Python Program that prints the rows of a given length from a matrix

Last Updated : 28 Apr, 2023

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 list
test_list = [[3, 4, 5, 6], [1, 4, 6], [2], [2, 3, 4, 5, 6], [7, 3, 1]]
 
# printing original list
print("The original list is : " + str(test_list))
 
# initializing K
K = 3
 
# list comprehension is used for extracting K len rows
res = [sub for sub in test_list if len(sub) == K]
 
# printing result
print("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 list
test_list = [[3, 4, 5, 6], [1, 4, 6], [2], [2, 3, 4, 5, 6], [7, 3, 1]]
 
# printing original list
print("The original list is : " + str(test_list))
 
# initializing K
K = 3
 
# filter() + lambda to filter out rows of len K
res = list(filter(lambda sub: len(sub) == K, test_list))
 
# printing result
print("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 list
test_list = [[3, 4, 5, 6], [1, 4, 6], [2], [2, 3, 4, 5, 6], [7, 3, 1]]
 
# printing original list
print("The original list is : " + str(test_list))
 
# initializing K
K = 3
 
# using for loop to filter out rows of len K
res = []
for sub in test_list:
    if len(sub) == K:
        res.append(sub)
 
# printing result
print("The filtered rows : " + str(res))
#This code is contributed by Vinay Pinjala.
 
  

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 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 list
test_list = [[3, 4, 5, 6], [1, 4, 6], [2], [2, 3, 4, 5, 6], [7, 3, 1]]
 
# printing original list
print("The original list is : " + str(test_list))
 
# initializing K
K = 3
 
# using map() and lambda to filter out rows of len K
res = list(filter(lambda x: len(x) == K, test_list))
 
# printing result
print("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]]

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.

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