Python program to remove rows with duplicate element in Matrix

Given Matrix, remove all rows which have duplicate elements in them.

Input : test_list = [[4, 3, 2], [7, 6, 7], [2, 4, 4], [8, 9, 9]] 
Output : [[4, 3, 2]] 
Explanation : [4, 3, 2] is the only unique row.

Input : test_list = [[4, 3, 3, 2], [7, 6, 7], [2, 4, 4], [8, 9, 9]] 
Output : [] 
Explanation : No unique row. 

Method 1 : Using list comprehension + set() + len()

In this, we extract only the rows which remain in the same length after converting it into a set.

Python3

# initializing list
test_list = [[4, 3, 2], [7, 6, 7], [2, 4, 5], [8, 9, 9]]

# printing original list
print("The original list is : " + str(test_list))

# set() removing all elements
# list comprehension used to filter
res = [sub for sub in test_list if len(set(sub)) == len(sub)]

# printing result
print("Rows after removal : " + str(res))

Output

The original list is : [[4, 3, 2], [7, 6, 7], [2, 4, 5], [8, 9, 9]]
Rows after removal : [[4, 3, 2], [2, 4, 5]]

Time Complexity: O(n) where n is the number of elements in the list “test_list”. 
Auxiliary Space: O(1) additional space is not needed.

Method #2 : Using filter() + lambda + set() + len()

In this, we perform the task of filtering using filter() + lambda function, and set and len() are used to check.

Python3

# initializing list
test_list = [[4, 3, 2], [7, 6, 7], [2, 4, 5], [8, 9, 9]]

# printing original list
print("The original list is : " + str(test_list))

# set() removing all elements
# filter() used to filter
res = list(filter(lambda ele: len(set(ele)) == len(ele), test_list))

# printing result
print("Rows after removal : " + str(res))

Output

The original list is : [[4, 3, 2], [7, 6, 7], [2, 4, 5], [8, 9, 9]]
Rows after removal : [[4, 3, 2], [2, 4, 5]]

Method #3: Using Counter() function

Python3

from collections import Counter
# initializing list
test_list = [[4, 3, 2], [7, 6, 7], [2, 4, 5], [8, 9, 9]]

# printing original list
print("The original list is : " + str(test_list))

for i in test_list.copy():
  freq = Counter(i)
  if(len(freq)!=len(i)):
    test_list.remove(i)


# printing result
print("Rows after removal : " + str(test_list))

Output

The original list is : [[4, 3, 2], [7, 6, 7], [2, 4, 5], [8, 9, 9]]
Rows after removal : [[4, 3, 2], [2, 4, 5]]

Time Complexity:O(N*N)
Auxiliary Space :O(N)

Method #4: Using for loop and an if condition

Python3

# initializing list
test_list = [[4, 3, 2], [7, 6, 7], [2, 4, 5], [8, 9, 9]]

# printing original list
print("The original list is : " + str(test_list))

res = []
for sub in test_list:
    if len(set(sub)) == len(sub):
        res.append(sub)

# printing result
print("Rows after removal : " + str(res))
#This code is contributed by Vinay Pinjala.

Output

The original list is : [[4, 3, 2], [7, 6, 7], [2, 4, 5], [8, 9, 9]]
Rows after removal : [[4, 3, 2], [2, 4, 5]]

Time Complexity:O(N*NlogN)
Auxiliary Space :O(N)

Method#5: Using Recursive method.

Algorithm:

1. Check if the list is empty. If it is, return an empty list.
2. Check if the first sub-list in the list has duplicate elements. If it doesn’t, include it in the result list and call the function recursively with the remaining sub-lists.
3. If it does have duplicate elements, skip it and call the function recursively with the remaining sub-lists.
4. Return the result list.

Python3

def remove_duplicate_rows(test_list):
    if not test_list:
        return []

    # check if the first sub-list has duplicate elements
    if len(set(test_list[0])) == len(test_list[0]):
        # if not, include it in the result and call the function recursively with the remaining sub-lists
        return [test_list[0]] + remove_duplicate_rows(test_list[1:])
    else:
        # if it has duplicate elements, skip it and call the function recursively with the remaining sub-lists
        return remove_duplicate_rows(test_list[1:])

# initializing list
test_list = [[4, 3, 2], [7, 6, 7], [2, 4, 5], [8, 9, 9]]

# printing original list
print("The original list is : " + str(test_list))

res = remove_duplicate_rows(test_list)

# printing result
print("Rows after removal : " + str(res))
#This code is contributed by tvsk.

Output

The original list is : [[4, 3, 2], [7, 6, 7], [2, 4, 5], [8, 9, 9]]
Rows after removal : [[4, 3, 2], [2, 4, 5]]

Time complexity: O(n^2), where n is the total number of elements in the list. This is because we need to compare each sub-list with every other sub-list to determine if it has duplicate elements.
Auxiliary Space: O(n), where n is the total number of elements in the list. This is because we are creating a new list to store the non-duplicate sub-lists.

Method 6: Using the built-in all() function:

Step-by-step approach:

1. Initialize an empty list to store the rows that don’t have duplicate elements.
2. Use a list comprehension to iterate over each row of the input list.
3. For each row, create a set and check if the length of the set is equal to the length of the row using the all() function.
   • If all() elements in the row are unique, append the row to the list from step 1.
4. Print the resulting list.

Below is the implementation of the above approach:

Python3

# initializing list
test_list = [[4, 3, 2], [7, 6, 7], [2, 4, 5], [8, 9, 9]]

# printing original list
print("The original list is : " + str(test_list))

# using all() function to check if all elements are unique
res = [sub for sub in test_list if all(sub.count(elem) == 1 for elem in sub)]

# printing result
print("Rows after removal : " + str(res))

Output

The original list is : [[4, 3, 2], [7, 6, 7], [2, 4, 5], [8, 9, 9]]
Rows after removal : [[4, 3, 2], [2, 4, 5]]

Time complexity: O(n*m), where n is the number of rows and m is the maximum length of a row.
Auxiliary space: O(k), where k is the number of rows that don’t have duplicate elements.