Python | Find most frequent element in a list
Last Updated :
03 May, 2023
Given a list, find the most frequent element in it. If there are multiple elements that appear maximum number of times, print any one of them.
Examples:
Input : [2, 1, 2, 2, 1, 3]
Output : 2
Input : ['Dog', 'Cat', 'Dog']
Output : Dog
Approach #1 : Naive Approach
This is a brute force approach in which we make use of for loop to count the frequency of each element. If the current frequency is greater than the previous frequency, update the counter and store the element.
Python3
def most_frequent(List):
counter = 0
num = List[0]
for i in List:
curr_frequency = List.count(i)
if(curr_frequency> counter):
counter = curr_frequency
num = i
return num
List = [2, 1, 2, 2, 1, 3]
print(most_frequent(List))
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Approach #2 : Pythonic Naive approach
Make a set of the list so that the duplicate elements are deleted. Then find the highest count of occurrences of each element in the set and thus, we find the maximum out of it.
Python3
def most_frequent(List):
return max(set(List), key = List.count)
List = [2, 1, 2, 2, 1, 3]
print(most_frequent(List))
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Approach #3 : Using Counter
Make use of Python Counter which returns count of each element in the list. Thus, we simply find the most common element by using most_common() method.
Python3
from collections import Counter
def most_frequent(List):
occurence_count = Counter(List)
return occurence_count.most_common(1)[0][0]
List = [2, 1, 2, 2, 1, 3]
print(most_frequent(List))
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Approach #4 : By finding mode
Finding most frequent element means finding mode of the list. Hence, we use mode method from statistics.
Python3
import statistics
from statistics import mode
def most_common(List):
return(mode(List))
List = [2, 1, 2, 2, 1, 3]
print(most_common(List))
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Approach #5 : Using Python dictionary
Use python dictionary to save element as a key and its frequency as the value, and thus find the most frequent element.
Python3
def most_frequent(List):
dict = {}
count, itm = 0, ''
for item in reversed(List):
dict[item] = dict.get(item, 0) + 1
if dict[item] >= count :
count, itm = dict[item], item
return(itm)
List = [2, 1, 2, 2, 1, 3]
print(most_frequent(List))
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Approach #6 : Using pandas library.
Incase of multiple values getting repeated. Print all values.
Python3
import pandas as pd
List = [2, 1, 2, 2, 1, 3, 1]
df=pd.DataFrame({'Number': List})
df1 = pd.DataFrame(data=df['Number'].value_counts(), columns=[['Number','Count']])
df1['Count']=df1['Number'].index
list(df1[df1['Number']==df1.Number.max()]['Count'])
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Approach #7: Using numpy library
Note: Install numpy module using command “pip install numpy”
Use numpy library to create an array of unique elements and their corresponding counts. Then, find the index of the maximum count and return the corresponding element from the unique array.
Python
import numpy as np
def most_frequent(List):
unique, counts = np.unique(List, return_counts=True)
index = np.argmax(counts)
return unique[index]
List = [2, 1, 2, 2, 1, 3]
print(most_frequent(List))
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Output:
2
Time complexity: O(n) for numpy unique function and argmax
Space complexity: O(n) for storing unique values
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