Best way to handle list.index(might-not-exist) in python?

Question

I have code which looks something like this:

thing_index = thing_list.index(thing)
otherfunction(thing_list, thing_index)

ok so that's simplified but you get the idea. Now thing might not actually be in the list, in which case I want to pass -1 as thing_index. In other languages this is what you'd expect index() to return if it couldn't find the element. In fact it throws a ValueError.

I could do this:

try:
    thing_index = thing_list.index(thing)
except ValueError:
    thing_index = -1
otherfunction(thing_list, thing_index)

But this feels dirty, plus I don't know if ValueError could be raised for some other reason. I came up with the following solution based on generator functions, but it seems a little complex:

thing_index = ( [(i for i in xrange(len(thing_list)) if thing_list[i]==thing)] or [-1] )[0]

Is there a cleaner way to achieve the same thing? Let's assume the list isn't sorted.

Answer 1

There is nothing "dirty" about using try-except clause. This is the pythonic way. ValueError will be raised by the .index method only, because it's the only code you have there!

To answer the comment:
In Python, easier to ask forgiveness than to get permission philosophy is well established, and no index will not raise this type of error for any other issues. Not that I can think of any.