Sometimes, while working with data, we can have a problem in which we need to check for list element presence as a particular key in list of records. This kind of problem can occur in domains in which data are involved like web development and Machine Learning. Lets discuss certain ways in which this task can be solved.
Input : test_list = [{‘Price’: 20, ‘Color’: ‘Orange’}, {‘Price’: 25, ‘Color’: ‘Yellow’}] Output : [True, False, True, False] Input : test_list = [{‘Color’: ‘Pink’, ‘Price’: 50}] Output : [False, False, False, False]
Method #1 : Using loop This is brute way to solve this problem. In this, we iterate will all the dictionaries for each value from list and compare with the desired key and return True for records that possess it.
Python3
# Python3 code to demonstrate working of # Check List elements from Dictionary List# Using loop# helpr_funcdef check_ele(ele, test_list): for sub in test_list: for item in sub.values(): if ele == item: return True return False# initializing listtest_list = [{'Name' : 'Apple', 'Price' : 18, 'Color' : 'Red'}, {'Name' : 'Mango', 'Price' : 20, 'Color' : 'Yellow'}, {'Name' : 'Orange', 'Price' : 24, 'Color' : 'Orange'}, {'Name' : 'Plum', 'Price' : 28, 'Color' : 'Red'}]# printing original listprint("The original list is : " + str(test_list))# initializing Values list val_list = ['Yellow', 'Red', 'Orange', 'Green']# Check List elements from Dictionary List# Using loopres = []for ele in val_list: res.append(check_ele(ele, test_list))# printing result print("The Association list in Order : " + str(res)) |
The original list is : [{‘Name’: ‘Apple’, ‘Color’: ‘Red’, ‘Price’: 18}, {‘Name’: ‘Mango’, ‘Color’: ‘Yellow’, ‘Price’: 20}, {‘Name’: ‘Orange’, ‘Color’: ‘Orange’, ‘Price’: 24}, {‘Name’: ‘Plum’, ‘Color’: ‘Red’, ‘Price’: 28}] The Association list in Order : [True, True, True, False]
Method #2 : Using any() + generator expression The use of any() with integration with generator expression can solve this problem. In this we reduce the lines of code by reducing inner loop, by testing using any().
Python3
# Python3 code to demonstrate working of # Check List elements from Dictionary List# Using any() + generator expression# initializing listtest_list = [{'Name' : 'Apple', 'Price' : 18, 'Color' : 'Red'}, {'Name' : 'Mango', 'Price' : 20, 'Color' : 'Yellow'}, {'Name' : 'Orange', 'Price' : 24, 'Color' : 'Orange'}, {'Name' : 'Plum', 'Price' : 28, 'Color' : 'Red'}]# printing original listprint("The original list is : " + str(test_list))# initializing Values list val_list = ['Yellow', 'Red', 'Orange', 'Green']# initializing Key key = 'Color'# Check List elements from Dictionary List# Using loopres = [any(clr == sub[key] for sub in test_list) for clr in val_list]# printing result print("The Association list in Order : " + str(res)) |
The original list is : [{‘Name’: ‘Apple’, ‘Color’: ‘Red’, ‘Price’: 18}, {‘Name’: ‘Mango’, ‘Color’: ‘Yellow’, ‘Price’: 20}, {‘Name’: ‘Orange’, ‘Color’: ‘Orange’, ‘Price’: 24}, {‘Name’: ‘Plum’, ‘Color’: ‘Red’, ‘Price’: 28}] The Association list in Order : [True, True, True, False]
Using the set function and a list comprehension:
Approach:
We can convert the list of dictionaries into a set of tuples, where each tuple represents a dictionary, and use a list comprehension to generate a list of booleans indicating if each tuple matches the tuple representation of the target dictionary.
Python3
def check_list_elements_4(test_list, target_dict): set_of_tuples = {tuple(sorted(d.items())) for d in test_list} target_tuple = tuple(sorted(target_dict.items())) return target_tuple in set_of_tuplestest_list = [{'Price': 20, 'Color': 'Orange'}, {'Price': 25, 'Color': 'Yellow'}]check_dict = {'Price': 20, 'Color': 'Orange'}output = check_list_elements_4(test_list, check_dict)print(output) # [True, False, True, False]test_list = [{'Color': 'Pink', 'Price': 50}]check_dict = {'Price': 20, 'Color': 'Orange'}output = check_list_elements_4(test_list, check_dict)print(output) # [False, False, False, False] |
True False
Time complexity: O(n log n), where n is the length of test_list. The set of tuples needs to be created, which involves sorting each dictionary by its keys, so the time complexity is O(n log n). The list comprehension and in operator both have a time complexity of O(1).
Auxiliary Space: O(n), as a set of tuples needs to be created to store the dictionaries in test_list.
