Number of Active Objects with the same Alterable Value

[kikiriki]

I have 100's of copies of active object named "ACTIVE01" on the screen.
Those object carries a different values in "Alterable Value A" (for 1 to 10).
I have a lot of calculations to do between cycles so I need the fastest way to get a number of objects that have the same value in "Alterable Value A"?

so:
how many "ACTIVE01" have "Alterable value A" = x

Thank you in advance! :-)

[MuddyMole]

The Select Object extension is the easiest and most efficient method, if it's compatible with whichever exporter(s) you are using.
Failing that, use a ForEach loop (a fastloop would work too, but it would be inefficient).

Another alternative is this:
+ Alt Val A of ACTIVE01 = X
-> Create Token at 0,0 from ACTIVE01
-> Set Counter to NoObjects("Token")
-> Destroy Token

That will create one Token for each ACTIVE01 with an Alt Val A of X, so you can then count how many were created, and then immediately destroy them all. It seems pretty inefficient to me, but it's been suggested before by someone (I forget who).

[SolarB]

another quite efficient method is this:

Alt Val A of 'Active' = X
+flag 0 is off
+pick one 'Active' at random
-> add 1 to counter
-> set flag 0 ('active') on

[MuddyMole]

Yeah, if you want to use a fastloop, that would be the way to do it (obviously you wouldn't do that outside of a fastloop, or it would be way too slow - taking several seconds) - but it's still simpler and quite a bit more efficient to use the ForEach or Select Object extensions.

eg.
+ Alt Val A of Active = X // Note that you only run 1 loop for each object that meets the condition (not 1 loop per object) and the loop itself doesn't have any conditions to test
-> Set Counter to 0
-> Start ForEach loop for Active

+ On ForEach loop for Active
-> Add 1 to Counter

[ProdigyX]

Easiest method is ForEach loop and compare the two values. If equal, add 1 to some counter value

[AyreGuitar]

There's a much better method: Still requires ForEach object, but you don't need to loop.
Simply add the Actives to a group and then get the number of members in the group.
You'll need to clear your group beforehand.

Here's an example to show how the counting can all be on one line (requires ForEach object)

[kikiriki]

Guys, thank you very, very much! :-)
I need to make some tests to see which way will be the best for my case. I will let you know!
Thank you again!

[gkinfinity]

@AyreGuitar: Thanks for the example, but I'm not getting consistent results with it. Sometimes when you toggle an object's Alt Val A the count doesn't change properly.

[ProdigyX]

This may be better suited to what you were looking for.

All of the counting in my example is done in one single frame loop.

[Skyhunter]

By the way, the foreach object will do a loop taking only the actives with a certain condition true to them, which is very handy, so you dont actually need any further conditions or an actuall loop taking place at all. I revised the ProdigyX example to show you what i mean:

https://dl.dropbox.com/u/14962632/Pu...ing%20rev2.mfa