比较两个列表并输出类似的匹配

Question

我有两个Python列表：

prefixList = ["12","9"]

files = ["12-a.csv","12-b.csv","9-t.txt","8-a.txt"]

并希望创建一个包含以前缀列表开头的文件列表的新列表，因此输出如下：

fileOutput = ["12-a.csv","12-b.csv","9-t.txt"]

Answer 1

您可以使用regex，并在prefixList中查找编号和搜索号。

prefixList = ["12","9"]

files = ["12-a.csv","12-b.csv","9-t.txt","8-a.txt"]

new_files = [file 
             for file in files 
             if(re.search(r'\d+', file).group(0) in prefixList)]

print(new_files)

输出：

['12-a.csv', '12-b.csv', '9-t.txt']