Python:如何在列表中查找与元素名称的一部分匹配的元素 [英] Python: how to find the element in a list which match part of the name of the element

问题描述

我有一个关键字列表，可从文件名列表中查找.例如，如果关键字为'U12'，我想查找包含'U12'的csv文件，该文件为'h_ABC_U12.csv'并打印出来.

I have a list of keyword to find from a list of file name. Example, if the keyword is 'U12', I want to find the csv file that contain 'U12' which is 'h_ABC_U12.csv'and print it out.

word = ['U12','U13','U14']
file_list = ['h_ABC_U12.csv','h_GGG_U13.csv','h_HVD_U14.csv','h_MMMB_U15.csv']

for x in range (len(word)):
    if word[x] in file_list:
       #print the file name

这是代码的一部分，但在进行某些搜索后无法继续.我需要与单词匹配的全名才能打印出来.

This is part of the code but unable to continue after some searches. I need the full name that match the word to print out.

推荐答案

我建议使用 os.path.splitext 获取不带扩展名的文件名.

I suggest using os.path.splitext to get the filename without extension.

>>> from os.path import splitext
>>> for f in file_list:
...     name = splitext(f)[0]
...     if any(name.endswith(tail) for tail in word):
...         print(name)
... 
h_ABC_U12
h_GGG_U13
h_HVD_U14

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