Develop Reference

InDesign extendScript: How do I transform the entire selection?

By using the simple array app.selection[x], you can apply a transformation to any object in the selection, independently. But how do I apply a transformation to the entire selection together?
For example: inside InDesign, I can select two side-by-side objects and flip them horizontally, causing them to switch places and flip.
Inside a script, I can target each object in the selection, but they will not switch places; they will remain in the same place and flip.
for ( var x = 0; x < app.selection.length; x++ ){
app.selection[x].absoluteFlip = Flip.HORIZONTAL;
}
I could possibly group the selection, apply a transformation, then ungroup when finished, but this seems like unnecessary bulk that could slow down the code. I can easily do it manually inside InDesign, so it should follow that there's some way to access app.selection as a single object instead of an array. Does such an object exist?

Not really a solution, but it's worth noting that I don't think absoluteFlip is the action being performed, but a state indicating if the item has ben flipped. It's writable so you can set the state, but I think what's happening when using the menu to flip is flipItem: http://jongware.mit.edu/idcs6js/pc_PageItem.html#flipItem,
in which you can set "around" coordinates.
Now getting the origin of the selection box isn't straightforward for some reason (or maybe it is but I don't know how), but you can either use first object coordinates to set the origin so you can flip it around different coordinates depending on order of selection. Or you can sort the array to find left most coordinates (or whichever is needed), like this:
var selection_array = app.selection;
selection_array.sort(function(a, b){return a.visibleBounds[1]-b.visibleBounds[1]})
var flip_origin = [selection_array[0].visibleBounds[1],selection_array[0].visibleBounds[0]]
for(i=0;i<app.selection.length;i++){
app.selection[i].flipItem(Flip.HORIZONTAL, flip_origin);
}
Not sure it's easier or faster than grouping and ungrouping though.

Consider resize. It has a "individual/global" parameter :
void resize (in: varies, from: varies, by: ResizeMethods, values: Array of varies[, resizeIndividually: bool=true][, consideringRulerUnits: bool=false])
Resize the page item.

raphael move set multiple times

I have a set containing a circle,rectangle and a text
I can move it to specific location (50 points to the right) like this:
object.entireSet.transform("T50,0");
And it works just OK
Now I want to move it again (50 points to the right again)
object.entireSet.transform("T50,0");
BUT the object stays on the same place. If I want to move it like I want, I have to rewrite the command like this
object.entireSet.transform("T100,0");
So my thought here is, that the raphael somehow remember the original point (0,0) of transformation and therefore (T50,0) will always move to the same point.
Do you know how to reset the transformation, so following code
object.entireSet.transform("T50,0"); //first move
object.entireSet.transform("T50,0"); //second move
will result in an object moved from original point (x,y) to point (x+50,y) and then to (x+100,y)?

You can find the solution in the documentation :
http://raphaeljs.com/reference.html#Element.transform
set.transform("T50,0");
set.transform("...t50,0"); // This appends the transformation to the first one
jsFiddle here : http://jsfiddle.net/vyFC6/1
EDIT : I realised you may need a bit more explainations to understand why your code isn't working.
It has to do with the SVG nature of Raphael. You might want to quickly learn the basics of SVG to understand better some of Raphael's functionnalities.
All the transform calls you do on a same element actually update the value of a string, that's used to... well transform it.
When you do this :
set.transform("T50,0");
set.transform("T50,0");
The value of the string is "T50,0" after the first call. You just rewrite it with the second call ==> its value is still "T50,0" in the end. This is why it doesn't change.
When you do this :
set.transform("T50,0");
set.transform("...t50,0");
The value of the string becomes more or less this : "T50,0t50,0" which means in Raphael : translate 50 on x and 0 on y, THEN 50 on x and 0 on y.
To make it clear i updated my fiddle. You can find in it different transform calls that i hope will help you understand how it works.

paper.js: getting raw pixel data for two layers/groups?

I am trying to get the raw pixel data from two "items" in paper.js. One is already a Raster object so that's not too bad. The problem is that I have another Group object containing a bunch of triangles and I want to capture the Raster data for that layer and then be able to compare it.
I have the following (highlighted lines) code:
https://gist.github.com/mtahmed/2b27c4c6aee42d3ac3fb#file-paper_update-js
It seems to always return 0 or some other odd unexpected number. Any hints/ideas?
Thanks! :)

It looks like you are always setting child_gene.visible = false, but never set it back to visible = true before you rasterize the layer in computeFitness(). I'm not sure that there's a need for juggling layers in each frame - it should work just as well without it.
Here's a simplified example that uses a square with a gradient as the target raster.

d3.v3 scatterplot with all circles the same radius

Every example I have found shows all of the scatter plot points to be of random radii. Is it possible to have them all the same size? If I try to statically set the radius all of the circles will be very small (I'm assuming the default radius). However, if I use Math.random() as in most examples there are circles large and small. I want all the circles to be large. Is there a way to do that? Here's the code snippet forming the graph data using Math.random() (this works fine for some reason):
function scatterData(xData, yData)
{
var data = [];
for (var i = 0; i < seismoNames.length; i++)
{
data.push({
key: seismoNames[i],
values: []
});
var xVals=""+xData[i];
xVals=xVals.split(",");
var yVals=""+yData[i];
yVals=yVals.split(",");
for (var j = 0; j < xVals.length; j++)
{
data[i].values.push({
x: xVals[j],
y: yVals[j],
size: Math.random()
});
}
}
return data;
}
Math.random() spits out values between 0 and 1 such as 0.164259538891095 and 0.9842195005008699. I have tried putting these as static values in the 'size' attribute, but no matter what the circles are always really small. Is there something I'm missing?

Update: The NVD3 API has changed, and now uses pointSize, pointSizeDomain, etc. instead of just size. The rest of the logic for exploring the current API without complete documentation still applies.
For NVD3 charts, the idea is that all adjustments you make can be done by calling methods on the chart function itself (or its public components) before calling that function to draw the chart in a specific container element.
For example, in the example you linked too, the chart function was initialized like this:
var chart = nv.models.scatterChart()
.showDistX(true)
.showDistY(true)
.color(d3.scale.category10().range());
chart.xAxis.tickFormat(d3.format('.02f'));
chart.yAxis.tickFormat(d3.format('.02f'));
The .showDistX() and .showDistY() turn on the tick-mark distribution in the axes; .color() sets the series of colours you want to use for the different categories. The next too lines access the default axis objects within the chart and set the number format to be a two-digit decimal. You can play around with these options by clicking on the scatterplot option from the "Live Code" page.
Unfortunately, the makers of the NVD3 charts don't have a complete documentation available yet describing all the other options you can set for each chart. However, you can use the javascript itself to let you find out what methods are available.
Inspecting a NVD3.js chart object to determine options
Open up a web page that loads the d3 and nvd3 library. The live code page on their website works fine. Then open up your developer's console command line (this will depend on your browser, search your help pages if you don't know how yet). Now, create a new nvd3 scatter chart function in memory:
var testChart = nv.models.scatterChart();
On my (Chrome) console, the console will then print out the entire contents of the function you just created. It is interesting, but very long and difficult to interpret at a glance. And most of the code is encapsulated so you can't change it easily. You want to know which properties you can change. So run this code in the next line of your console:
for (keyname in testChart){console.log(keyname + " (" + typeof(testChart[keyname]) + ")");}
The console should now print out neatly the names of all the methods and objects that you can access from that chart function. Some of these will have their own methods and objects you can access; discover what they are by running the same routine, but replacing the testChart with testChart.propertyName, like this:
for (keyname in testChart.xAxis){console.log(keyname + " (" + typeof(testChart.xAxis[keyname]) + ")");}
Back to your problem. The little routine I suggested above doesn't sort the property names in any order, but skimming through the list you should see three options that relate to size (which was the data variable that the examples were using to set radius)
size (function)
sizeDomain (function)
sizeRange (function)
Domain and range are terms used by D3 scales, so that gives me a hint about what they do. Since you don't want to scale the dots, let's start by looking at just the size property. If you type the following in the console:
testChart.size
It should print back the code for that function. It's not terribly informative for what we're interested in, but it does show me that NVD3 follows D3's getter/setter format: if you call .property(value) you set the property to that value, but if you call .property() without any parameters, it will return back the current value of that property.
So to find out what the size property is by default, call the size() method with no parameters:
testChart.size()
It should print out function (d) { return d.size || 1}, which tells us that the default value is a function that looks for a size property in the data, and if it doesn't exist returns the constant 1. More generally, it tells us that the value set by the size method determines how the chart gets the size value from the data. The default should give a constant size if your data has no d.size property, but for good measure you should call chart.size(1); in your initialization code to tell the chart function not to bother trying to determine size from the data and just use a constant value.
Going back to the live code scatterplot can test that out. Edit the code to add in the size call, like this:
var chart = nv.models.scatterChart()
.showDistX(true)
.showDistY(true)
.color(d3.scale.category10().range())
.size(1);
chart.xAxis.tickFormat(d3.format('.02f'));
chart.yAxis.tickFormat(d3.format('.02f'));
Adding that extra call successfully sets all the dots to the same size -- but that size is definitely not 1 pixel, so clearly there is some scaling going on.
First guess for getting bigger dots would be to change chart.size(1) to chart.size(100). Nothing changes, however. The default scale is clearly calculating it's domain based on the data and then outputting to a standard range of sizes. This is why you couldn't get big circles by setting the size value of every data element to 0.99, even if that would create a big circle when some of the data was 0.01 and some was 0.99. Clearly, if you want to change the output size, you're going to have to set the .sizeRange() property on the chart, too.
Calling testChart.sizeRange() in the console to find out the default isn't very informative: the default value is null (nonexistent). So I just made a guess that, same as the D3 linear scale .range() function, the expected input is a two-element array consisting of the max and min values. Since we want a constant, the max and min will be the same. So in the live code I change:
.size(1);
to
.size(1).sizeRange([50,50]);
Now something's happening! But the dots are still pretty small: definitely not 50 pixels in radius, it looks closer to 50 square pixels in area. Having size computed based on the area makes sense when sizing from the data, but that means that to set a constant size you'll need to figure out the approximate area you want: values up to 200 look alright on the example, but the value you choose will depend on the size of your graph and how close your data points are to each other.
--ABR
P.S. I added the NVD3.js tag to your question; be sure to use it as your main tag in the future when asking questions about the NVD3 chart functions.

The radius is measured in pixels. If you set it to a value less than one, yes, you will have a very small circle. Most of the examples that use random numbers also use a scaling factor.
If you want all the circles to have a constant radius you don't need to set the value in the data, just set it when you add the radius attribute.
Not sure which tutorials you were looking at, but start here: https://github.com/mbostock/d3/wiki/Tutorials
The example "Three little circles" does a good step-by-step of the different things you can do with circles:
http://mbostock.github.io/d3/tutorial/circle.html

Create SVGPoint inside an element with user coordinate

I have a small project (to learn SVG) running (using javascript).
I would like to be able to track a point in a shape with its own user coordinate system. My idea is to find the coordinates of the point within the shape, then create an SVGPoint, so that I can pass on that element. I have seen the method create SVGPoint in examples, but it seems it is used in the context of the 'SVG_root' (that is, document.documentElement.createSVGPoint() works).
When I use (in Firefox)
inSvgObj.createSVGPoint()
where inSVGObj is a element, the web console says "TypeError: inSvgObj.createSVGPoint is not a function". Is it possible to create an SVG point within the to subsequently set with values representing coordinates in that 's user coordinate system?
EDIT (after considernig Robert Longson's answer):
Given that SVGPoint is created only within an "SVG root" and that I have been unable to find a way to move that to within another element, I have found more convenient to use a different svg element type: SVGMatrix. In case it helps someone (as I have spent some time trying to deal with this),It is possible to manipulate analogue values inside an SVG Point by creating an SVGMatrix that would work as a simulated point (for the purposes of coordinates. To that endthe methods .createSVGMatrix(), getCTM() and.multiply() (this last from SVGMatrix) are used. To illustrate that, I will include a (js) function that takes 4 arguments: x-coordinate in user coordinate system (ucs) to transform, y-coordinate is that ucs, object whose ucs is the want we want to transform and an object in the ucs we want to transform to; and returns am object with thrre poperties the x-coordinate in the transformed ucs, its y-coordinate and 1 (for consistency with SVG Recommendations).
function coorUcsAToUcsB(ucsAx,ucsAy,svgObjUcsA,svgObjUcsB){
var ctmUcsA=svgObjUcsA.getCTM();
var ctmUcsB=svgObjUcsB.getCTM().inverse();
var mtx=document.getElementsByTagName('svg')[0].createSVGMatrix();
mtx.e=ucsAx;
mtx.f=ucsAy;
var simulSvgP=ctmUcsB.multiply(ctmUcsA.multiply(mtx)); //1
return {"x":simulSvgP.e,"y":simulSvgP.f,"z":1};
}
//1 this line creates an svg matrix with 1st and 2nd column at 0, 3rd with coordinates of ucsB from the analogue svg matrix with coordinates in ucsA - it takes the coordinates in ucsA to viewport's cs and from there to coordinates in ucsB. For the matrix operation explanation, see this.
Any comments, in particular having overlooked a existing method that does the same or any drawbacks, will be more than welcome.

You create the SVG Point using the root element creation but once you've done that you can set whatever values in it you want. When you assign those values to an object the object will interpret them in its coordinate system.