Opening files with the name of a variable

Question

I am using a Markov chain. When the chain arrives at a particular state, two files (a .png and an .mp3) need to open.

s is the current state of the chain, an integer from 1-59.

I can't seem to find how to open the file with the same number as 's'.

I'm sure it has something to do with %str formatting, but I can't seem to implement it.

img = Image.open('/.../.../s.png')
img.show()

@redFur Will only work if OP is using Python 3.6, for which I don't see any indication — DeepSpace, Mar 15 '18 at 19:52

score 1 · Answer 1 · answered Mar 15 '18 at 19:52

1

You should use the following line in your code:

img = Image.open('/.../.../{0}.png'.format(s))

answered Mar 15 '18 at 19:52

Laurent H.

score 0 · Answer 2 · answered Mar 15 '18 at 19:50

0

You can format a string using a variable like this

>>> s = 10
>>> '/path/to/file/{}.png'.format(s)
'/path/to/file/10.png'

answered Mar 15 '18 at 19:50

Brendan Abel

2 Answers2