有两个列表
Experimental = ["2311","1441","31","233122"]
Reference = ["asasa2311","dadewr1441","app"]
我想要做的是对于 Experimental 中的元素，如果它们与 Reference 中的元素中的任何子字符串匹配，然后将 Experimental 中的匹配元素放在一个新列表中。
所以新列表将是
["2311","1441","31"]
我尝试做的是
[i for i in Experimental if any(sub in i for sub in Reference)]
【问题讨论】：
"2311" 和"31" 都将匹配字符串"asasa2311"？在这种情况下，新列表将是 ["2311", "1441", "31"]。
您的问题陈述与预期输出不符。您说“子字符串”，但您的预期输出不包含31，即使它是asasa1311 的子字符串。这是一种方法：
>>> e = ["2311","1441","31","233122"]
>>> r = ["asasa2311","dadewr1441","app"]
>>> [eelem for relem in r for eelem in e if eelem in relem]
['2311', '31', '1441']
或者如果你想要预期的输出：
>>> e = ["2311","1441","31","233122"]
>>> r = ["asasa2311","dadewr1441","app"]
>>> [eelem for relem in r for eelem in e if relem.endswith(eelem)]
['2311', '1441']
或者，如果您想在第一场比赛中更有效率地突破，请使用生成器方法：
>>> e = ["2311","1441","31","233122"]
>>> r = ["asasa2311","dadewr1441","app"]
>>> [eelem for eelem in e if any(r if eelem in relem else False for relem in r)]
['2311', '1441', '31']
如果我们使用in 返回一个布尔值这一事实，或者更短一点：
>>> [eelem for eelem in e if any(eelem in relem for relem in r)]
['2311', '1441', '31']
【讨论】：
你应该在这里使用regex：
In [20]: e=["2311","1441","31","233122"]
In [21]: r=["asasa2311","dadewr1441","app"]
In [22]: [x for x in e if any(x in re.findall("\d+",y) for y in r)]
Out[22]: ['2311', '1441']
【讨论】：
您能否澄清这个问题，因为31 也产生了一个匹配，它是asasa1311 的子字符串。
In [516]: Experimental = ["2311","1441","31","233122"]
...: Reference = ["asasa2311","dadewr1441","app"]
...:
In [517]: [i for i in Experimental for x in Reference if i in x ]
Out[517]: ['2311', '1441', '31']
【讨论】：