Question

How do I find an object in a sequence satisfying a particular criterion?

List comprehension and filter go through the entire list. Is the only alternative a handmade loop?

mylist = [10, 2, 20, 5, 50]
find(mylist, lambda x:x>10) # Returns 20

Answer 1

Here’s the pattern I use:

mylist = [10, 2, 20, 5, 50]
found = next(i for i in mylist if predicate(i))

Or, in Python 2.4/2.5, next() is a not a builtin:

found = (i for i in mylist if predicate(i)).next()

Do note that next() raises StopIteration if no element was found. In most cases, that’s probably good. You asked for the first element, no such element exists, and so the program probably cannot continue.

If, on the other hand, you do know what to do in that case, you can supply a default to next():

conf_files = ['~/.foorc', '/etc/foorc']
conf_file = next((f for f in conf_files if os.path.exists(f)),
                 '/usr/lib/share/foo.defaults')