Python – Filter rows with Elements as Multiple of K

Given a Matrix, extract rows with elements multiple of K.

Input : test_list = [[5, 10, 15], [4, 8, 12], [100, 15], [5, 10, 23]], K = 4 
Output : [[4, 8, 12]] 
Explanation : All are multiples of 4.

Input : test_list = [[5, 10, 15], [4, 8, 11], [100, 15], [5, 10, 23]], K = 4 
Output : [] 
Explanation : No rows with all multiples of 4. 

Method #1 : Using list comprehension + all()

In this, we check for all elements to be multiple using all(), and iteration of rows occur using list comprehension.

Python3

# Python3 code to demonstrate working of
# Access element at Kth index in String
# Using list comprehension + all()

# initializing string list
test_list = [[5, 10, 15], [4, 8, 3], [100, 15], [5, 10, 23]]

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

# initializing K
K = 5

res = [sub for sub in test_list if all(ele % K == 0 for ele in sub)]

# printing result
print("Rows with K multiples : " + str(res))

Output

The original list is : [[5, 10, 15], [4, 8, 3], [100, 15], [5, 10, 23]]
Rows with K multiples : [[5, 10, 15], [100, 15]]

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 + all()

In this, we perform task of filtering using filter() and lambda function and task of checking for all elements in rows using all().

Python3

# Python3 code to demonstrate working of
# Access element at Kth index in String
# Using filter() + lambda + all()

# initializing string list
test_list = [[5, 10, 15], [4, 8, 3], [100, 15], [5, 10, 23]]

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

# initializing K
K = 5

# using all() to check for all elements being multiples of K
res = list(filter(lambda sub : all(ele % K == 0 for ele in sub), test_list))

# printing result
print("Rows with K multiples : " + str(res))

Output

The original list is : [[5, 10, 15], [4, 8, 3], [100, 15], [5, 10, 23]]
Rows with K multiples : [[5, 10, 15], [100, 15]]

Method #3:Using itertools.filterfalse() method

Python3

# Python3 code to demonstrate working of
# Access element at Kth index in String
import itertools

# initializing string list
test_list = [[5, 10, 15], [4, 8, 3], [100, 15], [5, 10, 23]]

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

# initializing K
K = 5

# using all() to check for all elements being multiples of K
res = list(itertools.filterfalse(lambda sub : not all(ele % K == 0 for ele in sub), test_list))

# printing result
print("Rows with K multiples : " + str(res))

Output

The original list is : [[5, 10, 15], [4, 8, 3], [100, 15], [5, 10, 23]]
Rows with K multiples : [[5, 10, 15], [100, 15]]

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

Method #4:Using the reduce() function and a lambda function with the all() function:

Algorithm :

1. Import the reduce() function from the functools module and define the test_list and K variables.
2. Use a lambda function with the filter() function to iterate through each sublist of test_list and check if all its elements are divisible by K.
3. Print the result.

Python3

from functools import reduce
# initializing string list
test_list = [[5, 10, 15], [4, 8, 3], [100, 15], [5, 10, 23]]
# printing original list
print("The original list is : " + str(test_list))

K = 5
res = list(filter(lambda x: reduce(lambda a, b: a and b, [i % K == 0 for i in x]), test_list))
# printing result
print("Rows with K multiples : " + str(res))
#This code is contributed by Jyothi pinjala.

Output

The original list is : [[5, 10, 15], [4, 8, 3], [100, 15], [5, 10, 23]]
Rows with K multiples : [[5, 10, 15], [100, 15]]

The time complexity : O(n*m), where n is the number of sublists in test_list and m is the maximum length of a sublist. This is because the filter() function and the lambda function inside it are applied to each sublist.

The space complexity : O(k), where k is the number of sublists that satisfy the condition of having all their elements divisible by K. This is because a new list is created to store the filtered sublists, and the maximum size of this list is k.