[SCRIPT] How to select most top/left object

[lfcorullon]

Hello gurus,

I need to select the most top/left object in a page.

The Y is the tiebreaker. Any suggestion?

I tried sorting by geometricBounds[0] and geometricBounds[1] but can not accomplish what I need.

Thanks in advance.

[lfcorullon · Accepted solution]

I have no idea why, but I got it.

Tired of playing with this sort function, I put this absurd and it works.

If anyone understand the logic behind this function, I'd love to know.

var doc = app.activeDocument;
var pg = app.activeWindow.activePage;
getFirstImg(pg);

function getFirstImg(pg) {
	var gbs = [];
	for (var i=0; i<pg.rectangles.length; i++) {
		if (pg.rectangles[i].itemLayer == img && pg.rectangles[i].graphics.length) {
			var recgb = pg.rectangles[i].geometricBounds;
			gbs.push([recgb[0] , recgb[1] , pg.rectangles[i].graphics[0].itemLink.name , pg.rectangles[i]]);
			}
		}
	gbs.sort(function(a, b) { return (a[0] - b[0]) + (a[1] - b[1]); });
	app.select(gbs[0][3]);
	}

[Peter Kahrel]

You create an array of subarrays, sort the array by top and left coordinates, then take the first item, which is gbs[0]. This array item is an array itself with the rectangle as the fourth element, which you address by gbs[0][3].

[Marc Autret]

Hi @lfcorullon 

In fact the elements are sorted by comparing their (top+left) sum. Just note that the sort key (a[0]–b[0])+(a[1]–b[1]) is nothing but (a[0]+a[1])–(b[0]+b[1]), that is, (top+left)₀ – (top+left)₁. This is on average a good approximation when the X and Y axes have the same priority. In the below example you can see that the winner is the rectangle that reaches the minimal sum (Σ=340.)

However, using the sort method for this purpose is clearly useless and expensive, since the minimum value can always be found in linear time O(N)—while sort relies on a loglinear algorithm, i.e. O(N×log(N)). I think you could just do as follows:

function selectFirstImage(/*?Layer|str*/ly,/*?Document*/doc,/*?Page*/pg,  t,a,b,c,i,iBest,xyMin)
//----------------------------------
// Select and return the (top+left)most image container on a page.
// `ly`   : Target layer or layer name (opt.)
// `doc`  : Target document (default=active.)
// `pg`   : Target page (default=active.)
// ---
// => SplineItems [OK]  |  false [KO]
{
    // Checkpoint -> doc, pg
    // ---
    doc || (doc=app.properties.activeDocument||0);
    if( !doc.isValid || 'Document' != doc.constructor.name ) return false;
    
    pg || ((pg=app.properties.activeWindow)&&(pg=pg.properties.activePage)) || (pg=doc.pages[0]);
    if( !pg.isValid || 'Page' != pg.constructor.name ) return false;
    
    // Layer filter (optional.)
    // ---
    'string'==typeof ly && ly.length && (ly=doc.layers.itemByName(ly));
    ( ly && 'Layer'==ly.constructor.name && ly.isValid ) || (ly=void 0);
    
    // Browse the SplineItems collection.
    // ---
    t = pg.splineItems.everyItem();
    if( !t.isValid ) return false;
    a = t.getElements();
    b = ly ? t.itemLayer : [];
    
    // Identify the object having the minimal (x+y) sum.
    // ---
    for
    (
        i=a.length, iBest=false, xyMin=1/0 ;
        i-- ;
        ly===b[i] && (t=a[i]).graphics.length && (t=t.geometricBounds)
        && xyMin > (t=t[0]+t[1]) && (xyMin=t, iBest=i)
    );
    
    return false !== iBest && (app.select(t=a[iBest]), t);
}

// Example:
selectFirstImage('Layer 1');

[lfcorullon]

Wow, thank you @Marc Autret!!!

Your knowledge makes em feel illiterate.

I tried your code in two different documents and nothing is being selected.

I just typed the layer name in the function call...

[lfcorullon]

Now I'm having the same difficult to select elements in row order.

Any help?

[brian_p_dts · Accepted solution]

For anyone wondering, here's the solution I came up with for Luis on a Facebook thread. The yThresh accounted for objects not perfectly aligned on the top (in mm), but could be set  to whatever threshold you felt appropriate. 

var apis = app.activeDocument.pages[0].pageItems.everyItem().getElements();
apis.sort(function (a,b) {
var agb = a.geometricBounds;
var bgb = b.geometricBounds;
var yThresh = Math.abs(agb[0] - bgb[0]);
if (yThresh > 5) {
if (agb[0] > bgb[0]) return 1;
else if (bgb[0] > agb[0]) return -1;
}
return agb[1] > bgb[1];
});