python - 比较两个列表中常见项目的最快方法

Question

我有两个这样的列表:

listt = [["a","abc","zzz","xxx","abc","abc"],["yyy","ggg","abc","cccc"]]

我有另一个这样的查询列表:

queryList = ["abc","cccc","abc","yyy"]

queryList 和 listt[0] 包含 2 个共同的 "abc"。

queryList & listt[1] 包含 1 "abc", 1 "cccc" & 1 "yyy" 共同。

所以我想要这样的输出:

[2,3] #2 = Total common items between queryList & listt[0]
      #3 = Total common items between queryList & listt[1]

我目前正在使用循环来执行此操作，但这似乎很慢。我将有数百万个列表，每个列表有数千个项目。

listt = [["a","abc","zzz","xxx","abc","abc"],["yyy","ggg","abc","cccc"]]
queryList = ["abc","cccc","abc","yyy"]

totalMatch = []
for hashtree in listt:
    matches = 0
    tempQueryHash = queryList.copy()
    for hash in hashtree:
        for i in range(len(tempQueryHash)):
            if tempQueryHash[i]==hash:
                matches +=1
                tempQueryHash[i] = "" #Don't Match the same block twice.
                break

    totalMatch.append(matches)
print(totalMatch)

Answer 1

嗯，我仍在学习 Python 的基本知识。但是根据this较旧的帖子，类似以下内容应该有效:

from collections import Counter
listt = [["a","abc","zzz","xxx","abc","abc"],["yyy","ggg","abc","cccc"]]
queryList = ["abc","cccc","abc","yyy"]
OutputList = [len(list((Counter(x) & Counter(queryList)).elements())) for x in listt]
# [2, 3]

我会留意其他方法...