Sequence find function in Python

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') 

Answer 2

Actually, in Python 3, at least, filter doesn't go through the entire list.

To double check:

def test_it(x):     print(x)     return x>10  var = next(filter(test_it, range(20))) 

In Python 3.2, that prints out 0-11, and assigns var to 11.

In 2.x versions of Python you may need to use itertools.ifilter.