Python: 获取目录中首个文件名

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头信息。在

例如：path = "Users/SDB/case_23/scan_1"

现在在子目录scan_1中，我想检查一些头信息。为此，如何获得子目录中第一个文件（按名称）的完整路径和名称？在

Answer 1

os.walk

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). ...

示例

dirs结构

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

dirs+文件结构

^{pr2}$

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

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