Python Program to assign each list element value equal to its magnitude order

0 0 2 minutes read

Given a list, the task is to write a Python program to assign each element value equal to its magnitude order.

Input : test_list = [8, 3, 5, 8, 1, 5, 4]

Output : [4, 1, 3, 4, 0, 3, 2]
Explanation : 8 is 5th maximum hence 4 [starting from 0th index] is assigned to it.

Input : test_list = [3, 2, 0]
Output : [2, 1, 0]
Explanation : 3 is 3rd maximum as 2 is assigned.

Method 1 : Using set() + zip() + dict() + list comprehension

In this, we perform task of mapping sorted index of set converted list to values using zip(), and then convert to dictionary of value mapped to values. Then list comprehension is used to ordering in list.

Python3

# Python3 code to demonstrate working of
# Relative Size Ordering 
# Using set() + zip() + sorted() + dict() + list comprehension
 
# initializing list
test_list = [8, 3, 5, 8, 1, 5, 4]
              
# printing original list
print("The original list is : " + str(test_list))
 
# assigning order to each value
ord_dict = dict(zip(list(set(test_list)),
                    range(len(set(test_list)))))
 
# mapping element with ordered value
res = [ord_dict[ele] for ele in test_list]
 
# printing result
print("Relative size ordered list : " + str(res))

Output:

The original list is : [8, 3, 5, 8, 1, 5, 4]
Relative size ordered list : [4, 1, 3, 4, 0, 3, 2]

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

Method 2 : Using sorted() + set() + index() + list comprehension

In this, we perform task of conversion to set and then sort, index mapping to order is done using index() and list comprehension.

Python3

# Python3 code to demonstrate working of
# Relative Size Ordering 
# Using sorted() + set() + index() + list comprehension
 
# initializing list
test_list = [8, 3, 5, 8, 1, 5, 4]
              
# printing original list
print("The original list is : " + str(test_list))
 
# getting order
ord_dict = list(set(test_list))
 
# mapping index
res = [ord_dict.index(ele) for ele in test_list]
 
# printing result
print("Relative size ordered list : " + str(res))

Output:

The original list is : [8, 3, 5, 8, 1, 5, 4]
Relative size ordered list : [4, 1, 3, 4, 0, 3, 2]

Time Complexity: O(n*nlogn) where n is the number of elements in the list “test_list”. The sorted() + set() + index() + list comprehension is used to perform the task and it takes O(n*nlogn) time.
Auxiliary Space: O(n) additional space of size O(n) is required