Python – 其他列表中元素的频率

给定两个列表，计算列表 2 的元素在列表 1 中出现的频率。

Input : test_list1 = [4, 6, 8, 9], test_list2 = [4, 6, 6, 5, 8, 10, 4, 9, 8, 10, 1]
Output : {4: 2, 6: 2, 8: 2, 9: 1}
Explanation : 4 occurs 2 times, 6 occurs 2 times in second list and so on.

Input : test_list1 = [4, 6, 8, 9], test_list2 = [4, 6, 5, 8, 10, 4, 9, 8, 10, 1]
Output : {4: 2, 6: 1, 8: 2, 9: 1}
Explanation : 4 occurs 2 times, 6 occurs 1 time in second list and so on.

方法 #1：使用字典理解 + count()

这是可以执行此任务的方式之一。在此，count() 用于提取列表 2 中元素的计数，字典理解用于编译结果并遍历列表 1 值。

Python3

# Python3 code to demonstrate working of 
# Frequency of elements from other list
# Using dictionary comprehension + count()
  
# initializing lists
test_list1 = [4, 6, 8, 9, 10]
test_list2 = [4, 6, 6, 5, 8, 10, 4, 9, 8, 10, 1]
  
# printing original lists
print("The original list 1 : " + str(test_list1))
print("The original list 2 : " + str(test_list2))
  
# count() used to perform count.
# returns 0 is no occurrence.
# ignores different new elements in list 2
res = {key : test_list2.count(key) for key in test_list1}
  
# printing result 
print("Lists elements Frequency : " + str(res))

输出

The original list 1 : [4, 6, 8, 9, 10]
The original list 2 : [4, 6, 6, 5, 8, 10, 4, 9, 8, 10, 1]
Lists elements Frequency : {4: 2, 6: 2, 8: 2, 9: 1, 10: 2}

方法 #2：使用 Counter() + setdefault() + 循环

这是可以执行此任务的另一种方式。在此，我们使用 Counter() 计算频率并使用循环分配值。对于不存在的值，setdefault() 用于将计数器初始化为 0。

Python3

# Python3 code to demonstrate working of 
# Frequency of elements from other list
# Using Counter() + setdefault() + loop
from collections import Counter
  
# initializing lists
test_list1 = [4, 6, 8, 9, 10]
test_list2 = [4, 6, 6, 5, 8, 10, 4, 9, 8, 10, 1]
  
# printing original lists
print("The original list 1 : " + str(test_list1))
print("The original list 2 : " + str(test_list2))
  
# Counter() used to perform count on list 2
res = dict(Counter(test_list2))
  
# filtering only list 1 keys
res = {key : val for key, val in res.items() if key in test_list1} 
  
for key in test_list1:
      
    # initializing elements not present in list 2 of list 1 to 0
    res.setdefault(key, 0) 
      
  
# printing result 
print("Lists elements Frequency : " + str(res))

输出

The original list 1 : [4, 6, 8, 9, 10]
The original list 2 : [4, 6, 6, 5, 8, 10, 4, 9, 8, 10, 1]
Lists elements Frequency : {4: 2, 6: 2, 8: 2, 10: 2, 9: 1}