Python中序列中重复次数第二多的单词

给定一个字符串序列，任务是找出给定序列中第二重复（或频繁）的字符串。 （考虑到没有两个词是第二多重复的，总会有一个词）。
例子：

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

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

此问题已有解决方案，请参考序列中重复次数第二多的单词链接。我们可以在Python中使用 Counter(iterator) 方法快速解决这个问题。
方法很简单——

1. 使用Counter(iterator)方法创建一个字典，其中包含单词作为键，频率作为值。
2. 现在获取字典中所有值的列表并按降序对其进行排序。从排序列表中选择第二个元素，因为它将是第二大的。
3. 现在再次遍历字典并打印其值等于第二大元素的键。

Python3

# 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)

输出：

bbb

替代实施：

Python3

# returns the second most repeated word
from collections import Counter
class Solution:
    def secFrequent(self, arr, n):
        all_freq = dict(Counter(arr))
        store = []
        for w in sorted(all_freq, key=all_freq.get):
            # if add key=all_freq.get will sort according to values
            # without key=all_freq.get will sort according to keys
            if w not in store:
                store.append(w)
             
        return store[-2]
 # driver code or main function
if __name__ == '__main__':
    t = int(input())
    for _ in range(t):
        n = int(input().strip())
        arr = input().strip().split(" ")
        ob = Solution()
        ans = ob.secFrequent(arr,n)
        print(ans)
     
     
contributed by Pratyush Pratap Singh

输出

bbb