Skip to main content

Explaining error when using Part inside Module in Dataset


I have a vague recollection of seen an explanation for this, but I can not find it, so it may be a false memory. Will delete if duplicated.


Let ds be a simple Dataset



ds = Dataset@Table[
<|"index" -> i, "data" -> RandomReal[1, 4]|>
, {i, 3}]

This does NOT work (error message: k is not a valid part specification, Tag: dataset)


ds[All, Module[{k = 2, l}, l = #data[[k]]; <|#, "part2" -> l|>] &]


Mathematica graphics




But these very similar expression do work fine


ds[All, Module[{k = 1, l}, <|#, "part2" -> #data[[k + 1]]|>] &] (* Example 1 *)

ds[All, <|#, "part2" -> Module[{k = 2, l}, l = #data[[k]]]|> &] (* Example 2 *)

ds[All, Module[{k = 1, l},l = Extract[#data, {k + 1}]; <|#, "part2" -> l|>] &] (* Example 3 *)


Mathematica graphics




Using Block instead of Module gives the same result.


Where does this error come from?


Using Mathematica 11.0.1 on Windows 7 64 bit.



Answer



Cause


The problem is caused by a type-inferencing failure. When presented with a complex expression, the type inferencer will sometimes give up and tag the expression as being of type UnknownType. But at other times, the inferencer will fail outright. Expressions of the form <|#, ... |> or Extract[...] are identified as UnknownType. In constrast, expressions like expr[[k]] fall into the "failure" category when k is a symbol.


The expression that fails contains a "naked" reference to #data[[k]]. In working examples #1 and #2, the problematic part reference is contained within an outer expression <|#, ... |> -- so the type inferencer gives up (yielding UnknownType) before the part reference can fail. Working example #3 omits the Part reference altogether and uses Extract instead (which also yields UnknownType).


Work-around


The usual work-around for Dataset type system failures applies... dodge the type system by using Query on the raw data and wrapping Dataset around the result for visualization:


ds //

Normal //
Query[All, Module[{k=2, l}, l=#data[[k]]; <|#, "part2"->l|>]&] //
Dataset

dataset screenshot


Analysis


current as of Mathematica v11.1.0


As part of the operation of a Dataset query, the system first performs some type checks to do some "sanity checks". In the case at hand, these checks fail outright:


ds[All, Module[{k = 2, l}, l = #data[[k]]; <|#, "part2" -> l|>] &]
(* Failure[...] *)


We can reduce this to a smaller example:


ds[All, Module[{k = 1}, #data[[k]]] &]
(* Failure[...] *)

Or even:


Dataset[{0}][Module[{k = 1}, #[[k]]] &]
(* Failure[...] *)

The cause is a failure of the type-inferencing function TypeApply:



Needs["TypeSystem`"]

TypeApply[{0}[[k]]&, {}]
(* FailureType[{Part, "Spec"}, <|"Type" -> Tuple[{Atom[Integer]}], "Part" -> k|>] *)

The full sequence of events looks like this:


stack excerpt


The component that is letting us down is TypePart. It can handle indices that are scalars, vectors or types:


TypePart[Tuple[{Atom[Integer]}], 1]
(* Atom[Integer] *)


TypePart[Tuple[{Atom[Integer]}], {1, 1}]
(* Tuple[{Atom[Integer], Atom[Integer]}] *)

TypePart[Tuple[{Atom[Integer]}],Atom[Integer]]
(* AnyType *)

But it fails outright when presented with a symbol:


TypePart[Tuple[{Atom[Integer]}], k]
(* FailureType[...] *)


I think it is arguable that this is a bug... a symbolic index should probably result in AnyType under the assumption that it will evaluate to a valid index at run time.


The working examples (1, 2, and 3) all "hide" the problematic part reference from the type inferencer by surrounding it by a troublesome, but not failing, outer expression:


TypeApply[<|#, "a" -> #a[[k]]|> &, DeduceType /@ {<|"a" -> {1, 2}|>}]
(* UnknownType *)

TypeApply[Extract[#a, {k + 1}] &, DeduceType /@ {<|"a" -> {1, 2}|>}]
(* UnknownType *)

The type inferencer is still confused, returning UnknownType, but at least it does not fail outright. The vague type information has no bearing upon the query evaluation process so we get the expected results. (UnknownType has been known to cause Dataset rendering issues from time-to-time.)



The Dataset type system tends to be (overly?) conservative when type-checking, but Query does no type-checking at all. So Query remains a viable work-around for these kinds of errors.


Comments

Popular posts from this blog

plotting - Filling between two spheres in SphericalPlot3D

Manipulate[ SphericalPlot3D[{1, 2 - n}, {θ, 0, Pi}, {ϕ, 0, 1.5 Pi}, Mesh -> None, PlotPoints -> 15, PlotRange -> {-2.2, 2.2}], {n, 0, 1}] I cant' seem to be able to make a filling between two spheres. I've already tried the obvious Filling -> {1 -> {2}} but Mathematica doesn't seem to like that option. Is there any easy way around this or ... Answer There is no built-in filling in SphericalPlot3D . One option is to use ParametricPlot3D to draw the surfaces between the two shells: Manipulate[ Show[SphericalPlot3D[{1, 2 - n}, {θ, 0, Pi}, {ϕ, 0, 1.5 Pi}, PlotPoints -> 15, PlotRange -> {-2.2, 2.2}], ParametricPlot3D[{ r {Sin[t] Cos[1.5 Pi], Sin[t] Sin[1.5 Pi], Cos[t]}, r {Sin[t] Cos[0 Pi], Sin[t] Sin[0 Pi], Cos[t]}}, {r, 1, 2 - n}, {t, 0, Pi}, PlotStyle -> Yellow, Mesh -> {2, 15}]], {n, 0, 1}]

plotting - Plot 4D data with color as 4th dimension

I have a list of 4D data (x position, y position, amplitude, wavelength). I want to plot x, y, and amplitude on a 3D plot and have the color of the points correspond to the wavelength. I have seen many examples using functions to define color but my wavelength cannot be expressed by an analytic function. Is there a simple way to do this? Answer Here a another possible way to visualize 4D data: data = Flatten[Table[{x, y, x^2 + y^2, Sin[x - y]}, {x, -Pi, Pi,Pi/10}, {y,-Pi,Pi, Pi/10}], 1]; You can use the function Point along with VertexColors . Now the points are places using the first three elements and the color is determined by the fourth. In this case I used Hue, but you can use whatever you prefer. Graphics3D[ Point[data[[All, 1 ;; 3]], VertexColors -> Hue /@ data[[All, 4]]], Axes -> True, BoxRatios -> {1, 1, 1/GoldenRatio}]

plotting - Adding a thick curve to a regionplot

Suppose we have the following simple RegionPlot: f[x_] := 1 - x^2 g[x_] := 1 - 0.5 x^2 RegionPlot[{y < f[x], f[x] < y < g[x], y > g[x]}, {x, 0, 2}, {y, 0, 2}] Now I'm trying to change the curve defined by $y=g[x]$ into a thick black curve, while leaving all other boundaries in the plot unchanged. I've tried adding the region $y=g[x]$ and playing with the plotstyle, which didn't work, and I've tried BoundaryStyle, which changed all the boundaries in the plot. Now I'm kinda out of ideas... Any help would be appreciated! Answer With f[x_] := 1 - x^2 g[x_] := 1 - 0.5 x^2 You can use Epilog to add the thick line: RegionPlot[{y < f[x], f[x] < y < g[x], y > g[x]}, {x, 0, 2}, {y, 0, 2}, PlotPoints -> 50, Epilog -> (Plot[g[x], {x, 0, 2}, PlotStyle -> {Black, Thick}][[1]]), PlotStyle -> {Directive[Yellow, Opacity[0.4]], Directive[Pink, Opacity[0.4]],