Fast way to read filename from directory?

Question

Given a local directory structure of /foo/bar, 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)?

Answer 1

1st element of os.listdir()

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

Answer 2

Well I know this code works...

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

Answer 3

you can also use glob

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