删除python中文件名的结尾

我需要删除以下文件名的结尾：

Testfile_20190226114536.CSV.986466.1551204043175

因此，CSV之后的任何内容都需要删除，因此我有一个名为：

Testfile_20190226114536.CSV

就这么简单:

s = 'Testfile_20190226114536.CSV.986466.1551204043175'
suffix = '.CSV'

s[:s.rindex(suffix) + len(suffix)]
=> 'Testfile_20190226114536.CSV'

您可以使用re.sub：

假设file_name=Testfile_20190226114536.CSV.986466.1551204043175

简单的方法就是这样

<> P>你的所有文件中间都有这个CSV？

您可以尝试拆分并加入您的姓名，如下所示：

name = "Testfile_20190226114536.CSV.986466.1551204043175"
print ".".join(name.split(".")[0:2])

---

以下是查看发生了什么的步骤

>>> filename = 'Testfile_20190226114536.CSV.986466.1551204043175'

# split the string into a list at '.'
>>> l = filename.split('.')

>>> print(l)
['Testfile_20190226114536', 'CSV', '986466', '1551204043175']

# index the list to get all the elements before and including 'CSV'
>>> filtered_list = l[0:l.index('CSV')+1]

>>> print(filtered_list)
['Testfile_20190226114536', 'CSV']

# join together the elements of the list with '.'
>>> out_string = '.'.join(filtered_list)
>>> print(out_string)

Testfile_20190226114536.CSV

以下是完整的功能：

def filter_filename(filename):
    l = filename.split('.')
    filtered_list = l[0:l.index('CSV')+1]
    out_string = '.'.join(filtered_list)
    return out_string

>>> filter_filename('Testfile_20190226114536.CSV.986466.1551204043175')
'Testfile_20190226114536.CSV'

---

到目前为止，您尝试了什么？除了substring.CSV之外，您是否提前知道有关文件名的任何信息。发生在名称中的某个位置？我将使用.CSV而不是CSV：

name = "Testfile_20190226114536.CSV.986466.1551204043175"
print ".".join(name.split(".")[0:2])

>>> filename = 'Testfile_20190226114536.CSV.986466.1551204043175'

# split the string into a list at '.'
>>> l = filename.split('.')

>>> print(l)
['Testfile_20190226114536', 'CSV', '986466', '1551204043175']

# index the list to get all the elements before and including 'CSV'
>>> filtered_list = l[0:l.index('CSV')+1]

>>> print(filtered_list)
['Testfile_20190226114536', 'CSV']

# join together the elements of the list with '.'
>>> out_string = '.'.join(filtered_list)
>>> print(out_string)

Testfile_20190226114536.CSV

def filter_filename(filename):
    l = filename.split('.')
    filtered_list = l[0:l.index('CSV')+1]
    out_string = '.'.join(filtered_list)
    return out_string

>>> filter_filename('Testfile_20190226114536.CSV.986466.1551204043175')
'Testfile_20190226114536.CSV'