Fast way to read filename from directory?

Given a local directory structure of /foo/bar, 开发者_JAVA技巧and assuming that a given path contains exactly one file (filename and content does not matter), what is a reasonably fast way to get the filename of that single file (NOT the file content)?

1st element of os.listdir()

import os
os.listdir('/foo/bar')[0]

Well I know this code works...

for file in os.listdir('.'):
    #do something

you can also use glob

import glob
print glob.glob("/path/*")[0]

os.path.basename will return the file name for you

so you can use it for the exact one file by adding your file path :

os.path.basename("/foo/bar/file.file")

or you can run through the files in the folder and read all names

file_src = "/foo/bar/"
for x in os.listdir(file_src):
        print(os.path.basename(x))