Figured it out:
var filteredUsers =
userList.Where(
user=>
user.RoleTypes.Any(
userForRole => userRoleList.Contains
(userForRole)));
On Feb 3, 11:48 am, "[email protected]"
<[email protected]> wrote:
> Unfortunately I am implementing an interface that was designed up
> stream, so I really can't change the design.
>
> I was able to do a hybrid approach:
>
> foreach (var user in userList)
> {
> if (user.RoleTypes.Intersect(RoleList).Count() > 0)
> {
> filteredUsers.Add(user);
> }
> }
>
> Thanks for the feedback.
>
> On Feb 3, 1:28 am, Joe Enos <[email protected]> wrote:
>
> > Whenever I see roles like this, the first thing I think about is using
> > bit logic - I've never actually implemented it, but I know it's a
> > powerful technique - assigning powers of 2 to each role, then you can
> > use bit functions to quickly determine whether or not a user has a
> > specific role. Not really an answer to your question, but just a
> > thought...
>
> > Also, I could be wrong, but I think querying objects using Linq to
> > objects provides no performance benefit versus doing the looping
> > yourself - so there's no real benefit other than code style - unless
> > you're planning on doing a lot of Linq in your project, my opinion is
> > that good old-fashioned loops are easier to read and debug than a
> > single implementation of a Linq function.
>
> > On Feb 2, 2:43 pm, "[email protected]" <[email protected]>
> > wrote:
>
> > > I have a function that takes an array of integers(roles) and returns a
> > > set of objects that contain one or more of the roles within a child
> > > collection:
>
> > > public Users[] GetUsersByRoles(int[] Roles)
> > > {
>
> > > }
>
> > > public class Users
> > > {
> > > public int[] roles;
> > > public string blah;
> > > ...
>
> > > }
>
> > > If I pass 2,4,6 then I would return any users that has at least one of
> > > the elements.
>
> > > Is there any way that Linq to Objects can do the heavy lifting for
> > > me? I would rather not have to loop through each user and query their
> > > roles.
>
> > > Thanks,
>
> > > Brian
