remove keys on property A except where property B has keys

[Xavier Gomez]

There is a misprint inside the secondary loop, it should be: if (t == BKeyTimes[i]) { (==, not =).

Additionnaly, the loop logic isnt very good: as it is now, it will remove a keyframe on A as soon as that keyframe’s time doesnt match ALL keyframes times on B, which is impossible except in rare situations (only one key on B).
You’d rather want to keep the key on A if its time matches ONE key time on B, and remove otherwise.

Here is another approach, simpler to write down, but which might be slow if many keyframes, because of the use of nearestKeyIndex:

function sameTimes(t1, t2){
return Math.abs(t2-t1)<;=0.0005;
};
function removeAllKeys(prop){
while(prop.numKeys>0) prop.removeKey(prop.numKeys);
};

function removeKeys(A, B){
if (B.numKeys===0) return removeAllKeys(A);
var idxA, idxB, tA, tB;
for (idxA=A.numKeys; idxA>0; idxA--){
tA = A.keyTime(idxA);
idxB = B.nearestKeyIndex(tB);
tB = B.keyTime(idxB);
if (!sameTimes(tA, tB)) A.removeKey(idxA);
};
};

Xavier

[Steve Sierra]

Hello Xavier,

Thanks for your reply !
I haven’t had the time to thourougly check out your proposition, but I can’t seem to get it to work…
I believe there was a “;” too many in the first function, but I think there might be a problem with the last function as well.
Can IdxB use tB to define itself while tB uses idxB ? Doesn’t one have to be defined first to define the second ?

Thanks again for your answer, I’ll fiddle with it tomorrow and keep you posted !

Cheers !;)

[Xavier Gomez]

Yes you’re right for both.
The “;” come from the auto-formatting from the browser (i tried to fix but failed).
And idxB = B.nearestKeyIndex(tB); should indeed be:

idxB = B.nearestKeyIndex(tA);

I havent tried to check whether it works or not. It should, be can be slow if many keyframes.
Your way, collecting keyframes first and compare afterwards, should be faster in that case.

Xavier

[Steve Sierra]

Yep ! That works Xavier ???? Thank you very much…

Meanwhile, I did a less elegant version of my own… It has more functions, but doeasn’t use nearestKeyIndex.
Could you tell me which one is quickest ?

Thanks again !
S.

PS: There seems to be an error in the pasted code in the first function… it’s : for(var t = 1; t <= Len; t++)

var B = app.project.item(1).layer("KeyLayer Comp 1").property("Time Remap");
var A= app.project.item(1).layer("Shape Layer 1").property("Contents").property("Group 1").property("Contents").property("Path 1").property("Path");
var ATimes = [];
var BTimes = [];
var BAIndexes = [];

//--------------------------------------------------------------------------------------------------------------------------
app.beginUndoGroup("S_RemoveKeys");
//--------------------------------------------------------------------------------------------------------------------------

BTimes = getKeyTimes(B);
ATimes = getKeyTimes(A);
removePostKeys(A, B, ATimes, BTimes);
removeMiddleKeys(A, B, ATimes, BTimes);
removePreKeys(A, ATimes, BTimes);

//--------------------------------------------------------------------------------------------------------------------------
app.endUndoGroup();
//--------------------------------------------------------------------------------------------------------------------------

function getKeyTimes(prop){
var Len = prop.numKeys;
var keyTimes = [];
for(var t = 1; t &lt;= Len; t++){
var e = prop.keyTime(t);
keyTimes.push(e);
}
return keyTimes;
}
function removePostKeys(A, B, ATimes, BTimes){
for(var i = B.numKeys - 1; i > 0; i--){
var c = BTimes[i];
for(var t = A.numKeys; t > 0; t--){
var a = ATimes[t-1];
if(a !== c ){
A.removeKey(t);
}else{
break;
}
}break;
}
}

function removeMiddleKeys(A, B, ATimes, BTimes){
for(var j = B.numKeys - 1; j > 0; j--){
for(var i = j; i > 0; i--){
var c = BTimes[i];
var q = BTimes[i - 1];
for(var t = A.numKeys - 1; t > 0; t--){
var a = ATimes[t];
if(a > q && a < c ){
A.removeKey(t + 1);
}else{
if(a == q){
break;
}
}
}
}break;
}
}
function removePreKeys(A, ATimes, BTimes){
var c = BTimes[1];
for(var t = A.numKeys; t > 0; t--){
var a = ATimes[t];
if(a < c ){
A.removeKey(t);
}
}
}


[Steve Sierra]

Hello again,

Could someone tell me how to mark a post as resolved ?
Thanks !