统计lis中搜索字符串的出现次数

Question

我有以下清单：

data_items = ['abc','123data','dataxyz','456','344','666','777','888','888', 'abc', 'xyz']

我有一个搜索项目列表：

^{pr2}$

我想使用搜索列表迭代匹配项的data_项，并构建一个为每个匹配项提供计数的基本结构。e、 g

counts = ['abc':'2', '123':'1', 'xyz':'2'.........]

最好的办法是什么？在

Answer 1

您可以使用re.search和collections.Counter，例如：

import re
from collections import Counter

data_items = ['abc','123data','dataxyz','456','344','666','777','888','888', 'abc', 'xyz']
search = ['abc','123','xyz','456']

to_search = re.compile('|'.join(sorted(search, key=len, reverse=True)))
matches = (to_search.search(el) for el in data_items)
counts = Counter(match.group() for match in matches if match)
# Counter({'abc': 2, 'xyz': 2, '123': 1, '456': 1})

Answer 2

counts={}
for s in search:
    lower_s=s.lower()  
    counts[lower_s]=str(data_items.count(lower_s))

如果你不介意用字典的话（因为你说的是结构，这是一个更好的选择）。在

Answer 3

看来你也需要部分匹配。下面的代码是直观的，但可能不是有效的。也假设你对dict结果没问题。在

>>> data_items = ['abc','123data','dataxyz','456','344','666','777','888','888', 'abc', 'xyz']
>>> search = ['abc','123','xyz','456']
>>> result = {k:0 for k in search}
>>> for item in data_items:
        for search_item in search:
            if search_item in item:
                result[search_item]+=1
>>> result
{'123': 1, 'abc': 2, 'xyz': 2, '456': 1}