En Thu, 23 Oct 2008 05:23:51 -0200, Andreas Müller <[EMAIL PROTECTED]>
escribió:
(Python 2.2.3 if this is relevant :-)
I have a list of objects with, lets say, the attributes "ID", "x" and
"y". Now I want to find the index of list element with ID=10.
Of course I can loop through the list manually, but is there a
construct like
list.find (10, key='ID')
If you define __eq__ for your objects you may use list.index:
class Obj:
def __init__(self, id_, x=0, y=0):
self.id = id_
self.x = x
self.y = y
def __eq__(self, other):
if not isinstance(other, Obj): raise NotImplementedError
return self.id==other.id
py> a = Obj(10)
py> b = Obj(11)
py> c = Obj(12)
py> alist = [a,b,c]
py> print alist.index(Obj(11))
1
py> print alist.index(Obj(20))
Traceback (most recent call last):
File "<stdin>", line 1, in <module>
ValueError: list.index(x): x not in list
Note that this is a "global" change - two separate instances with the same
ID were "not equal" before, and are "equal" now, everywhere, not just
regarding list.index
Of course you always have the old "for" loop solution (`enumerate` would
make things a lot nicer, but it requires 2.3):
def index_by_id(alist, id_):
for i in range(len(alist)):
if alist[i].id==id_:
return i
--
Gabriel Genellina
--
http://mail.python.org/mailman/listinfo/python-list
