[racket] Regexp: Ways to reduce number of list elements returned?

From: Greg Hendershott (greghendershott at gmail.com)
Date: Mon May 13 11:37:08 EDT 2013

I've taken to using `match` for this. Your example:

(regexp-match #rx"\"(.*)" "a\"b")

I would write as:

(match "a\"b"
  [(regexp "\"(.*)" (list _ x)) x])
; "b"

Handling the no-match case by returning #f:

(match "ab"
  [(regexp "\"(.*)" (list _ x)) x]
  [_ #f])
; #f

Laurent's approach of defining a function works well, too. But I often
find myself matching multiple items, and use the list to bind to
multiple variables:

(match "x=y"
  [(pregexp "(.+)\\s*=\\s*(.+)" (list _ k v)) (values k v)]
  [_ #f])
; "x"
; "y"

So I like match as it works in the general case.

Note that there's `pregexp` as well as `regexp`, like I used here.

On Fri, May 10, 2013 at 11:27 AM, Don Green <infodeveloperdon at gmail.com> wrote:
> Regexp question:
> Is there a way using regexp only to return a list with a single element?
> I could use a Racket list function such as caar to return the second element
> but I am wondering if there is a way to do this using regexp only.
> For example this regexp-match function generates the "b" that I want but it
> is the second element in the list.  It would be ideal, from my perspective,
> if that was all it generated. Is there a way to write the regexp-match
> expression so that '("b") is output?
> I get this:
> (regexp-match #rx"\"(.*)" "a\"b") ; => '("\"b" "b")
> I'd prefer:
> (regexp...                      "a\"b" ) ; => '("b")
> Thanks.
> Don Green
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