Given a Matrix, the following articles shows how to extract all the rows with a specified length.
Input : test_list = [[3, 4, 5, 6], [1, 4, 6], [2], [2, 3, 4, 5, 6], [7, 3, 1]], K = 3
Output : [[1, 4, 6], [7, 3, 1]]
Explanation : Extracted lists have length of 3.
Input : test_list = [[3, 4, 5, 6], [1, 4, 6], [2], [2, 3, 4, 5, 6], [7, 3, 1]], K = 4
Output : [[3, 4, 5, 6]]
Explanation : Extracted lists have length of 4.
Method 1 : Using list comprehension and len()
In this, we perform the task of getting length using len() and list comprehension does the task of filtering all the rows which have a specified length.
Python3
# initializing listtest_list = [[3, 4, 5, 6], [1, 4, 6], [2], [2, 3, 4, 5, 6], [7, 3, 1]]# printing original listprint("The original list is : " + str(test_list))# initializing KK = 3# list comprehension is used for extracting K len rowsres = [sub for sub in test_list if len(sub) == K]# printing resultprint("The filtered rows : " + str(res)) |
Output:
The original list is : [[3, 4, 5, 6], [1, 4, 6], [2], [2, 3, 4, 5, 6], [7, 3, 1]]
The filtered rows : [[1, 4, 6], [7, 3, 1]]
Time Complexity: O(n*m)
Auxiliary Space: O(k)
Method 2 : Using filter(), lambda and len()
In this, we perform the task of filtering using filter() and lambda. len() is used for finding length of rows.
Python3
# initializing listtest_list = [[3, 4, 5, 6], [1, 4, 6], [2], [2, 3, 4, 5, 6], [7, 3, 1]]# printing original listprint("The original list is : " + str(test_list))# initializing KK = 3# filter() + lambda to filter out rows of len Kres = list(filter(lambda sub: len(sub) == K, test_list))# printing resultprint("The filtered rows : " + str(res)) |
Output:
The original list is : [[3, 4, 5, 6], [1, 4, 6], [2], [2, 3, 4, 5, 6], [7, 3, 1]]
The filtered rows : [[1, 4, 6], [7, 3, 1]]
Time Complexity: O(n)
Auxiliary Space: O(n)
Method 3: Using a for loop
Python3
# initializing listtest_list = [[3, 4, 5, 6], [1, 4, 6], [2], [2, 3, 4, 5, 6], [7, 3, 1]]# printing original listprint("The original list is : " + str(test_list))# initializing KK = 3# using for loop to filter out rows of len Kres = []for sub in test_list: if len(sub) == K: res.append(sub)# printing resultprint("The filtered rows : " + str(res))#This code is contributed by Vinay Pinjala. |
The original list is : [[3, 4, 5, 6], [1, 4, 6], [2], [2, 3, 4, 5, 6], [7, 3, 1]] The filtered rows : [[1, 4, 6], [7, 3, 1]]
Time Complexity: O(n)
Auxiliary Space: O(n)
Method 4: Using the map() function
we can use the map() function along with a lambda function to create a new list containing only the sublists that have length K.
Python3
# initializing listtest_list = [[3, 4, 5, 6], [1, 4, 6], [2], [2, 3, 4, 5, 6], [7, 3, 1]]# printing original listprint("The original list is : " + str(test_list))# initializing KK = 3# using map() and lambda to filter out rows of len Kres = list(filter(lambda x: len(x) == K, test_list))# printing resultprint("The filtered rows : " + str(res)) |
The original list is : [[3, 4, 5, 6], [1, 4, 6], [2], [2, 3, 4, 5, 6], [7, 3, 1]] The filtered rows : [[1, 4, 6], [7, 3, 1]]
The time complexity of this method is also O(n), where n is the number of sublists in the original list.
The auxiliary space required for this method is O(1), as it does not create any new lists.
