Question

我有一系列相同长度的24个字符的字符串，如

my_series = ['ThisIsASentenceXXXXXXXXX', 'SoIsThisXXXXXXXXXXXXXXXX', 'YouGetThePointXXXXXXXXXX']

我有一个列表，其长度等于4个字符，如

my_list = ['This', 'XXXX', 'GetT']

我想将my_list中的每个条目与my_series中每个条目中的每个4个字符的块进行比较，并返回找到列表字符串的my_series项。

例如，对于my_list中的字符串'This'，我希望my_series项目1和2返回，而'XXXX'my_series项目1,2,3将被返回。

Answer 1

以下生成器将创建一个二维列表。每个列表将包含任何匹配项，并且它的位置将与my_list索引匹配。

n_list = [[x for x in my_series if item in x] for item in my_list]

输出：

[[＆＃39; ThisIsASentenceXXXXXXXXX＆＃39 ;,＆＃39; SoIsThisXXXXXXXXXXXXXXXX＆＃39;]，[＆＃39; ThisIsASentenceXXXXXXXXX＆＃39 ;,＆＃39; SoIsThisXXXXXXXXXXXXXXXX＆＃39 ;,＆＃39; YouGetThePointXXXXXXXXXX＆＃39 ;]，[＆＃39; YouGetThePointXXXXXXXXXX＆＃39;]]

因此n_list[0]包含my_list[0]等的匹配项... 我希望这能帮到你！

Answer 2

您需要将my_series中的每个条目拆分为6个块：

serires_chunks = [(s[0:4], s[4:8], s[8:12], s[12:16], s[16:20], s[20:24])
                  for s in my_series]

然后你可以迭代这个块来找到匹配的项目：

for item in my_list:
    for index, chunks in enumerate(serires_chunks):
        for place, chunk in enumerate(chunks, 1):
            if item == chunk:
                location = my_series[index]
                print("Found '{item}' in '{location}' at {place}".format(**locals()))

您将获得：

Found 'This' in 'ThisIsASentenceXXXXXXXXX' at 1
Found 'This' in 'SoIsThisXXXXXXXXXXXXXXXX' at 2
Found 'XXXX' in 'ThisIsASentenceXXXXXXXXX' at 5
Found 'XXXX' in 'ThisIsASentenceXXXXXXXXX' at 6
Found 'XXXX' in 'SoIsThisXXXXXXXXXXXXXXXX' at 3
Found 'XXXX' in 'SoIsThisXXXXXXXXXXXXXXXX' at 4
Found 'XXXX' in 'SoIsThisXXXXXXXXXXXXXXXX' at 5
Found 'XXXX' in 'SoIsThisXXXXXXXXXXXXXXXX' at 6
Found 'XXXX' in 'YouGetThePointXXXXXXXXXX' at 5
Found 'XXXX' in 'YouGetThePointXXXXXXXXXX' at 6

检查python系列是否包含另一个列表中的任何字符串

2 个答案: