Beeping Alarm Function

Question

I'm trying to create an alarm function with Python3. The below code does work but it doesn't seem like this is the most optimal way of achieving something like this. Is there a more pythonic way?

#!/usr/local/bin/python3
import os
import time
def alarm():
 os.system('amixer -D pulse sset Master 30%') # >> Set initial volume
 for beep in range(0, 20):
 time.sleep(.002)
 os.system('play --no-show-progress --null --channels 1 synth %s sine %f' % (.08, 2500))
alarm()

Answer 1

• Pythonic approach

  For sound playback in Python, see this discussion.

• Avoid os.system

  If you are forced to use it, supply a fully qualified path to the executable. You never know what weird program called play may be in the search path prior to an intended one.

• Modularize

  As written, the code is not reusable. An attempt to import it would result in an immediate alarm sound. Put the call to alarm in an

  if __name__ == '__main__':
  

  clause.

• Avoid magic numbers

  All the parameters (beep duration, volume, etc) better be passed via a command line (see sys.argv and optparse module).

• Dummy variable

  is customarily denoted as _:

   for _ in range(20):
  

Answer 2

This is the old way of formatting strings:

'play --no-show-progress --null --channels 1 synth %s sine %f' % (.08, 2500)

The current Python method is to use str.format.

'play --no-show-progress --null --channels 1 synth {} sine {}'.format(.08, 2500)

It's not a huge advantage here, though you'll notice it's type agnostic. But it has many advantages available so it's good to get used to using it.