Python – 提取频率大于 K 的元素

给定一个 List，提取所有频率大于 K 的元素。

Input : test_list = [4, 6, 4, 3, 3, 4, 3, 4, 3, 8], K = 3
Output : [4, 3]
Explanation : Both elements occur 4 times.

Input : test_list = [4, 6, 4, 3, 3, 4, 3, 4, 6, 6], K = 2
Output : [4, 3, 6]
Explanation : Occur 4, 3, and 3 times respectively.

方法 #1：使用 count() + 循环

在此，我们使用 count() 来获取频率，并使用循环来迭代 List 的每个元素。

Python3

# Python3 code to demonstrate working of 
# Extract elements with Frequency greater than K
# Using count() + loop
  
# initializing list
test_list = [4, 6, 4, 3, 3, 4, 3, 7, 8, 8]
  
# printing string
print("The original list : " + str(test_list))
  
# initializing K 
K = 2
  
res = [] 
for i in test_list: 
      
    # using count() to get count of elements
    freq = test_list.count(i) 
      
    # checking if not already entered in results
    if freq > K and i not in res: 
        res.append(i)
  
# printing results 
print("The required elements : " + str(res))

输出

The original list : [4, 6, 4, 3, 3, 4, 3, 7, 8, 8]
The required elements : [4, 3]

方法 #2：使用列表理解 + Counter()

在此，我们使用 Counter() 执行计数任务，迭代部分在列表推导中完成。

Python3

# Python3 code to demonstrate working of 
# Extract elements with Frequency greater than K
# Using list comprehension + Counter()
from collections import Counter
  
# initializing list
test_list = [4, 6, 4, 3, 3, 4, 3, 7, 8, 8]
  
# printing string
print("The original list : " + str(test_list))
  
# initializing K 
K = 2
  
# using list comprehension to bind result
res = [ele for ele, cnt in Counter(test_list).items() if cnt > K]
  
# printing results 
print("The required elements : " + str(res))

输出

The original list : [4, 6, 4, 3, 3, 4, 3, 7, 8, 8]
The required elements : [4, 3]