Iain King wrote:
> Iain King wrote:
> > William Meyer wrote:
> > > hi,
> > >
> > > I need to get the index of an object in a list. I know that no two
> > > objects
> > > in the list are the same, but objects might evaluate as equal. for example
> > >
> > > list = [obj1, obj2, obj3, obj4, obj5]
> > > for object in list:
> > > objectIndex = list.index(object)
> > > print objectIndex
> > >
> > > prints 0, 1, 2, 3, 2 instead of 0, 1, 2, 3, 4 because obj3 == obj5. I
> > > could loop
> > > through the list a second time comparing id()'s
> > >
> > > for object in list:
> > > objectIndex = 0
> > > for i in list:
> > > if id(object) == id(i):
> > > break
> > > objectIndex += 1
> > > print objectIndex
> > >
> > > but that seems like a real ugly pain. Somewhere, someplace python is
> > > keeping
> > > track of the current index in list, does anyone know how to access it? Or
> > > have
> > > any other suggestions?
> >
> > Um, one of us is being really really dense today :) I hope it's not
> > me...
> > what's wrong with:
> >
> > i = 0
> > for object in list:
> > objectIndex = i
> > print objectIndex
> > i += 1
> >
> > Iain
>
> Reading it again, I'm thinking it probably is me...
>
> If you aren't looking them up sequentially then I think your second
> example is the only way. You can make it a little prettier by using
> 'object is i' rather than 'id(object) == id(i)'.
> I think python only stores lists one way - i.e. each index maps to it's
> value, but no backwards trace is kept from value to index.
>
> Iain
OTOH, if memory is not an issue, you can create a lookup yourself:
def createLookup(l):
d = {}
for index in xrange(len(l)):
objID = id(l[index])
d[objID] = index
return d
lookup = createLookup(list)
for i in list:
objectIndex = lookup[id(i)]
print objectIndex
Iain
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