List python 2.7中列表之间公共元素的频率

我有两份清单：

list1 = ['red','blue', 'yellow'] #unique

list2 = ['green','purple','red','yellow','purple', 'red'] #duplicates

['red', 2],['blue',0],['yellow',1]

我找到了多个Q&A，用于查找两个列表之间的唯一公共元素或列表中元素的频率：

import numpy as np

match  = np.in1d(list1,list2, assume_unique=False) 

match = list(set(list1).intersection(set(list2)))

match = set(list1) & set(list2)

[element for element in list1 if element in list2]

list2, common = list1[:], [ e for e in list1 if e in list2 and (list2.pop(list2.index(e)))]

from scipy.stats import itemfreq

count = itemfreq(res)

list.count

但我需要的是两个列表之间公共元素（如果有）的频率：

list1 = ['red','blue', 'yellow'] #unique

list2 = ['green','purple','red','yellow','purple', 'red'] #duplicates

['red', 2],['blue',0],['yellow',1]

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一些理解可以简明扼要地做到这一点：

list1 = [1,2,3]
list2 = [3,4,5,3,3]

itemcounter = {item:count for item, count in [(item, list2.count(item)) for item in set(list1)]}
final_counts = [(item, count) for item, count in itemcounter.iteritems()]

print(final_counts)

…产生了以下结果：

[(1, 0), (2, 0), (3, 3)]

---

Python 2.7有一个计数器类，可用于：

from collections import Counter

list1 = ['red', 'blue', 'yellow']
list2 = ['green', 'purple', 'red', 'yellow', 'red']
set_list1 = set(list1)
list2_in_list1 = [x for x in list2 if x in set_list1]
count = Counter(list2_in_list1)
print count
# gives: Counter({'red': 2, 'yellow': 1})

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将列表1转换为集合是出于性能原因（检查集合中的项目比检查列表快）；对于短列表，您可以跳过它。

您可以使用列表中可用的count语句

list1 = ['red','blue', 'yellow']
list2 = ['green','purple','red','yellow','purple', 'red']
for color in list1:
    count = list2.count(color)
    print [color, count],

这将打印出以下内容：

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琼尼的的确更快，但斯科特的更简洁