Python program to find tuples which have all elements divisible by K from a list of tuples

Last Updated : 21 Mar, 2023

Given a list of tuples. The task is to extract all tuples which have all elements divisible by K.

Input : test_list = [(6, 24, 12), (60, 12, 6), (12, 18, 21)], K = 6
Output : [(6, 24, 12), (60, 12, 6)]
Explanation : Both tuples have all elements multiple of 6.

Input : test_list = [(6, 24, 12), (60, 10, 5), (12, 18, 21)], K = 5
Output : [(60, 10, 5)]
Explanation : Multiple of 5 tuples extracted.

Method #1 : Using list comprehension + all()

In this, we test for all elements to be K multiple using all() for filtering purpose, and list comprehension is used for the iteration of all tuples.

Python3

# Python3 code to demonstrate working of
# K Multiple Elements Tuples
# Using list comprehension + all()
 
# initializing list
test_list = [(6, 24, 12), (7, 9, 6), (12, 18, 21)]
 
# printing original list
print("The original list is : " + str(test_list))
 
# initializing K
K = 6
 
# all() used to filter elements
res = [sub for sub in test_list if all(ele % K == 0 for ele in sub)]
 
# printing result
print("K Multiple elements tuples : " + str(res))

Output

The original list is : [(6, 24, 12), (7, 9, 6), (12, 18, 21)]
K Multiple elements tuples : [(6, 24, 12)]

Time Complexity: O(n*m)
Auxiliary Space: O(n)

Method #2 : Using filter() + lambda + all()

Step-by-step approach:

In this, perform task of filtering using filter() + lambda, and logic provided by all().
The program initializes a list called test_list containing three tuples, each with three integer values.
The program prints the original list by concatenating a string with the string representation of test_list.
The program initializes an integer variable K.
The program uses the filter() function to filter the tuples in test_list based on the condition that all elements in the tuple are divisible by K.
The program defines a lambda function that takes a sublist (tuple) as an argument and returns True if all elements in the sublist are divisible by K.
The filter() function iterates through each tuple in test_list, applies the lambda function to each tuple, and keeps the tuples that satisfy the condition.
The res variable stores the filtered tuples in a list.
Prints the result by concatenating a string with the string representation of res.

Below is the implementation of the above approach:

Python3

# Python3 code to demonstrate working of
# K Multiple Elements Tuples
# Using filter() + lambda + all()
 
# initializing list
test_list = [(6, 24, 12), (7, 9, 6), (12, 18, 21)]
 
# printing original list
print("The original list is : " + str(test_list))
 
# initializing K
K = 6
 
# filter() + lambda for filter operation
res = list(filter(lambda sub: all(ele % K == 0 for ele in sub), test_list))
 
# printing result
print("K Multiple elements tuples : " + str(res))

Output

The original list is : [(6, 24, 12), (7, 9, 6), (12, 18, 21)]
K Multiple elements tuples : [(6, 24, 12)]

Time Complexity: O(n*m)
Auxiliary Space: O(n)

Method #3 : Using list comprehension + not any()

In this, we test for all elements to be K multiple using not any () for filtering purpose, and list comprehension is used for iteration of all tuples.

Python3

# Initialize list
test_list = [(6, 24, 12), (7, 9, 6), (12, 18, 21)]
 
# Initialize K
K = 6
 
# list comprehension
result = [tup for tup in test_list if not any(x % K != 0 for x in tup)]
# printing original list
print("The original list is : " + str(test_list))
# Print the list of tuples whose elements are multiples of K
print("K Multiple elements tuples : " + str(result))

Output

The original list is : [(6, 24, 12), (7, 9, 6), (12, 18, 21)]
K Multiple elements tuples : [(6, 24, 12)]

Time Complexity: O(n*m), where n is the number of tuples in the list, and m is the number of elements in each tuple.
Auxiliary Space: O(n), where n is the number of tuples in the list.

Method #4: Using for loops

Python3

# Python3 code to demonstrate working of
# K Multiple Elements Tuples
 
# initializing list
test_list = [(6, 24, 12), (7, 9, 6), (12, 18, 21)]
 
# printing original list
print("The original list is : " + str(test_list))
 
# initializing K
K = 6
 
res = []
for i in test_list:
    c=0
    for j in i:
        if(j%K==0):
            c+=1
    if(c==len(i)):
        res.append(i)
 
# printing result
print("K Multiple elements tuples : " + str(res))

Output

The original list is : [(6, 24, 12), (7, 9, 6), (12, 18, 21)]
K Multiple elements tuples : [(6, 24, 12)]

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

Method #5: Using Recursive method.

Python3

# Python3 code to demonstrate working of
# K Multiple Elements Tuples
 
def even_tuple(lst,K,newlst=[],start=0):
  if start==len(lst):
    return newlst
  for i in lst[start]:
    if i%K!=0:
      return even_tuple(lst,K,newlst,start+1)
  else:
    newlst.append(lst[start])
  return even_tuple(lst,K,newlst,start+1)
# initializing list
test_list = [(6, 24, 12), (7, 9, 6), (12, 18, 21)]
 
# printing original list
print("The original list is : " + str(test_list))
 
# initializing K
K = 6
 
res = even_tuple(test_list,K)
 
# printing result
print("K Multiple elements tuples : " + str(res))
#this code contributed by tvsk

Output

The original list is : [(6, 24, 12), (7, 9, 6), (12, 18, 21)]
K Multiple elements tuples : [(6, 24, 12)]

Time Complexity: O(n*m), where n is length of the list and m is length of each tuple.
Auxiliary Space: O(n)

Method 6: Using the filter() function, without using lambda and all()

The implementation defines a function is_multiple_of_K() that checks if all elements in a tuple are multiples of K, and then applies the filter() function to keep only the tuples that satisfy this condition. Finally, it converts the filtered result into a list and prints it.

Python3

test_list = [(6, 24, 12), (7, 9, 6), (12, 18, 21)]
K = 6
 
# define a function to check if all elements in a tuple are multiples of K
def is_multiple_of_K(tup):
    return all(ele % K == 0 for ele in tup)
 
# use filter() to keep only tuples that satisfy the condition
res = list(filter(is_multiple_of_K, test_list))
 
# printing original list
print("The original list is : " + str(test_list))
 
print("K Multiple elements tuples : " + str(res))

Output

The original list is : [(6, 24, 12), (7, 9, 6), (12, 18, 21)]
K Multiple elements tuples : [(6, 24, 12)]

Time complexity: O(n), where n is the number of tuples in the test_list.
Auxiliary space: O(1), as the function only uses a constant amount of space to store the K value.

Method #7: Using numpy array and np.all()

Python3

import numpy as np
 
# initializing the list
test_list = [(6, 24, 12), (7, 9, 6), (12, 18, 21)]
 
# printing the original list
print("The original list is : " + str(test_list))
 
# initializing K
K = 6
 
# converting the list to a numpy array
arr = np.array(test_list)
 
# using np.all() to filter elements
res = arr[np.all(arr % K == 0, axis=1)]
 
# printing the result
print("K Multiple elements tuples : " + str(res))

Output:

K Multiple elements tuples : [[ 6 24 12]]

Time Complexity: O(nm), where n is the number of tuples in the input list, and m is the number of elements in each tuple.
Auxiliary Space: O(nm)