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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.

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  • $\begingroup$ @gwr sorry, I don't see your point. That function acting at that data should work, for example, Block[{k = 2, l}, l = #data[[k]]; <|#, "part2" -> l|>] & /@ Normal@ds works too. $\endgroup$
    – rhermans
    Commented Mar 20, 2017 at 16:06
  • $\begingroup$ @gwr Thanks for your comments, but the problem is not to find something that works, I offered myself three examples that do work. The question was Where does this error come from? $\endgroup$
    – rhermans
    Commented Mar 20, 2017 at 16:25
  • $\begingroup$ Looking at what is going on with trace I note some rather complex evaluation with HoldComplete ... and Dataset-validity checking going on. Indeed beats me and for all this I usually always use Associationand lists of them to do a Query saving Dataset only for display purposes. Good luck then. $\endgroup$
    – gwr
    Commented Mar 20, 2017 at 16:28
  • 1
    $\begingroup$ Related: (87360) $\endgroup$
    – gwr
    Commented Mar 20, 2017 at 16:40
  • 3
    $\begingroup$ similar, alas: (85114), (85270), (88503), (126010), (130330), (137129) $\endgroup$
    – WReach
    Commented Mar 22, 2017 at 5:27

1 Answer 1

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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.

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