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

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.

# 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.

# 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

# 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.

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.

# 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))

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.