Sometimes, we are encountered such problems in which we need to find the maximum of each column in a matrix i.e maximum of each index in list of lists. This kind of problem is quite common and useful in competitive programming. Let’s discuss certain ways in which this problem can be solved.

Method #1: Using for loop+ max()

# Python3 code to demonstrate
# Maximum of each Column

# initializing list
test_list = [[3, 7, 6], [1, 3, 5], [9, 3, 2]]

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


# Maximum of each Column
res=[]
for i in range(0,len(test_list)):
    x=[]
    for j in range(0,len(test_list[i])):
        a=test_list[j][i]
        x.append(a)
    res.append(max(x))
        
# print result
print("The Maximum of each index list is : " + str(res))

# Python3 code to demonstrate

# Maximum of each Column

​

# initializing list

test_list = [[3, 7, 6], [1, 3, 5], [9, 3, 2]]

​

# printing original list

print("The original list : " + str(test_list))

​

​

# Maximum of each Column

res=[]

for i in range(0,len(test_list)):

    x=[]

    for j in range(0,len(test_list[i])):

        a=test_list[j][i]

        x.append(a)

    res.append(max(x))

# print result

print("The Maximum of each index list is : " + str(res))

The original list : [[3, 7, 6], [1, 3, 5], [9, 3, 2]]
The Maximum of each index list is : [9, 7, 6]

Time Complexity: O(n),The above code iterates through the list once, hence the time complexity is linear, i.e. O(n).
Space Complexity: O(n),The algorithm uses an additional list to store the result, thus consuming linear space which is O(n).

Method #2: Using max() + list comprehension + zip()
The combination of above methods are required to solve this particular problem. The max function is used to get the required maximum value and zip function provides the combination of like indices and then list is created using list comprehension. 

# Python3 code to demonstrate
# Maximum of each Column
# using max() + list comprehension + zip()

# initializing list
test_list = [[3, 7, 6], [1, 3, 5], [9, 3, 2]]

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

# using max() + list comprehension + zip()
# Maximum of each Column
res = [max(idx) for idx in zip(*test_list)]

# print result
print("The Maximum of each index list is : " + str(res))

# Python3 code to demonstrate

# Maximum of each Column

# using max() + list comprehension + zip()

​

# initializing list

test_list = [[3, 7, 6], [1, 3, 5], [9, 3, 2]]

​

# printing original list

print("The original list : " + str(test_list))

​

# using max() + list comprehension + zip()

# Maximum of each Column

res = [max(idx) for idx in zip(*test_list)]

​

# print result

print("The Maximum of each index list is : " + str(res))

The original list : [[3, 7, 6], [1, 3, 5], [9, 3, 2]]
The Maximum of each index list is : [9, 7, 6]

Time Complexity: O(n*n), where n is the number of elements in the list “test_list”.
Auxiliary Space: O(n), where n is the number of elements in the list “test_list”.

Method #3: Using map() + max() + zip()
This works in an almost similar way as the above method, but the difference is just that we use map function to build the max element list rather than using list comprehension. 

# Python3 code to demonstrate 
# Maximum index value 
# using max() + map() + zip() 

# initializing list 
test_list = [[3, 7, 6], [1, 3, 5], [9, 3, 2]] 

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

# using max() + map() + zip() 
# Maximum index value 
res = list(map(max, zip(*test_list))) 

# print result 
print("The Maximum of each index list is : " + str(res)) 

# Python3 code to demonstrate 

# Maximum index value 

# using max() + map() + zip() 

​

# initializing list 

test_list = [[3, 7, 6], [1, 3, 5], [9, 3, 2]] 

​

# printing original list 

print("The original list : " + str(test_list)) 

​

# using max() + map() + zip() 

# Maximum index value 

res = list(map(max, zip(*test_list))) 

​

# print result 

print("The Maximum of each index list is : " + str(res))

The original list : [[3, 7, 6], [1, 3, 5], [9, 3, 2]]
The Maximum of each index list is : [9, 7, 6]

Time Complexity: O(n*n), where n is the number of elements in the list “test_list”.
Auxiliary Space: O(n), where n is the number of elements in the list “test_list”.

Method #4: Using numpy()

Note: Install numpy module using command "pip install numpy"

Another approach to find the maximum of each column in a matrix can be using the numpy library.

#Importing numpy library
import numpy as np

#Initializing list
test_list = [[3, 7, 6], [1, 3, 5], [9, 3, 2]]

#Converting list to 2-D numpy array
arr = np.array(test_list)

#Finding maximum of each column
result = np.amax(arr, axis=0)

#Printing result
print("The Maximum of each index list is:", result)

#This code is contributed by Edula Vinay Kumar Reddy

#Importing numpy library

import numpy as np

​

#Initializing list

test_list = [[3, 7, 6], [1, 3, 5], [9, 3, 2]]

​

#Converting list to 2-D numpy array

arr = np.array(test_list)

​

#Finding maximum of each column

result = np.amax(arr, axis=0)

​

#Printing result

print("The Maximum of each index list is:", result)

​

#This code is contributed by Edula Vinay Kumar Reddy

Output:

The Maximum of each index list is: [9 7 6]
 

Time Complexity: O(m * n), where m is the number of rows and n is the number of columns in the matrix.
Auxiliary Space: O(n), where n is the number of elements in the result array.

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