[Dan Ebberts]

Actually, I don’t think you even need the markers. If you set the precomp up so it is the exact length you need to encompass however many 5-second clips you have, then time remap it in the main comp and drag the outpoint to the end of the main comp, this time remapping expression should take care of it:

dur = 5; // play each clip for 5 seconds
seed = 1103; // change this to get different results
nTot = Math.floor(source.duration/dur);
n = Math.floor(time/dur)
seedRandom(seed+n,true);
tStart = Math.floor(random(nTot))*dur;
tStart + time%dur

[Dan Ebberts]

I think your 2D circle layer (not parented to any other layer and anchor point centered) will need a position expression like this:

L = thisComp.layer("Line graph");
path = L.content("Shape 1").content("Path 1").path;
pct = L.content("Trim Paths 1").start/100;
L.toComp(path.pointOnPath(pct));

[Dan Ebberts]

Something like this time remapping expression maybe:

dur = 5; // play each clip 5 seconds
seed = 1103; // change this to get different result
m = marker;
val = time;
if (m.numKeys > 0){
  period = Math.floor(time/dur);
  t = time%dur;
  seedRandom(seed+period,true);
  n = Math.floor(random(m.numKeys)) + 1;
  val = m.key(n).time + t;
}
val

[Dan Ebberts]

I think the whole problem gets simpler if you switch from rectangular coordinates to polar coordinates (radius and angle). Then a circular random distribution centered around the comp center becomes something like this:

c = [thisComp.width,thisComp.height]/2; // center
rMax = 500; // max radius
seedRandom(index,true);
a = random(Math.PI*2);
r = random(rMax);
x = r*Math.cos(a);
y = r*Math.sin(a);
c + [x,y]

It gets more complicated if you have a rectangular exclusion zone in the center of the field, but by using a little trig, you can avoid the while loop that can hang AE:

c = [thisComp.width,thisComp.height]/2; // center
w = 400; // width of rectangular exclusion area
h = 300; // height of rectangular exclusion area
rMax = 500; // max radius (must be big enough to enclose the rectangle)
seedRandom(index,true);
a = random(Math.PI/2);
aMid = Math.atan2(h/2,w/2);
if (a < aMid){
  rMin = (w/2)/Math.cos(a);
}else{
  rMin = (h/2)/Math.cos(Math.PI/2 - a);
}
r = random(rMin,rMax);
x = r*Math.cos(a);
y = r*Math.sin(a);
n = Math.floor(random(4));
if (n == 1){
  x = -x;
}else if (n == 2){
  x = -x;
  y = -y;
}else if (n == 3){
  y = -y;
}
c + [x,y]

[Dan Ebberts]

This may not be what you’re after, as it requires the checkbox to be keyframed to on when you want the child’s x position to follow changes in the parent’s x position:

p = thisComp.layer("parent").transform.xPosition;
cb = effect("Checkbox Control")("Checkbox");
accum = value-value;
if (cb.numKeys > 0){
  n = 0;
  n = cb.nearestKey(time).index;
  if (cb.key(n).time > time) n--;
  if (n > 1){
    if (cb.key(n).value) accum += p - p.valueAtTime(cb.key(n).time);
    for (i = n-1; i > 0; i--){
      if (cb.key(i).value) accum += p.valueAtTime(cb.key(i+1).time) - p.valueAtTime(cb.key(i).time);
    }
    if (cb.key(1).value) accum += p.valueAtTime(cb.key(1).time) - p.valueAtTime(0);
  }
}else{
  if (cb.value) accum = p - p.valueAtTime(0);
}
value + accum

[Dan Ebberts]

This should work, but there might be a more efficient way to do it, depending on what’s driving your value (for this example, I’m assuming the value is contained in a slider):

s = effect("Slider Control")("Slider");
f = timeToFrames(time);
above = s.valueAtTime(0) >= 1;
n = above ? 1 : 0;
for (i = 1; i <= f; i++){
  if (above){
    if (s.valueAtTime(framesToTime(i)) < 1){
      above = false;
    }
  }else{
    if (s.valueAtTime(framesToTime(i)) >= 1){
      n++;
      above = true;
    }
  }
}
n

[Dan Ebberts]

That’s considerably more complicated and will bog down in a longer comp because of the additional processing that needs to happen at each successive frame. Try this:

numFrames = 5; // change colors every 5 frames
c1 = effect("Color Control 1")("Color");
c2 = effect("Color Control 2")("Color");
c3 = effect("Color Control 3")("Color");
c4 = effect("Color Control 4")("Color");
c5 = effect("Color Control 5")("Color");
colors = [c1,c2,c3,c4,c5];
f = timeToFrames(time);
curPeriod = Math.floor(f/numFrames);
colorIdx = [];
seedRandom(0,true);
for (i = 0; i < value.length; i++){
  colorIdx.push(Math.floor(random(colors.length)));
}
i = 1;
while (i <= curPeriod){
  seedRandom(i,true);
  for (j = 0; j < value.length; j++){
    temp = colorIdx[j] + Math.floor(random(colors.length-1)) + 1;
    colorIdx[j] = temp%colors.length;
  }
  i++;
}
myExpr = "style";
for (i = 0; i < value.length; i++){
  myExpr += ".setFillColor(colors[colorIdx[" + i + "]]," + i + ",1)";
}
eval(myExpr)

[Dan Ebberts]

Do you mean so that each letter has a color different than the one next to it, or so that a letter doesn’t get the same color for two consecutive periods?

[Dan Ebberts]

Something like this should work. You might have to fiddle with it a bit:

s = effect("Slider Control")("Slider");
fadeDuration = .333;
t = s.key(textIndex+1).time;
ease(time,t,t+fadeDuration,100,0);

[Dan Ebberts]

It works for me if I change your fadeDuration from 10 to .333.