hexbin as a transform

[Fil]

am i on the right track?

(I can't see how to access Z )

[mbostock]

I don’t think this approach will work… it’ll lead to the same duplicate computation of channels problem we had in the mbostock/layout branch. I need to think about this more.

[mbostock]

I think probably the hexbin will need to declare a transform, too, and redefine any channels that are needed by outputs so that they are cached. It can then reference those values in the initializer.

[mbostock]

To elaborate, I think we’ll need something more like this:

{
  const z = "species";
  const [Z, setZ] = Plot.channel(z);
  return Plot.plot({
    grid: true,
    marks: [
      Plot.dot(penguins, {
        x: "culmen_depth_mm",
        y: "culmen_length_mm",
        z: Z, // optionally make the z channel available to dot (no effect here)
        symbol: "hexagon",
        transform(data, facets) {
          setZ(Plot.valueof(data, z)); // compute the z channel from the data
          return {data, facets};
        },
        initialize([index], {x: {value: X}, y: {value: Y}}, {x, y}) {
          console.log(Z.transform()); // access the previously-computed z values
          const bins = Hexbin().x(i => x(X[i])).y(i => y(Y[i])).radius(20)(index);
          return {
            facets: [d3.range(bins.length)],
            channels: {
              x: {value: bins.map(bin => bin.x)},
              y: {value: bins.map(bin => bin.y)},
              r: {value: bins.map(bin => bin.length), radius: 20, scale: "r"}
            }
          };
        }
      })
    ]
  });
}

To break it down:

1. The options transform would (optionally) declare the z channel as a “lazy” channel via Plot.channel.
2. The mark transform would compute the z channel. (This part is composable with other transforms.)
3. The mark initializer could then access the previously-computed z channel values.

Because the channels are computed in a mark transform, they’ll have access to the (possibly transformed) data, so we won’t need to pass that data separately to the initializer. And because the options transform can declare the channel as an option, it won’t be computed twice.

[Fil]

superseded by #810 #819