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
# Python3 code to demonstrate# Maximum of each Column# initializing listtest_list = [[3, 7, 6], [1, 3, 5], [9, 3, 2]]# printing original listprint("The original list : " + str(test_list))# Maximum of each Columnres=[]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 resultprint("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
# Python3 code to demonstrate# Maximum of each Column# using max() + list comprehension + zip()# initializing listtest_list = [[3, 7, 6], [1, 3, 5], [9, 3, 2]]# printing original listprint("The original list : " + str(test_list))# using max() + list comprehension + zip()# Maximum of each Columnres = [max(idx) for idx in zip(*test_list)]# print resultprint("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
# 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.
Python3
#Importing numpy libraryimport numpy as np#Initializing listtest_list = [[3, 7, 6], [1, 3, 5], [9, 3, 2]]#Converting list to 2-D numpy arrayarr = np.array(test_list)#Finding maximum of each columnresult = np.amax(arr, axis=0)#Printing resultprint("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.
