As a sort of dare from our friendly neighbours in the Illustrator forum, I wrote this script that might be useful for InDesign users as well.
It creates a pie chart by asking for the series of values, and then slicing up a circle that you must have selected. Fun thing is, it doesn't "just" work with circles; you can select about anything, enter your values at the prompt (comma separated), and get the correct result! (Which depends on the shape; of course it's only mathematically correct when using a circle.)
//DESCRIPTION:Make me a pie// A Jongware script, 25-mar-2012 if (app.selection.length == 1){ val = prompt ("Values: ", "1,2,3"); if (val != null) { values = val.split(","); total = 0; for (i=0; i<values.length; i++) { values[i] = Number(values[i]); total += values[i]; } for (i=0; i<values.length; i++) { values[i] = values[i]*2*Math.PI/total; } circleCenter = [ (app.selection[0].geometricBounds[3]+app.selection[0].geometricBounds[1])/2, (app.selection[0].geometricBounds[2]+app.selection[0].geometricBounds[0])/2 ]; circleRadius = Math.max ( (app.selection[0].geometricBounds[3]-app.selection[0].geometricBounds[1])/2, (app.selection[0].geometricBounds[2]-app.selection[0].geometricBounds[0])/2); startangle = 0; for (i=0; i<values.length; i++) { endangle = startangle + values[i]; // warning: math! (though this is SIMPLE compared to what I've bin doing today) segstart = [ circleCenter[0]+ 2*circleRadius*Math.sin(startangle), circleCenter[1]- 2*circleRadius*Math.cos(startangle) ]; segend = [ circleCenter[0]+ 2*circleRadius*Math.sin(endangle), circleCenter[1]- 2*circleRadius*Math.cos(endangle) ]; cutoff = app.activeDocument.graphicLines.add(); // create surrounding border path = [ circleCenter, segstart ]; for (j=startangle; j<endangle; j += (values[i]/16)) path = path.concat ([[ circleCenter[0]+2*circleRadius*Math.sin(j),circleCenter[1]-2*circleRadius*Math.cos(j)]] ); path = path.concat ([segend]); cutoff.paths[0].entirePath = path; cutoff.paths[0].pathType = PathType.CLOSED_PATH; cutoff.intersectPath(app.selection[0].duplicate()).fillTint = (i+1)*100/values.length; startangle = endangle; } }}