检查python系列是否包含其他列表中的任何字符串

我有一系列24个字符的等长字符串，如

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

我有一个列表，也有4个字符的等长字符串，比如

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

我想将my_列表中的每个条目与my_系列中每个条目中的每个4个字符的块进行比较，并返回列表字符串所在的my_系列项目

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例如，对于my_列表中的字符串“This”，我希望返回my_系列项目1和2，对于“XXXX”，将返回my_系列项目1、2、3

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，并且其位置将与my_列表索引匹配

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

输出：

[['ThisisSentencexxxxxxxxx'、['ThisisSentencexxxxxxxxx'、['ThisisSentencexxxxxxxxx'、['ThisisSentencexxxxxxxxxxxxx'、['YouGetThePointxxxxxxxxx']、['YouGetThePointxxxxxxxxxxxxx']

因此

n\u列表[0]

包含

my\u列表[0]

等的匹配项。。。

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我希望这对你有帮助

您需要将

my_系列

中的每个条目分为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，您可以迭代此块以查找匹配的项：

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

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你试过什么来达到这个目的吗？也许您可以为我们提供一个您尝试过的示例以及任何错误消息。否则，请参阅“不会有太多帮助”，但类似于“如果我的_系列[0]中的_列表[0]：我不相信这是OP想要的输出。