Python | Test list element similarity

Last Updated : 23 Apr, 2023

Given a list, your task is to determine the list is K percent same i.e has element that is populated more than K % times.
Given below are few methods to solve the task.

Method #1 Using collections.Counter

Python3

# Python3 code to demonstrate
# to check whether the list
# K percent same or not
from collections import Counter
 
# initializing list
ini_list1 = [1, 2, 3, 1, 1, 1, 1, 1, 3, 2]
 
# printing initial list
print ("Initial list", ini_list1)
 
# initializing K
K = 60
 
# code to check whether list is K % same or not
i, freq = Counter(ini_list1).most_common(1)[0]
 
if len(ini_list1)*(K / 100) <= freq:
    print("True")
else:
    print("False")

Output:

Initial list [1, 2, 3, 1, 1, 1, 1, 1, 3, 2] 
True

Method #2: Using dictionary and its values

Python3

# Python3 code to demonstrate
# to check whether the list
# K percent same or not
from collections import Counter, defaultdict
 
# initializing list
ini_list1 = [1, 2, 3, 1, 1, 1, 1, 1, 3, 2]
 
# printing initial list
print ("Initial list", ini_list1)
 
# initializing K
K = 60
 
# code to check whether list is K % same or not
freq = defaultdict(int)
for x in ini_list1:
    freq[x] += 1
freq = freq.values()
if max(freq) >= (K / 100) * sum(freq):
    print ("True")
else:
    print ("False")

Output:

initial list [1, 2, 3, 1, 1, 1, 1, 1, 1, 1]
True

Method #3: Using List comprehension and set()

Python3

# Python3 code to demonstrate
# to check whether the list
# K percent same or not
   
# initializing list
ini_list1 = [1, 2, 3, 1, 1, 1, 1, 1, 3, 2]
   
# printing initial list
print ("Initial list", ini_list1)
   
# initializing K
K = 60
   
# code to check whether list is K % same or not
count = len([x for x in set(ini_list1) if ini_list1.count(x) >= (K/100) * len(ini_list1)])
 
if count > 0:
    print("True")
else:
    print("False")
#This code is contributed by Edula Vinay Kumar Reddy

Output

Initial list [1, 2, 3, 1, 1, 1, 1, 1, 3, 2]
True

In method #3, we use list comprehension and set() to check if the given list is K percent same i.e has element that is populated more than K % times.

First, the program creates a set of all the unique elements in the list, then using list comprehension, it counts the number of occurrences of each unique element in the original list. The program then checks if the count of any of the unique elements is greater than or equal to (K/100) * len(ini_list1), this check is performed for all the unique elements in the list. If the count of any of the elements is greater than the required value, the program returns True, else it returns False.

Method #4: Using a for loop:

Python3

def is_k_percent_same(lst, k):
    count = {}
    for num in lst:
        if num in count:
            count[num] += 1
        else:
            count[num] = 1
    most_common = max(count.values())
    if most_common >= len(lst) * (k/100):
        return True
    else:
        return False
ini_list1 = [1, 2, 3, 1, 1, 1, 1, 1, 3, 2]
# printing initial list
print ("Initial list", ini_list1)
K = 60
print(is_k_percent_same(ini_list1, K))
#This code is contributed by Jyothi pinjala.

Output

Initial list [1, 2, 3, 1, 1, 1, 1, 1, 3, 2]
True

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

Method #5: Using numpy.bincount()

Import the numpy module.
Pass the list ‘lst’ to numpy.bincount() to obtain the counts of each element in ‘lst’.
Get the maximum count from the bincount array.
Calculate the threshold value (len(lst) * (k/100)).
Check if the maximum count is greater than or equal to the threshold value.
Return True if the condition is satisfied, False otherwise.

Python3

import numpy as np
 
def is_k_percent_same(lst, k):
    bincount = np.bincount(lst)
    most_common = np.max(bincount)
    threshold = len(lst) * (k/100)
    if most_common >= threshold:
        return True
    else:
        return False
 
ini_list1 = [1, 2, 3, 1, 1, 1, 1, 1, 3, 2]
# printing initial list
print ("Initial list", ini_list1)
K = 60
print(is_k_percent_same(ini_list1, K))

OUTPUT :
Initial list [1, 2, 3, 1, 1, 1, 1, 1, 3, 2]
True

Time Complexity: O(n)
Auxiliary Space: O(max(lst)) (for creating the bincount array)

Python | Lowercase first character of String

manjeet_04

Improve

Article Tags :

Practice Tags :

python

Python | Test list element similarity

Python3

Python3

Python3

Python3

Python3

Similar Reads

Thank You!

What kind of Experience do you want to share?