Python: 如何只获取目录中的第一个文件名

Question

我有各种目录的路径。现在我想查看每个目录中第一个文件的 header 信息。

例如：path = "Users/SDB/case_23/scan_1"

现在在子目录scan_1中，我想查看一些header信息。为此，我怎样才能获得子目录中第一个文件的完整路径和名称（按名称）？

Answer 1

os.walk(top, topdown=True, onerror=None, followlinks=False) Generate the file names in a directory tree by walking the tree either top-down or bottom-up. For each directory in the tree rooted at directory top (including top itself), it yields a 3-tuple (dirpath, dirnames, filenames). ...

例子

目录结构

$ tree -d
.
└── users
    └── sdb
        └── case_23
            └── scan_1

目录+文件结构

$ tree
.
├── a.txt
├── b.txt
├── c.txt
└── users
    ├── a.txt
    ├── b.txt
    ├── c.txt
    └── sdb
        ├── a.txt
        ├── b.txt
        ├── c.txt
        └── case_23
            ├── d.txt
            ├── e.txt
            ├── f.txt
            └── scan_1
                ├── a.txt
                ├── b.txt
                └── c.txt

python代码

>>> import os
>>> rootdir = '/tmp/so'
>>> # print full path for first file in rootdir and for each subdir
... for topdir, dirs, files in os.walk(rootdir):
...     firstfile = sorted(files)[0]
...     print os.path.join(topdir, firstfile)
... 
/tmp/so/a.txt
/tmp/so/users/a.txt
/tmp/so/users/sdb/a.txt
/tmp/so/users/sdb/case_23/d.txt
/tmp/so/users/sdb/case_23/scan_1/a.txt

Python: 如何只获取目录中的第一个文件名

Python: how to obtain only first filename in the directory

python

filenames

path