Python | Count occurrences of an element in a list

Given a list in Python and a number x, count number of occurrences of x in the given list.

Examples:

Input : lst = [15, 6, 7, 10, 12, 20, 10, 28, 10]
         x = 10
Output : 3 
10 appears three times in given list.

Input : lst = [8, 6, 8, 10, 8, 20, 10, 8, 8]
        x = 16
Output : 0

Method 1 (Simple approach)
We keep a counter that keeps on increasing if the esquired element is found in the list.

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# Python code to count the number of occurrences
def countX(lst, x):
    count = 0
    for ele in lst:
        if (ele == x):
            count = count + 1
    return count
  
# Driver Code
lst = [8, 6, 8, 10, 8, 20, 10, 8, 8]
x = 8
print('{} has occurred {} times'.format(x, countX(lst, x)))

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Output:
8 has occurred 5 times



Method 2 (Using count())

The idea is to use list method count() to count number of occurrences.











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# Python code to count the number of occurrences 


def countX(lst, x): 


    return lst.count(x) 


  


# Driver Code 


lst = [8, 6, 8, 10, 8, 20, 10, 8, 8] 


x = 8


print('{} has occurred {} times'.format(x, countX(lst, x))) 



















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Output:
8 has occurred 5 times 



Method 2 (Using Counter())

Counter method returns a dictionary with occurrences of all elements as a key-value pair, where key is the element and value is the number of times that element has occurred.











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from collections import Counter 


  


# declaring the list 


l = [1, 1, 2, 2, 3, 3, 4, 4, 5, 5] 


  


# driver program 


x = 3


d = Counter(l) 


print('{} has occurred {} times'.format(x, d[x])) 



















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Output:
3 has occurred 2 times






    

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