None
try:
    if K in test_list and K in test_dict:
        res = test_dict.get(K)
except KeyError:
    pass

# printing result
print("Extracted Value : " + str(res))

Output

The original list : ['Gfg', 'is', 'Good', 'for', 'Geeks']
The original Dictionary : {'Gfg': 2, 'is': 4, 'Best': 6}
Extracted Value : 2

Time complexity: O(n), Where n is the length of test_list.
Auxiliary space: O(1) for storing the result and constant extra space for the try-except block.

Method #8: Using a for loop to iterate through the list and dictionary

Step-by-step approach:

• Initialize the list, dictionary, and key as given in the problem statement:
• Use a for loop to iterate through the list and dictionary:
  • Checks each item in the list. If an item is equal to the key, it uses the get() method to retrieve the corresponding value from the dictionary. 
  • If the key is not found, res is set to None. The loop stops as soon as a matching item is found.
• Print the extracted value:

Python

test_list = ["Gfg", "is", "Good", "for", "Geeks"]
test_dict = {"Gfg" : 2, "is" : 4, "Best" : 6}
K = "Gfg"

# Method 9: Using a for loop to iterate through the list and dictionary
res = None
for item in test_list:
    if item == K:
        res = test_dict.get(K)
        break

print("Method 9: Extracted Value : " + str(res))

Output

Method 9: Extracted Value : 2

Time complexity: O(n), where n is the length of the list
Auxiliary space: O(1)