Python Program that filters out non-empty rows of a matrix

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Given Matrix, the following article shows how to filter all the Non-Empty rows of a matrix. In simpler terms, the codes provided below return a matrix after removing empty rows from it.

Input : test_list = [[4, 5, 6, 7], [], [], [9, 8, 1], []] 
Output : [[4, 5, 6, 7], [9, 8, 1]] 
Explanation : All empty rows are removed.

Input : test_list = [[4, 5, 6, 7], [], [9, 8, 1], []] 
Output : [[4, 5, 6, 7], [9, 8, 1]] 
Explanation : All empty rows are removed.

Method 1: Using list comprehension and len()

In this we check each row for its length, if its length is greater than 0 then that row is added to result.

Python3

# initializing list
test_list = [[4, 5, 6, 7], [], [], [9, 8, 1], []]
 
# printing original lists
print("The original list is : " + str(test_list))
 
# checking for row lengths using len()
res = [row for row in test_list if len(row) > 0]
 
# printing result
print("Filtered Matrix " + str(res))

Output

The original list is : [[4, 5, 6, 7], [], [], [9, 8, 1], []]
Filtered Matrix [[4, 5, 6, 7], [9, 8, 1]]

Time Complexity: O(n*m)
Auxiliary Space: O(k)

Method 2 : Using filter(), lambda and len()

In this, we filter rows w.r.t lengths using filter() and lambda function. The len() is used to get the length.

Python3

# Initializing list
test_list = [[4, 5, 6, 7], [], [], [9, 8, 1], []]
 
# Printing original lists
print("The original list is : " + str(test_list))
 
# Checking for row lengths 
# using len() filtering using filter() + lambda
res = list(filter(lambda row: len(row) > 0, test_list))
 
# Printing result
print("Filtered Matrix " + str(res))

Output

The original list is : [[4, 5, 6, 7], [], [], [9, 8, 1], []]
Filtered Matrix [[4, 5, 6, 7], [9, 8, 1]]

Time Complexity: O(n) where n is the number of elements in the list “test_list”. The time complexity of the filter() and lambda function is O(n)
Auxiliary Space: O(1), no extra space is required

Method 3 : Using find() method

Python3

#Filter non empty rows
 
# Initializing list
test_list = [[4, 5, 6, 7], [], [], [9, 8, 1], []]
 
# Printing original lists
print("The original list is : " + str(test_list))
 
# Empty list 
res=[]
 
for i in test_list:
    if str(i).find("[]")==-1:
        res.append(i)
 
# Printing result
print("Filtered Matrix " + str(res))

Output

The original list is : [[4, 5, 6, 7], [], [], [9, 8, 1], []]
Filtered Matrix [[4, 5, 6, 7], [9, 8, 1]]

Method 4 : Using remove() method

Python3

#Python Program that filters out non-empty rows of a matrix
# initializing list
test_list = [[4, 5, 6, 7], [], [], [9, 8, 1], []]
 
# printing original lists
print("The original list is : " + str(test_list))
 
while([] in test_list):
    test_list.remove([])
# printing result
print("Filtered Matrix " + str(test_list))

Output

The original list is : [[4, 5, 6, 7], [], [], [9, 8, 1], []]
Filtered Matrix [[4, 5, 6, 7], [9, 8, 1]]

Method 5 : Using pop() method

Python3

# Initializing list
test_list = [[4, 5, 6, 7], [], [], [9, 8, 1], []]
 
# Printing original lists
print("The original list is : " + str(test_list))
 
# Removing empty rows using while loop
i = 0
while i < len(test_list):
    if not test_list[i]:
        test_list.pop(i)
    else:
        i += 1
 
# Printing result
print("Filtered Matrix: " + str(test_list))

Output

The original list is : [[4, 5, 6, 7], [], [], [9, 8, 1], []]
Filtered Matrix: [[4, 5, 6, 7], [9, 8, 1]]

Time complexity: O(nm), where n is the number of rows and m is the maximum number of elements in a row.
Auxiliary space O(1), since we are modifying the original list in place without creating any additional data structures.