d3.js v4: How to access parent group's datum index?

Well, sometimes an answer doesn't provide a solution, because the solution may not exist. This seems to be the case.

According to Bostock:

I’ve merged the new bilevel selection implementation into master and also simplified how parents are tracked by using a parallel parents array.

A nice property of this new approach is that selection.data can evaluate the values function in exactly the same manner as other selection functions: the values function gets passed {d, i, nodes} where this is the parent node, d is the parent datum, i is the parent (group) index, and nodes is the array of parent nodes (one per group). Also, the parents array can be reused by subselections that do not regroup the selection, such as selection.select, since the parents array is immutable.

This change restricts functionality—in the sense that you cannot access the parent node from within a selection function, nor the parent data, nor the group index — but I believe this is ultimately A Good Thing because it encourages simpler code.

(emphasis mine)

Here's the link: https://github.com/d3/d3-selection/issues/47

So, it's not possible to get the index of the parent's group using selection (the parent's group index can be retrieved using selection.data, as this snippet bellow shows).

var testData = [
[
  {x: 1, y: 40},
  {x: 2, y: 43},
  {x: 3, y: 12},
  {x: 6, y: 23}
], [
  {x: 1, y: 12},
  {x: 4, y: 18},
  {x: 5, y: 73},
  {x: 6, y: 27}
], [
  {x: 1, y: 60},
  {x: 2, y: 49},
  {x: 3, y: 16},
  {x: 6, y: 20}
 ] 
];

var svg = d3.select("body")
	.append("svg")
  .attr("width", 300)
  .attr("height", 300);
  
var g = svg.selectAll(".groups")
    .data(testData)
    .enter()
    .append("g");
    
var rects = g.selectAll("rect")
    .data(function(d, i , j) { console.log("Data: " + JSON.stringify(d), "\nIndex: " + JSON.stringify(i), "\nNode: " + JSON.stringify(j)); return d})
    .enter()
    .append("rect");

<script src="https://d3js.org/d3.v4.min.js"></script>

My workaround is somewhat similar to Dinesh Rajan's, assuming the parent index is needed for attribute someAttr of g.nestedElt:

v3:

svg.selectAll(".someClass")
    .data(nestedData)
  .enter()
  .append("g")
    .attr("class", "someClass")
  .selectAll(".nestedElt")
    .data(Object)
  .enter()
  .append("g")
    .attr("class", "nestedElt")
    .attr("someAttr", function(d, i, j) {

    });

v4:

svg.selectAll(".someClass")
    .data(nestedData)
  .enter()
  .append("g")
    .attr("class", "someClass")
    .attr("data-index", function(d, i) { return i; }) // make parent index available from DOM
  .selectAll(".nestedElt")
    .data(Object)
  .enter()
  .append("g")
    .attr("class", "nestedElt")
    .attr("someAttr", function(d, i) {
      var j = +this.parentNode.getAttribute("data-index");
    });

I ended up defining an external variable "j" and then increment it whenever "i" is 0

example V3 snippet below.

rowcols.enter().append("rect")
 .attr("x", function (d, i, j) { return CalcXPos(d, j); })
 .attr("fill", function (d, i, j) { return GetColor(d, j); })

and in V4, code converted as below.

var j = -1;
rowcols.enter().append("rect") 
 .attr("x", function (d, i) { if (i == 0) { j++ }; return CalcXPos(d, j); })
 .attr("fill", function (d, i) { return GetColor(d, j); })

If j is the nodeList...

• j[i] is the current node (eg. the td element),
• j[i].parentNode is the level-1 parent (eg. the row element),

• j[i].parentNode.parentNode is the level-2 parent (eg. the table element),

• j[i].parentNode.parentNode.childNodes is the array of level-1 parents (eg. array of row elements) including the original parent.

So the question is, what is the index of the parent (the row) with respect to it's parent (the table)?

We can find this using Array.prototype.indexOf like so...

k = Array.prototype.indexOf.call(j[i].parentNode.parentNode.childNodes,j[i].parentNode);

You can see in the snippet below that the row is printed in each td cell when k is returned.

var testData = [
[
  {x: 1, y: 1},
  {x: 1, y: 2},
  {x: 1, y: 3},
  {x: 1, y: 4}
], [
  {x: 2, y: 1},
  {x: 2, y: 2},
  {x: 2, y: 3},
  {x: 2, y: 4}
], [
  {x: 3, y: 4},
  {x: 3, y: 4},
  {x: 3, y: 4},
  {x: 3, y: 4}
 ]
];

var tableData =
  d3.select('body').selectAll('table')
    .data([testData]);

var tables =
  tableData.enter()
  .append('table');

var rowData =
  tables.selectAll('table')
    .data(function(d,i,j){
      return d;
    });

var rows =
  rowData.enter()
  .append('tr');

var eleData =
  rows.selectAll('tr')
    .data(function(d,i,j){
      return d;
    });

var ele =
  eleData.enter()
  .append('td')
    .text(function(d,i,j){
      var k = Array.prototype.indexOf.call(j[i].parentNode.parentNode.childNodes,j[i].parentNode);
      return k;
    });

<script src="https://d3js.org/d3.v4.min.js"></script>

Reservations

This approach is using DOM order as a proxy for data index. In many cases, I think this is a viable band-aid solution if this is no longer possible in D3 (as reported in this answer).

Some extra effort in manipulating the DOM selection to match data might be needed. As an example, filtering j[i].parentNode.parentNode.childNodes for <tr> elements only in order to determine the row -- generally speaking the childNodes array may not match the selection and could contain extra elements/junk.

While this is not a cure-all, I think it should work or could be made to work in most cases, presuming there is some logical connection between DOM and data that can be leveraged which allows you to use DOM child index as a proxy for data index.

Here's an example of how to use the selection.each() method. I don't think it's messy, but it did slow down the render on a large matrix. Note the following code assumes an existing table selection and a call to update().

update(matrix) {
        var self = this;
        var tr = table.selectAll("tr").data(matrix);

        tr.exit().remove();

        tr.enter().append("tr");

        tr.each(addCells);

        function addCells(data, rowIndex) {
            var td = d3.select(this).selectAll("td")
                .data(function (d) {
                    return d;
                });
            td.exit().remove();

            td.enter().append("td");

            td.attr("class", function (d) {
                return d === 0 ? "dead" : "alive";
            });

            td.on("click", function(d,i){
                matrix[rowIndex][i] = d === 1 ? 0 : 1; // rowIndex now available for use in callback.                   
            });
        }

        setTimeout(function() {
            update(getNewMatrix(matrix))
        }, 1000);
    },

Assume you want to do a nested selectiom, and your data is some array where each element in turn contains an array, let's say "values". Then you have probably some code like this:

var aInnerSelection = oSelection.selectAll(".someClass") //
    .data(d.values) //
    ...

You can replace the array with the values by a new array, where you cache the indices within the group.

var aInnerSelection = oSelection.selectAll(".someClass") //
    .data(function (d, i) {
        var aData = d.values.map(function mapValuesToIndexedValues(elem, index) {
                return {
                    outerIndex: i,
                    innerIndex: index,
                    datum: elem
                };
            })
            return aData;
    }, function (d, i) {
        return d.innerIndex;
    }) //
    ...

Assume your outer array looks like this: [{name "X", values: ["A", "B"]}, {name "y", values: ["C", "D"]}
With the first approach, the nested selection brings you from here

                  d                                i
------------------------------------------------------------------
root dummy X      {name "X", values: ["A", "B"]}    0
     dummy Y      {name "Y", values: ["C", "D"]}    1

to here.

             d       i
------------------------------------------------------------------
root X  A    "A"     0
        B    "B"     1
     Y  C    "C"     2
        D    "D"     3

With the augmented array, you end up here instead:

             d                                              i
------------------------------------------------------------------
root X  A    {datum: "A", outerIndex: 0, innerIndex: 0}     0
        B    {datum: "B", outerIndex: 0, innerIndex: 1}     1
     Y  C    {datum: "C", outerIndex: 1, innerIndex: 0}     2
        D    {datum: "D", outerIndex: 1, innerIndex: 1}     3

So you have within the nested selections, in any function(d,i), all information you need.

Here's a snippet I crafter after re-remembering this usage of .each for nesting, I thought it may be useful to others who end up here. This examples creates two layers of circles, and the parent group index is used to determine the color of the circles - white for the circles in the first layer, and black for the circles in the top layer (only two layers in this case).

const nested = nest().key(layerValue).entries(data);

let layerGroups = g.selectAll('g.layer').data(nested);
layerGroups = layerGroups.enter().append('g').attr('class', 'layer')
  .merge(layerGroups);

layerGroups.each(function(layerEntry, j) {   
  const circles = select(this)
    .selectAll('circle').data(layerEntry.values);
  circles.enter().append('circle')
    .merge(circles)
      .attr('cx', d => xScale(xValue(d)))
      .attr('cy', d => yScale(yValue(d)))
      .attr('r', d => radiusScale(radiusValue(d)))
      .attr('fill', j === 0 ? 'white' : 'black'); // <---- Access parent index.
});