Python | Difference in Record Lists
Last Updated :
05 May, 2023
Sometimes, while working with data, we may have a problem in which we require to find the difference records between two lists that we receive. This is a very common problem and records usually occurs as a tuple. Let’s discuss certain ways in which this problem can be solved.
Method #1 : Using list comprehension
List comprehension can be opted as method to perform this task in one line rather than running a loop to find the difference elements. In this, we just iterate for single list and check if any element occurs in other one.
Python3
# Python3 code to demonstrate working of
# Difference in Record Lists
# Using list comprehension
# Initializing lists
test_list1 = [('gfg', 1), ('is', 2), ('best', 3)]
test_list2 = [('i', 3), ('love', 4), ('gfg', 1)]
# Printing original lists
print("The original list 1 is : " + str(test_list1))
print("The original list 2 is : " + str(test_list2))
# Intersection in Tuple Records Data
# Using list comprehension
res = [ele1 for ele1 in test_list1 for ele2 in test_list2 if ele1 == ele2]
res = list(set(res) ^ set(test_list1))
# Printing result
print("The difference of data records is : " + str(res))
Output : The original list 1 is : [('gfg', 1), ('is', 2), ('best', 3)]
The original list 2 is : [('i', 3), ('love', 4), ('gfg', 1)]
The difference of data records is : [('best', 3), ('is', 2)] Time Complexity: O(n*n), where n is the number of elements in the list “test_list”.
Auxiliary Space: O(n*n), where n is the number of elements in the list “test_list”.
Method #2: Using set.intersection() + set operations
This task can also be performed in a smaller way using the generic set intersection. In this, we first convert the list of records to a set and then perform its intersection using intersection(). Then the result is computed by taking the uncommon element of the result from the first list.
Python3
# Python3 code to demonstrate working of
# Difference in Record Lists
# Using set.intersection() + set operations
# Initializing lists
test_list1 = [('gfg', 1), ('is', 2), ('best', 3)]
test_list2 = [('i', 3), ('love', 4), ('gfg', 1)]
# printing original lists
print("The original list 1 is : " + str(test_list1))
print("The original list 2 is : " + str(test_list2))
# Difference in Record Lists
# Using set.intersection() + set operations
res = list(set(test_list1).intersection(set(test_list2)))
res = list(set(res) ^ set(test_list1))
# printing result
print("The difference of data records is : " + str(res))
Output : The original list 1 is : [('gfg', 1), ('is', 2), ('best', 3)]
The original list 2 is : [('i', 3), ('love', 4), ('gfg', 1)]
The difference of data records is : [('best', 3), ('is', 2)]
Method #3: Here is another approach to finding the difference between two lists of records, using the filter() method.
Python3
# Python3 code to demonstrate working of
# Difference in Record Lists
# Using filter() method
# Initializing lists
test_list1 = [('gfg', 1), ('is', 2), ('best', 3)]
test_list2 = [('i', 3), ('love', 4), ('gfg', 1)]
# printing original lists
print("The original list 1 is : " + str(test_list1))
print("The original list 2 is : " + str(test_list2))
# Difference in Record Lists
# Using filter() method
res = list(filter(lambda x: x not in set(test_list2), set(test_list1)))
# printing result
print("The difference of data records is : " + str(res))
OutputThe original list 1 is : [('gfg', 1), ('is', 2), ('best', 3)]
The original list 2 is : [('i', 3), ('love', 4), ('gfg', 1)]
The difference of data records is : [('best', 3), ('is', 2)]
Time Complexity: O(n), where n is the length of the test_list1
Auxiliary Space: O(n), as a new list res of length n is created.
Method 4: use a for loop and conditional statements.
Steps:
- Initialize two lists of records.
- Create an empty list to store the difference.
- Use a for loop to iterate through each record in the first list.
- Use a nested for loop to iterate through each record in the second list.
- Check if the record in the first list matches the record in the second list.
- If the records match, break out of the nested loop and move to the next record in the first list.
- If the records do not match, append the record to the difference list and move to the next record in the first list.
- Print the difference list.
Python3
# Python3 code to demonstrate working of
# Difference in Record Lists
# Using for loop and conditional statements
# Initializing lists
test_list1 = [('gfg', 1), ('is', 2), ('best', 3)]
test_list2 = [('i', 3), ('love', 4), ('gfg', 1)]
# printing original lists
print("The original list 1 is : " + str(test_list1))
print("The original list 2 is : " + str(test_list2))
# Difference in Record Lists
# Using for loop and conditional statements
diff_list = []
for record1 in test_list1:
found_match = False
for record2 in test_list2:
if record1 == record2:
found_match = True
break
if not found_match:
diff_list.append(record1)
# printing result
print("The difference of data records is : " + str(diff_list))
OutputThe original list 1 is : [('gfg', 1), ('is', 2), ('best', 3)]
The original list 2 is : [('i', 3), ('love', 4), ('gfg', 1)]
The difference of data records is : [('is', 2), ('best', 3)]
Time Complexity: O(n*m), where n and m are the lengths of the two lists respectively.
Auxiliary Space: O(k), where k is the length of the difference list.
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