Utilities

Accessing sub-objects

Get layer's pixel size

function getSize(layer){
var src = layer.sourceRectAtTime();
var s = layer.transform.scale
var size = [src.width*s[0],src.height*s[1]]/100;
return size;
}

Get layer's top left coordinates

function getSrcRectTopLeft(layer){
s = layer.sourceRectAtTime();
localPos = [s.left,s.top];
worldPos = layer.toWorld(localPos);
return worldPos;
}

Get effect index number

// effect index
thisProperty.propertyGroup(1).propertyIndex;

Get number of effects applied to a layer

// referencing locally
thisLayer("ADBE Effect Parade").numProperties
thisLayer("Effects").numProperties
​
// referencing other layer
thisComp.layer("layer")("ADBE Effect Parade").numProperties
thisComp.layer("layer")("Effects").numProperties
​
// other method
thisProperty.propertyGroup(2).numProperties

Working with Array

  • https://dmitripavlutin.com/operations-on-arrays-javascript/

​Shuffling elements​

function shuffle(a) {
var j, x, i;
for (i = a.length - 1; i > 0; i--) {
j = Math.floor(Math.random() * (i + 1));
x = a[i];
a[i] = a[j];
a[j] = x;
}
return a;
}

Check how many array is empty

const arr = [[], [1,4,5], [], [4,6,7];
​
// method 1: conditional -> sum
numbers.map(x=> x.length>0: 1:0).reduce((a, b) => a + b, 0)
​
// method 2
arr.filter(x => x.length>0).length // get array of true results -> length
​

Finding minimum or maximum element

In ES6, you can use the ... operator to spread an array and take the minimum or maximum element.
var myArray = [1, 2, 3, 4, 99, 20];
​
var maxValue = Math.max(...myArray); // 99
var minValue = Math.min(...myArray); // 1

Get length of each element

// get length of each array element
elemLength = str.map(s=>s.length)
// method 1: using mapget character index
let array = [1,5,6,8,10],sum;
array = array.map(elem => sum = (sum || 0) + elem);
​
// method 2: using reduce and map
let array = [280,430,408,430,408];
array = array.map((elem, index) => array.slice(0,index + 1).reduce((a, b) => a + b));
​
// best method: using double arrow functions
var array =[280, 430, 408, 430, 408]
result = array.map((s => a => s += a)(0));
result;

Comparing a value to elements, filtering conditions

let array = [1, 3, 5, 7, 10];
v = 6;
​
// method 1: conditional -> sum of array
row = (array.map(c => v >= c ? 1 : 0)).reduce((a, b) => a + b, 0)
​
// method 2: get index of true results, then get max value
row = Math.max(...array.map(c => v>=c? array.indexOf(c):0))
​
// best method: filter
row = array.filter(c => v >= c).length

Increment by Index

  • Usage: tiling, valueAtTime offset
function indexInc(mainlayer, offset, randRange) {
var startIndex = mainlayer.layer(index - 1).index
var myIndex = index - startIndex;
try {
var inc = offset * random(randRange[0], randRange[0])* myIndex;
} catch (e) {
inc = offset * myIndex;
}
return inc;
}
​
// usage for tile y
mainLayer.transform.position + [0+indexInc(mainLayer,50]

Loops

Through every frame

for(t = 0; t < thisComp.duration; t = t + thisComp.frameDuration){
// statements to execute
}

Unsorted

//If you want it relative to the comp’s [0,0] (upper left corner)
​
[length(xyArray),radiansToDegrees(Math.atan2(xyArray[1],xyArray[0]))]

Explanation

Using simple trigonometry:
Imagine you have the point [960, 540];
Those 2 values, 960 and 540, represent two adjacent sides of a triangle.
Now, if you recall back to 9th Grade, you may remember that there is a simple equation to get the third side of that triangle. AΒ² + BΒ² = CΒ². So, we can find the third side (the magnitude) by using an equation like:
var mag = Math.sqrt(Math.exp(position[0], 2) + Math.exp(position[1], 2)); Essentially saying take the square root of XΒ² + YΒ².
Fortunately, AE gives us the handy little β€˜length’ operator, which simply does all of these computations for us.
β€”
Another similar equation can be used to find the angle of a triangle when you know two other sides, you may remember SOH CAH TOA.
Imagine the angle at the top left of your comp, now notice that you have the opposite side (the Y coord) and the adjacent side (the X coord). Since TOA uses both opposite and adjacent sides, we can get that angle using an equation like this:
tan(theta) = opposite/adjacent
tan(theta) = 540/960
theta = tanβˆ’1(5625)
​
//or, in code
var angle = Math.atan(position[1] / position[0]);
var angle_deg = radiansToDegrees(angle);

Null / Camera Rigs

Add sliders with specific affix names to property

expression
function
var n = thisLayer("Effects").numProperties
var fxstr = "z_add"
for(i=1;i<=n;i++){
if(effect(i).name.includes(str)){
value+= effect(i)(1);
}
}
function sliderAdd(fxName){
n = thisLayer("Effects").numProperties;
for(i=1;i<=n;i++){
if(effect(i).name.indexOf(fxName)==0){
value+= effect(i)(1);
}
}
return value;
};
​
sliderAdd("Slider");

Autonomous agents

Auto-capture ball by Aaron Cobb

var ball = thisLayer;
var cup = thisComp.layer("Cup");
​
var captureDuration = .25; //time for ball to reach center of cup once capture beings
var captureRadius = 100; //radius around anchor point of cup at which ball will be captured.
​
var captureTime = thisComp.duration; //time at which capture begins, default to end of comp
var currentDistance;
​
for(t = 0; t < captureTime; t = t + thisComp.frameDuration){ //loop through frames
currentDistance = length(ball.toComp(ball.anchorPoint.valueAtTime(t), t), cup.toComp(cup.anchorPoint.valueAtTime(t), t));
if(currentDistance < captureRadius) captureTime = t; //if inside capture radius exit the loop
}
// execute
ease(time, captureTime, captureTime + captureDuration, value, cup.toComp(cup.anchorPoint.value));

Compounding ease() interpolation (source)

You can't control the influence directly, but you can compound the ease such as below.
if (numKeys > 1){
t1 = key(1).time;
t2 = key(2).time;
v1 = [0,540];
v2 = [1080,540];
t = easeOut(time,t1,t2,0,1); // using a normalized value to drive the 2nd ease
easeOut(t,0,1,v1,v2);
}else
value