Python | Remove last element from each row in Matrix
Last Updated :
12 Apr, 2023
Sometimes, while working with Matrix data, we can have a stray element attached at rear end of each row of matrix. This can be undesired at times and wished to be removed. Let’s discuss certain ways in which this task can be performed.
Method #1: Using loop + del + list slicing The combination of the above functionalities can be used to perform this task. In this, we run a loop for each row in the matrix and remove the rear element using del.
Python3
test_list = [[1, 3, 4], [2, 4, 6], [3, 8, 1]]
print("The original list : " + str(test_list))
for ele in test_list:
del ele[-1]
print("Matrix after removal of rear element from rows : " + str(test_list))
|
Output :
The original list : [[1, 3, 4], [2, 4, 6], [3, 8, 1]]
Matrix after removal of rear element from rows : [[1, 3], [2, 4], [3, 8]]
Time complexity: O(n*m).
Auxiliary space: O(1).
Method #2: Using list comprehension + list slicing The combination of the above functions can also be used to perform this task. In this, we just iterate for each row and delete the rear element using list slicing.
Python3
test_list = [[1, 3, 4], [2, 4, 6], [3, 8, 1]]
print("The original list : " + str(test_list))
res = [ele[:-1] for ele in test_list]
print("Matrix after removal of rear element from rows : " + str(res))
|
Output :
The original list : [[1, 3, 4], [2, 4, 6], [3, 8, 1]]
Matrix after removal of rear element from rows : [[1, 3], [2, 4], [3, 8]]
Time complexity: O(nm), where n is the number of rows and m is the number of columns in the matrix.
Auxiliary space: O(nm), as a new matrix is created to store the modified rows.
Method #3 : Using pop() method
Python3
test_list = [[1, 3, 4], [2, 4, 6], [3, 8, 1]]
print("The original list : " + str(test_list))
res=[]
for i in test_list:
i.pop()
res.append(i)
print("Matrix after removal of rear element from rows : " + str(res))
|
Output
The original list : [[1, 3, 4], [2, 4, 6], [3, 8, 1]]
Matrix after removal of rear element from rows : [[1, 3], [2, 4], [3, 8]]
Time complexity: O(nm), where n is the number of rows and m is the number of columns in the matrix.
Auxiliary space: O(nm), as a new matrix is created to store the modified rows.
Method #4: Using map() function
The code uses the map() function to remove the last element from each row in the matrix. map() function applies a given function to all the items of an input_list, and returns a list containing all the results.
Here, the lambda function lambda x: x[:-1] is applied to each row of the matrix (which is represented by a nested list) using map() function. The lambda function takes a single argument x, which is a row of the matrix and returns the same row but with the last element removed, by using slicing x[:-1]. This way, all the rows of the matrix are processed by the lambda function and a new list is returned by the map() function, which contains all the rows with the last element removed.
The time complexity of this method is O(n), where n is the total number of elements in the matrix. This is because the map() function iterates over all the elements of the matrix and applies the lambda function to each element, which takes constant time.
The Auxiliary space of this method is O(n) where n is the total number of elements in the matrix. This is because a new list is returned by the map() function, which is equal to the size of the input matrix.
Python3
test_list = [[1, 3, 4], [2, 4, 6], [3, 8, 1]]
print("The original list : " + str(test_list))
res = list(map(lambda x: x[:-1], test_list))
print("Matrix after removal of rear element from rows : " + str(res))
|
Output
The original list : [[1, 3, 4], [2, 4, 6], [3, 8, 1]]
Matrix after removal of rear element from rows : [[1, 3], [2, 4], [3, 8]]
Method#5: Using Recursive method.
The remove_last_elem_recursive function takes a matrix as input, along with an optional index variable i (which is initially set to 0). The function removes the last element from the row at index i and then recursively calls itself with an updated value of i.
Python3
def remove_last_elem_recursive(matrix, i=0):
if i == len(matrix):
return
del matrix[i][-1]
remove_last_elem_recursive(matrix, i+1)
test_list = [[1, 3, 4], [2, 4, 6], [3, 8, 1]]
print("The original list : " + str(test_list))
remove_last_elem_recursive(test_list)
print("Matrix after removal of rear element from rows : " + str(test_list))
|
Output
The original list : [[1, 3, 4], [2, 4, 6], [3, 8, 1]]
Matrix after removal of rear element from rows : [[1, 3], [2, 4], [3, 8]]
The time complexity of this recursive method is also O(n), where n is the total number of elements in the matrix. This is because the function visits each element of the matrix exactly once.
The auxiliary space is O(n) as well, because the function modifies the matrix in place and does not create any additional data structures.
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