Q:

How do I open a file using python?

I am trying to open a random mp3 file using python so that I can play a random song from my music playlist.
these are two variations of the code that I have tried
file = str('C:\file.mp3')
open(file)

or
open('C:\file.mp3')

When I run the program, I expect to see the Microsoft mp3 player to open and for the file to play but instead I get a pop-up message saying "Unicode error: 'unicodeescape' codec can't decode bytes in position 2-3: truncated \UXXXXXXXX escape".
I have never used open() before so if it is something very simple then I'm sorry.

A:

Always use raw strings for Windows file paths (and regexes). Otherwise, a folder whose name begins with U (say, C:\Users) ends up looking like the beginning of a Unicode escape; many other characters end up being interpreted as escapes as well (e.g. f is escaped to the form feed character).
Using raw strings (prefixed with r) means backslashes only escape the quote character, nothing else. So:
with open(r'C:\file.mp3') as f:

works, where:
with open('C:\file.mp3') as f:

wouldn't. That said, none of this would open the file in your music player. open opens the raw file to read data from it (odds are an mp3 file will error if you try to read it in text mode, unless you're using one of the more permissive codecs, like latin-1). If you want to launch a file via its default handler, that's what os.startfile is for, e.g.:
os.startfile(r'C:\file.mp3')

