Question

我有以下列表：

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

我有一个搜索项列表：

search = ['abc','123','xyz','456']

我想使用搜索列表迭代data_items以进行匹配，并构建一个为每个匹配提供计数的基本结构。 e.g。

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

最好的方法是什么？

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

看起来你也需要部分匹配。下面的代码很直观，但可能效率不高。并假设你对dict结果没问题。

>>> 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}

Answer 3

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

如果您对使用字典感到满意（因为您说结构，它是更好的选择）。

计算列表中搜索字符串的出现次数

3 个答案: