Python – Test if custom keys equal to K in dictionary
Last Updated :
05 Apr, 2023
Given dictionary and custom keys list, check if all those custom keys equals K.
Input : test_dict = {“Gfg” : 5, “is” : 8, “Best” : 10, “for” : 10, “Geeks” : 10}, cust_keys = [“is”, “for”, “Geeks”], K = 10
Output : False
Explanation : “is” is having 8 as value not 10, hence False
Input : test_dict = {“Gfg” : 5, “is” : 10, “Best” : 10, “for” : 10, “Geeks” : 10}, cust_keys = [“is”, “for”, “Geeks”], K = 10
Output : True
Explanation : All keys have 10 as values.
Method #1 : Using loop
This is brute way in which this task can be performed. In this we iterate each custom key and check if all are equal to K by keeping track using boolean variable.
Python3
test_dict = {"Gfg" : 5, "is" : 8, "Best" : 10, "for" : 8, "Geeks" : 8}
print("The original dictionary is : " + str(test_dict))
cust_keys = ["is", "for", "Geeks"]
K = 8
res = True
for key in cust_keys:
if test_dict[key] != K:
res = False
break
print("Are all custom keys equal to K : " + str(res))
|
Output
The original dictionary is : {'Gfg': 5, 'is': 8, 'Best': 10, 'for': 8, 'Geeks': 8}
Are all custom keys equal to K : True
Method #2 : Using all() + generator expression
The combination of above functionalities can be used to solve this problem. In this, we use all() to check for all the values and generator expression performs required iteration.
Python3
test_dict = {"Gfg" : 5, "is" : 8, "Best" : 10, "for" : 8, "Geeks" : 8}
print("The original dictionary is : " + str(test_dict))
cust_keys = ["is", "for", "Geeks"]
K = 8
res = all(test_dict[key] == K for key in cust_keys)
print("Are all custom keys equal to K : " + str(res))
|
Output
The original dictionary is : {'Gfg': 5, 'is': 8, 'Best': 10, 'for': 8, 'Geeks': 8}
Are all custom keys equal to K : True
Method #3: Using Set intersection to check if all the custom keys in the dictionary have the value K
Step-by-step algorithm:
- Create a set of intersection of cust_keys and test_dict keys.
- Check if the intersection set is equal to the set of cust_keys.
- If the intersection set is not equal to the set of cust_keys, return False.
- If the intersection set is equal to the set of cust_keys, iterate over each key in cust_keys.
- Check if the value of the key in test_dict is equal to K.
- If the value of the key in test_dict is not equal to K, return False.
- If all the values in test_dict for the custom keys are equal to K, return True.
Python3
test_dict = {"Gfg" : 5, "is" : 8, "Best" : 10, "for" : 8, "Geeks" : 8}
cust_keys = ["is", "for", "Geeks"]
K = 8
print("The original dictionary is : " + str(test_dict))
res = set(cust_keys).intersection(test_dict.keys()) == set(cust_keys) and all(test_dict[key] == K for key in cust_keys)
print("Are all custom keys equal to K : " + str(res))
|
Output
The original dictionary is : {'Gfg': 5, 'is': 8, 'Best': 10, 'for': 8, 'Geeks': 8}
Are all custom keys equal to K : True
Complexity Analysis:
Time Complexity: O(n), where n is the length of the cust_keys list.
Auxiliary Space: O(m), where m is the number of keys in the test_dict dictionary.
Method #4: Using the filter() function and a lambda function:
- Initialize the integer variable K with the value to be compared.
- Define a lambda function that takes a key and checks if the value corresponding to the key in the dictionary is equal to K.
- Use the filter() function with the defined lambda function and the list of custom keys cust_keys to filter out the keys whose values are not equal to K.
- Convert the filtered result obtained in step 6 to a list and check if its length is equal to the length of the list cust_keys. If the lengths are equal, it means all the custom keys have values equal to K, so set res to True.
- Otherwise, set res to False.
- Print the result.
Python3
test_dict = {"Gfg" : 5, "is" : 8, "Best" : 10, "for" : 8, "Geeks" : 8}
print("The original dictionary is : " + str(test_dict))
cust_keys = ["is", "for", "Geeks"]
K = 8
res = all(filter(lambda x: test_dict[x] == K, cust_keys))
print("Are all custom keys equal to K : " + str(res))
|
Output
The original dictionary is : {'Gfg': 5, 'is': 8, 'Best': 10, 'for': 8, 'Geeks': 8}
Are all custom keys equal to K : True
Time Complexity: O(m), where m is the number of keys in the test_dict dictionary.
Auxiliary Space: O(1), as no extra space is required.
Method #5 : Using count() and len() methods:
Approach:
- Create an empty list x.
- Initiate a for loop to append the values of custom keys to a list x.
- Check whether the count of K in x is equal to the length of x(using count()). If yes, then print “Are all custom keys equal to K: True“, Otherwise, print “Are all custom keys equal to K: False“.
Below is the implementation of the above approach:
Python3
test_dict = {"Gfg": 5, "is": 8, "Best": 10, "for": 8, "Geeks": 8}
print("The original dictionary is : " + str(test_dict))
cust_keys = ["is", "for", "Geeks"]
K = 8
x = []
for i in cust_keys:
x.append(test_dict[i])
res = x.count(K) == len(x)
print("Are all custom keys equal to K : " + str(res))
|
Output
The original dictionary is : {'Gfg': 5, 'is': 8, 'Best': 10, 'for': 8, 'Geeks': 8}
Are all custom keys equal to K : True
Time Complexity: O(N) N – length of cust_keys
Auxiliary Space: O(1)
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