Given a dictionary, extract the largest N dictionaries keys in descending order.
Input : test_dict = {6 : 2, 8: 9, 3: 9, 10: 8}, N = 3
Output : [10, 8, 6]
Explanation : Max. N keys extracted in descending order.Input : test_dict = {6 : 2, 8: 9, 3: 9, 10: 8}, N = 2
Output : [10, 8]
Explanation : Max. N keys extracted in descending order.
Method #1 : Using sorted() + lambda + reverse
The combination of the above functions can be used to solve this problem. In this, we perform sort by keys using sorted() + lambda and reverse is used to perform reverse sort for getting required ordering.
Python3
# Python3 code to demonstrate working of # Extract top N Dictionaries by Key # Using sorted() + lambda + reverse # initializing dictionary test_dict = {6 : 2, 8: 9, 3: 9, 10: 8} # printing original dictionary print("The original dictionary is : " + str(test_dict)) # initializing N N = 4 res = [] # 0 in lambda used for keys, list sliced till N for top N values for key, val in sorted(test_dict.items(), key = lambda x: x[0], reverse = True)[:N]: res.append(key) # printing result print("Top N keys are: " + str(res)) |
The original dictionary is : {6: 1, 8: 9, 3: 9, 10: 8}
Top N keys are: [10, 8, 6, 3]
Method #2 : Using nlargest() + lambda
This is yet another way in which this task can be performed. In this, maximum values are extracted using nlargest(), and lambda function is used to drive the keys extraction and comparison logic.
Python3
# Python3 code to demonstrate working of # Extract top N Dictionaries by Key # Using nlargest() + lambda from heapq import nlargest # initializing dictionary test_dict = {6 : 2, 8: 9, 3: 9, 6: 1, 10: 8} # printing original dictionary print("The original dictionary is : " + str(test_dict)) # initializing N N = 4 res = [] # Using nlargest() to get maximum keys for key, val in nlargest(N, test_dict.items(), key = lambda ele: ele[0]): res.append(key) # printing result print("Top N keys are: " + str(res)) |
The original dictionary is : {6: 1, 8: 9, 3: 9, 10: 8}
Top N keys are: [10, 8, 6, 3]
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