In Python, sometimes we require to get only some of the dictionary required to solve the problem. This problem is quite common in web development that we require to get only the selective dictionary satisfying some given criteria. Let’s discuss certain ways in which this problem can be solved.
Method #1: Using list comprehension
Python3
# Python3 code to demonstrate # filtering of a list of dictionary # on basis of condition # initialising list of dictionary ini_list = [{'a':1, 'b':3, 'c':7}, {'a':3, 'b':8, 'c':17}, {'a':78, 'b':12, 'c':13}, {'a':2, 'b':2, 'c':2}] # printing initial list of dictionary print ("initial_list", str(ini_list)) # code to filter list # where c is greater than 10 res = [d for d in ini_list if d['c'] > 10] # printing result print ("resultant_list", str(res)) |
initial_list [{'a': 1, 'b': 3, 'c': 7}, {'a': 3, 'b': 8, 'c': 17}, {'a': 78, 'b': 12, 'c': 13}, {'a': 2, 'b': 2, 'c': 2}]
resultant_list [{'a': 3, 'b': 8, 'c': 17}, {'a': 78, 'b': 12, 'c': 13}]
Time complexity: O(n), where n is the number of key-value pairs in the dictionary.
Auxiliary space: O(n), to store the keys and values in dictionary.
Method #2: Using lambda and filter
Python3
# Python3 code to demonstrate # filtering of list of dictionary # on basis of condition # initialising list of dictionary ini_list = [{'a':1, 'b':3, 'c':7}, {'a':3, 'b':8, 'c':17}, {'a':78, 'b':12, 'c':13}, {'a':2, 'b':2, 'c':2}] # printing initial list of dictionary print ("initial_list", str(ini_list)) # code to filter list # where c is less than 10 res = list(filter(lambda x:x["c"] > 10, ini_list )) # printing result print ("resultant_list", str(res)) |
initial_list [{'a': 1, 'b': 3, 'c': 7}, {'a': 3, 'b': 8, 'c': 17}, {'a': 78, 'b': 12, 'c': 13}, {'a': 2, 'b': 2, 'c': 2}]
resultant_list [{'a': 3, 'b': 8, 'c': 17}, {'a': 78, 'b': 12, 'c': 13}]
Method #3: Using dict comprehension and list comprehension
Python3
# Python3 code to demonstrate # filtering of list of dictionary # on basis of condition # initialising list of dictionary ini_list = [{'a':1, 'b':3, 'c':7}, {'a':3, 'b':8, 'c':17}, {'a':78, 'b':12, 'c':13}, {'a':2, 'b':2, 'c':10}] # printing initial list of dictionary print ("initial_list", str(ini_list)) # code to filter list # where c is more than 10 res = [{ k:v for (k, v) in i.items()} for i in ini_list if i.get('c') > 10] # printing result print ("resultant_list", str(res)) |
initial_list [{'a': 1, 'b': 3, 'c': 7}, {'a': 3, 'b': 8, 'c': 17}, {'a': 78, 'b': 12, 'c': 13}, {'a': 2, 'b': 2, 'c': 10}]
resultant_list [{'a': 3, 'b': 8, 'c': 17}, {'a': 78, 'b': 12, 'c': 13}]
Time Complexity: The time complexity of this program is O(n), where n is the number of dictionaries in ini_list.
Auxiliary Space: The auxiliary space used by this program is O(n), where n is the number of dictionaries in ini_list.
Using a for loop:
Approach:
- Create a new empty dictionary
- Iterate over the key-value pairs in the original dictionary using a for loop
- If the key is greater than k, add the key-value pair to the new dictionary
- Return the new dictionary
Python3
d = {'a': 1, 'b': 2, 'c': 3, 'd': 4}k = 'c'new_d = {}for key, value in d.items(): if key > k: new_d[key] = valueprint(new_d) |
{'d': 4}
Time complexity: O(n), where n is the number of key-value pairs in the original dictionary
Space complexity: O(n), where n is the number of key-value pairs in the new dictionary
