Dividing aggregate by aggregate (D3 / DC/ Crossfilter / Reductio)

Question

Fiddle for Reference: https://jsfiddle.net/33228p1d/4/

Referencing Cost Average chart on the page. Chart rendered with the following code:

var cpa = ndx.dimension(function (d) {
  return d.location;
});
var cpaGroup = cpa.group().reduceSum(function (d) {
  return d.total;
});

And the chart itself: (styling left off for brevity)

cpaChart
    .dimension(cpa)
    .group(cpaGroup)
    .label(function (d) {
        return d.key + " $" + d.value;
    })
    .title(function (d) {
        return d.value;
    })
    .elasticX(true)
    .ordering(function (d) { return -d.value })
    .xAxis().ticks(4);

Right now, it is showing the total for Front and Back locations accurately.

What I want it to do is show total divided by number of unique PO's for that location to show the average cost per location.

So, for example, for Back I have two unique purchase orders totaling $396. What I would like to see is 396 / 2 for an average cost per location of 198.

score 2 · Accepted Answer · answered Apr 26 '16 at 21:51

If you're still willing to use Reductio, track sum and exceptionCount on the same group and then divide them in the valueAccessor accessor:

  var cpaGroup = reductio()
    .sum('total')
    .exception(function(d) { return d.po; })
      .exceptionCount(true)(cpa.group());

  cpaChart
        .width(400)
        .height(200)
        .margins({ top: 20, left: 10, right: 10, bottom: 20 })
        .dimension(cpa)
        .group(cpaGroup)
        .ordinalColors(['#3182bd', '#6baed6', '#9ecae1', '#c6dbef', '#dadaeb'])
        .label(function (d) {
            return d.key + " $" + (Math.round(d.value.sum / d.value.exceptionCount));
        })
        .title(function (d) {
            return d.value;
        })
        .valueAccessor(function(d) {
            return d.value.sum / d.value.exceptionCount;
        })
        .elasticX(true)
        .ordering(function (d) { return -d.value })
        .xAxis().ticks(4);

Example: https://jsfiddle.net/7wbcref9/

Dividing aggregate by aggregate (D3 / DC/ Crossfilter / Reductio)

1 Answers1