Python – Extract Key’s value from Mixed Dictionaries List
Last Updated :
21 Apr, 2023
Given a list of dictionaries, with each dictionary having different keys, extract value of key K.
Input : test_list = [{“Gfg” : 3, “b” : 7}, {“is” : 5, ‘a’ : 10}, {“Best” : 9, ‘c’ : 11}], K = ‘b’
Output : 7
Explanation : Value of b is 7.
Input : test_list = [{“Gfg” : 3, “b” : 7}, {“is” : 5, ‘a’ : 10}, {“Best” : 9, ‘c’ : 11}], K = ‘c’
Output : 11
Explanation : Value of c is 11.
Method #1 : Using list comprehension
This is one of the ways in which this task can be performed. In this, we iterate for each dictionary inside list, and check for key in it, if present the required value is returned.
Python3
test_list = [{"Gfg" : 3, "b" : 7},
{"is" : 5, 'a' : 10},
{"Best" : 9, 'c' : 11}]
print("The original list : " + str(test_list))
K = 'Best'
res = [sub[K] for sub in test_list if K in sub][0]
print("The extracted value : " + str(res))
|
Output
The original list : [{'Gfg': 3, 'b': 7}, {'is': 5, 'a': 10}, {'Best': 9, 'c': 11}]
The extracted value : 9
Time Complexity: O(n)
Auxiliary Space: O(n)
Method #2 : Using update() + loop
This is yet another way in which this task can be performed. In this, we update each dictionary into each other. Forming one large dictionary, and then the value is extracted from this dictionary.
Python3
test_list = [{"Gfg" : 3, "b" : 7},
{"is" : 5, 'a' : 10},
{"Best" : 9, 'c' : 11}]
print("The original list : " + str(test_list))
K = 'Best'
res = dict()
for sub in test_list:
res.update(sub)
print("The extracted value : " + str(res[K]))
|
Output
The original list : [{'Gfg': 3, 'b': 7}, {'is': 5, 'a': 10}, {'Best': 9, 'c': 11}]
The extracted value : 9
Time complexity: O(n*n), where n is the length of the dictionary list. The update() + loop takes O(n*n) time
Auxiliary Space: O(1), constant extra space is required
Method #3: Using map() + lambda function
We are using the map() function along with a lambda function to extract the value of the key K from each dictionary in the test_list.
First, we use the filter() function to filter out the dictionaries in the test_list that contain the key K. Then, we use the map() function to extract the value of K from each dictionary in the filtered list. Finally, we convert the result of map() into a list and select the first element, which is the value of K in the first dictionary that contains it.
Python3
test_list = [{"Gfg" : 3, "b" : 7},
{"is" : 5, 'a' : 10},
{"Best" : 9, 'c' : 11}]
print("The original list : " + str(test_list))
K = 'Best'
res = list(map(lambda x: x[K], filter(lambda x: K in x, test_list)))[0]
print("The extracted value : " + str(res))
|
Output
The original list : [{'Gfg': 3, 'b': 7}, {'is': 5, 'a': 10}, {'Best': 9, 'c': 11}]
The extracted value : 9
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.
Method 4: Use a for loop to iterate over each dictionary and check if the key exists in the dictionary
The program first initializes a list of dictionaries and a key K. It then uses filter() and lambda to create a new list containing only the dictionaries that contain the key K. Next, it uses map() and another lambda function to extract the value associated with K from each of these dictionaries. Finally, the program converts the resulting map object to a list and retrieves the first element, which should be the extracted value.
Python3
test_list = [{"Gfg" : 3, "b" : 7},
{"is" : 5, 'a' : 10},
{"Best" : 9, 'c' : 11}]
print("The original list : " + str(test_list))
K = 'Best'
res = None
for dictionary in test_list:
if K in dictionary:
res = dictionary[K]
break
print("The extracted value : " + str(res))
|
Output
The original list : [{'Gfg': 3, 'b': 7}, {'is': 5, 'a': 10}, {'Best': 9, 'c': 11}]
The extracted value : 9
The time complexity of this approach is O(n), where n is the number of dictionaries in the list.
The auxiliary space is O(1), as we are not creating any new data structures to store the intermediate results.
Method #5: Using try-except block
This program extracts the value of a given key from a list of dictionaries using the try-except block. It iterates through the list of dictionaries and uses a try-except block to extract the value of the given key. If the key is not present in any of the dictionaries, it returns None.
Python3
test_list = [{"Gfg" : 3, "b" : 7},
{"is" : 5, 'a' : 10},
{"Best" : 9, 'c' : 11}]
K = 'Best'
res = None
for dictionary in test_list:
try:
res = dictionary[K]
except KeyError:
pass
else:
break
print("The original list : " + str(test_list))
print("The extracted value : " + str(res))
|
Output
The original list : [{'Gfg': 3, 'b': 7}, {'is': 5, 'a': 10}, {'Best': 9, 'c': 11}]
The extracted value : 9
Time complexity: The time complexity of the above code is O(n), where n is the number of dictionaries in the list.
Auxiliary space: The auxiliary space used by the above code is O(1), as we are using constant space to store the key and value extracted from the list.
Method #6: Using dictionary comprehension
- Create a dictionary comprehension to extract the values of key ‘K’ from each dictionary in the list.
- Use the if statement to check if the key ‘K’ is present in the dictionary.
- Return the value of key ‘K’ if it is present in the dictionary, otherwise return None.
- Convert the resulting dictionary into a list of values.
Python3
test_list = [{"Gfg" : 3, "b" : 7},
{"is" : 5, 'a' : 10},
{"Best" : 9, 'c' : 11}]
K = 'Best'
res_dict = {d.get(K): None for d in test_list if K in d}
res_list = [v for v in res_dict if v is not None]
print("The extracted value : " + str(res_list[0]))
|
Output
The extracted value : 9
Time complexity: O(n)
Auxiliary space: O(n)
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