Python | Transpose elements of two dimensional list

Given a two-dimensional list of integers, write a Python program to get the transpose of given list of lists.
In Python, a matrix can be interpreted as a list of lists. Each element is treated as a row of the matrix. For example m = [[10, 20], [40, 50], [30, 60]] represents a matrix of 3 rows and 2 columns. First element of the list – m[0] and element in first row, first column – m[0][0].

Example:

Input :  l1 = [[4, 5, 3, 9],
               [7, 1, 8, 2],
               [5, 6, 4, 7]]

Output : lt = [[4, 7, 5],
               [5, 1, 6],
               [3, 8, 4],
               [9, 2, 7]]

Method #1: Using loops 

Python

# Python program to get transpose
# elements of two dimension list
def transpose(l1, l2):

    # iterate over list l1 to the length of an item
    for i in range(len(l1[0])):
        # print(i)
        row =[]
        for item in l1:
            # appending to new list with values and index positions
            # i contains index position and item contains values
            row.append(item[i])
        l2.append(row)
    return l2

# Driver code
l1 = [[4, 5, 3, 9], [7, 1, 8, 2], [5, 6, 4, 7]]
l2 = []
print(transpose(l1, l2))

Output

[[4, 7, 5], [5, 1, 6], [3, 8, 4], [9, 2, 7]]

Time Complexity: O(n*n) where n is the number of elements in the dictionary. The loops is used to perform the task and it takes O(n*n) time.
Auxiliary Space: O(1) constant additional space is required

Method #2: Using List comprehensions 

Python

# Python program to get transpose
# elements of two dimension list
def transpose(l1, l2):

    # we have nested loops in comprehensions
    # value of i is assigned using inner loop
    # then value of item is directed by row[i]
    # and appended to l2
    l2 =[[row[i] for row in l1] for i in range(len(l1[0]))]
    return l2

# Driver code
l1 = [[4, 5, 3, 9], [7, 1, 8, 2], [5, 6, 4, 7]]
l2 = []
print(transpose(l1, l2))

Output

[[4, 7, 5], [5, 1, 6], [3, 8, 4], [9, 2, 7]]

Time complexity: The nested loop used in the list comprehension has a time complexity of O(n^2) where n is the size of the input list
Auxiliary space: The space used in the function is the space required to create a new list l2, which is the size of the input list l1 in reverse order. 

Method #3: Using numpy 

Python3

# Python program to get transpose
# elements of two dimension list
import numpy

l1= [[4, 5, 3, 9], [7, 1, 8, 2], [5, 6, 4, 7]]
print(numpy.transpose(l1))

Output:

[[4 7 5]
 [5 1 6]
 [3 8 4]
 [9 2 7]]

Method #4: Using zip function 

Python

# Python program to get transpose
# elements of two dimension list
def transpose(l1, l2):

    # star operator will first
    # unpack the values of 2D list
    # and then zip function will
    # pack them again in opposite manner
    l2 = list(map(list, zip(*l1)))
    return l2

# Driver code
l1 = [[4, 5, 3, 9], [7, 1, 8, 2], [5, 6, 4, 7]]
l2 = []
print(transpose(l1, l2))

# code contributed by
# chaudhary_19
# Mayank Chaudhary
# modified by Chirag Shilwant

Output

[[4, 7, 5], [5, 1, 6], [3, 8, 4], [9, 2, 7]]