如何只获取没有扩展名的文件名？

Question

假设您有这些文件路径，您希望从中获取不带扩展名的文件名：

                       relfilepath
0                  20210322636.pdf
12              factuur-f23622.pdf
14                ingram micro.pdf
19    upfront.nl domein - Copy.pdf
21           upfront.nl domein.pdf
Name: relfilepath, dtype: object

我想到了以下内容，但这给了我一个问题，即第一项变成了输出“20210322636.0”的数字。

from pathlib import Path


for i, row in dffinalselection.iterrows():
    dffinalselection['xmlfilename'][i] = Path(dffinalselection['relfilepath'][i]).stem
    dffinalselection['xmlfilename'] = dffinalselection['xmlfilename'].astype(str)

这是错误的，因为它应该是“20210322636”

请帮忙！

Answer 1

如果列值始终是 filename/filepath，则在 . 上从右边拆分它，将 maxsplit 参数设置为 1，并取拆分后的第一个值。

>>> df['relfilepath'].str.rsplit('.', n=1).str[0]

0                  20210322636
12              factuur-f23622
14                ingram micro
19    upfront.nl domein - Copy
21           upfront.nl domein
Name: relfilepath, dtype: object

Answer 2

你做的是正确的，但是你对数据帧的操作不正确。

from pathlib import Path


for i, row in dffinalselection.iterrows():
    dffinalselection['xmlfilename'][i] = Path(dffinalselection['relfilepath'][i]).stem # THIS WILL NOT RELIABLY MUTATE THE DATAFRAME
    dffinalselection['xmlfilename'] = dffinalselection['xmlfilename'].astype(str) # THIS OVERWROTE EVERYTHING

相反，只需执行：

from pathlib import Path

dffinalselection['xmlfilename'] = ''
for row in dffinalselection.itertuples():
    dffinalselection.at[row.index, 'xmlfilename']= Path(row.relfilepath).stem

或者，

dffinalselection['xmlfilename'] = dffinalselection['relfilepath'].apply(lambda value: Path(value).stem)

如何只获取没有扩展名的文件名？

How to get only filename without extension?

python

pandas

stem