http://www.perlmonks.org?node_id=1045611


in reply to Re^3: Understanding the Schwartzian transform.
in thread Understanding the Schwartzian transform.

but, definitely, the "output_array = map {block} sort {block} map {block} input array" is the canonical for of the Schwartzian Transform.

Sorry, that is like the generic form of Guttman-Rosler Transforms (see links) except without the secret ingredient :)(normalize/restore)

Schwartzian is 

my @sorted = map  { restore_from_cache    }
             sort { by_cached_expensive   }
             map  { store_cache_expensive } @original;

[download]

Replies are listed 'Best First'.
Re^5: Understanding the Schwartzian transform.
by Laurent_R (Canon) on Jul 22, 2013 at 15:54 UTC

    Sorry, but except that I did not detail what goes into the code blocks (as this has been detailed well enough the the previous posts), I do not see any difference between your version of the Schwartzian Transform and mine.

[reply]
      Because in GRT, you just change the data so they can be sorted lexicographically as strings, while in Schwartzian transform, you create a structure. You can simplify both to map {} sort {} map {} list (in GRT, sort block can be omitted), their difference can be seen only if you give more details.
      لսႽ† ᥲᥒ⚪⟊Ⴙᘓᖇ Ꮅᘓᖇ⎱ Ⴙᥲ𝇋ƙᘓᖇ
[reply]

        Agreed, but: 1. the OP did not ask for the difference between Guttman-Rosler and Schwartzian transforms, but only if MJD's code he had was a canonical ST. And this example is a canonical ST, which it what I said. The fact that I somewhat summarized for brevity (and because it could be other code for a different probklem and still be a ST) does not invalidate my point and certainly does not make the OP's initial code nor the summary that I made a GR transform. As you said, a GR transform would have omitted the sort block. I reallyt think that the OP question was whether the standard ST consists of the map-sort-map construct.

[reply]

In Section Seekers of Perl Wisdom