How to open a random folder within a directory

Question

I was wondering how I would use a batch file or Python to open a random folder from a selection of many folders within a directory?

1) get list of all folder within a directory 2) random.choice() 3) open directory 4) PROFIT! — vchyzhevskyi, Mar 27 '13 at 20:44
@coirius - really? It should be 4) ... 5) PROFIT! The "..." is the most important step of all! — MattDMo, Mar 27 '13 at 20:48

garnertb · Answer 1 · 2013-03-27T20:50:56.067

2

>>> import random
>>> import os
>>> files = os.listdir('/tmp')
>>> dirs = [f for f in files if os.path.isdir(f)]
>>> random.sample(dirs,1)
['tempdir']

edited Mar 27 '13 at 20:50

answered Mar 27 '13 at 20:45

garnertb

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score 0 · Answer 2 · answered Mar 27 '13 at 22:14

In cmd you could do it like this:

@echo off

setlocal EnableDelayedExpansion

set root=C:\base\folder

for /f %%d in ('dir /b /a:d "%root%" ^| find /c /v ""') do set count=%%d

set /a num=%RANDOM% %% %count%

for /f "skip=%num% tokens=*" %%d in ('dir /b /a:d "%root%"') do (
  set folder=%%~fd
  goto :FIN
)

:FIN
echo %folder%

endlocal

How to open a random folder within a directory

2 Answers2