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Second most repeated word in a sequence in Python
  • Difficulty Level : Basic
  • Last Updated : 19 Nov, 2020

Given a sequence of strings, the task is to find out the second most repeated (or frequent) string in the given sequence. (Considering no two words are the second most repeated, there will be always a single word).

Examples:

Input : {"aaa", "bbb", "ccc", "bbb", 
         "aaa", "aaa"}
Output : bbb

Input : {"geeks", "for", "geeks", "for", 
          "geeks", "aaa"}
Output : for

This problem has existing solution please refer Second most repeated word in a sequence link. We can solve this problem quickly in Python using Counter(iterator) method.

Approach is very simple –



  1. Create a dictionary using Counter(iterator) method which contains words as keys and it’s frequency as value.
  2. Now get a list of all values in dictionary and sort it in descending order. Choose second element from the sorted list because it will be the second largest.
  3. Now traverse dictionary again and print key whose value is equal to second largest element.




# Python code to print Second most repeated 
# word in a sequence in Python 
from collections import Counter 
    
def secondFrequent(input): 
    
    # Convert given list into dictionary 
    # it's output will be like {'ccc':1,'aaa':3,'bbb':2} 
    dict = Counter(input
    
    # Get the list of all values and sort it in ascending order 
    value = sorted(dict.values(), reverse=True
    
    # Pick second largest element 
    secondLarge = value[1
    
    # Traverse dictionary and print key whose 
    # value is equal to second large element 
    for (key, val) in dict.items(): 
        if val == secondLarge: 
            print (key) 
            return
    
# Driver program 
if __name__ == "__main__"
    input = ['aaa','bbb','ccc','bbb','aaa','aaa'
    secondFrequent(input)


Output:

bbb


Alternate Implementation :




# Second most repeated word in a sequence in Python
  
def secondFrequent(input):
    from collections import Counter
  
    # this sorts from most common to least common to least common
    c = Counter(input
       
    # c.most_common()[1] prints ('bbb',2) 
    # c.most_common()[1][0] prints output: bbb
    print(c.most_common()[1][0])
  
# Driver code
input = ['aaa','bbb','ccc','bbb','aaa','aaa']
secondFrequent(input)


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