Saturday, February 02, 2008

Munging some XML with F# & Linq

I recently wanted to change the format of a test script, and canonicalize a flattened listing of all the regression test models. So I set myself the goal of doing it in F#, which would force me to use some of the language features I'd only read about, and possibly prod me to use some LINQ as well. I could have done this transformation manually in about an hour, I admit. And I learned that if I really want to transform a lot of XML to XML, I'd rather do it via XML, i.e. XSLT. However, Microsoft doesn't offer XSLT 2.0 compliancy, so I have less to gain from learning to do it that way (meaning, I'm stuck with the .NET XsltCompiledTransform implementation at work). What I would have needed is the tokenize function (to split up pathnames on the '\\' character), XSLT functions that could return Boolean values, and the ability to select groups. Anyway, that's what I did below.


open System
open System.Xml
open System.Linq
open System.Xml.Linq
open Microsoft.FSharp.Linq
open Microsoft.FSharp.Xml.Linq
open Microsoft.FSharp.Linq.SequenceOps
open Microsoft.FSharp.Xml.Linq.SequenceOps

let doc = XDocument.Load @"knownSolutions.xml"

// Simply tagged aliases for these various tuples. I want to create
// a tree of these, then finally convert them back to XML.
type Node =
| File of string * XElement
| Directory of string * Node list

// the temporary data structure: I'll create a list of these from
// the initial XML, then recursively group them into directories.
type Path = string list * XElement

let make_path (e : XElement) =
e.Element(xname "File") |> element_to_string |> String.split [ '\\' ], e

// I want to split up the path names, while carrying the original
// element around until I'm ready to use it.
let paths = doc.Root.Elements(xname "Model")
|> make_path
|> Seq.to_list // There's no Seq.partition, so...

let is_file (p, _) =
(List.length p) = 1

let strip (p, e) =
// Strip the next part of the path, and return this tuple. p, e

let separate (l : Path list) =
List.partition is_file l

let rec collect (l : Path list) =
match l with
| [] -> []
| _ -> let files, directories = separate l
let x = directories
|> Seq.groupBy (fun (l, _) -> List.hd l)
|> make_directories
|> Seq.to_list
(make_files files) @ x
and make_directories (l, r) =
Directory ( l, r |> strip |> Seq.to_list |> collect )
and make_files paths =
paths |> (fun (l, e) -> File (List.hd l, e))

let collected = paths |> collect

let make_xml l =
// Yes, you need the "let rec" here, too. That took me a while to
// figure out. Otherwise these nested functions can't call each other.
let rec make_file name (xml : XElement) =
// If you want to mix XElement and XAttribute objects, you'll have
// to upcast them to XObject. F#'s type inference won't try to find the
// lowest common denominator between two classes in a heterogeneous
// collection, as far as I can tell.
XElement(xname "File", xargs [XAttribute(xname "Name", name); XAttribute(xname "ObjectiveValue", xml.Element(xname "Solution") |> element_to_string)])
and make_directory name nodes =
let element = XElement(xname "Directory", xargs (nodes |> make_xml'))
element.SetAttributeValue(xname "Name", name)
and make_xml' e =
match e with
| File (name, element) -> make_file name element
| Directory (name, element) -> make_directory name element
match l with
| [] -> []
| _ -> make_xml' l

let xml = XElement(xname "KnownSolutions", make_xml collected)
let s = xml.ToString()

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