Hi,
I almost don’t know any Scheme but I think this could be done with a
map, a fold, and some function composition. In Haskell, you would do this:
validate = foldr (&&) True . map ((== 2) . length)
validateShort = and . map ((== 2) . length)
The function and is a shortcut defined in Haskell’s Prelude. I don’t
know how, but something similar must be possible in Scheme ;)
Malte
On 15.05.2014 10:09, Urs Liska wrote:
Hi all,
I am working on a Scheme function and would like to check if I have
found the best solution for a specific subpart. Somehow it looks more
complicated than necessary.
The function needs to test if each element of a given list is a
(sub)list with exactly two elements. So
'((1 2)(3 4))
would return #t while
'((5 6)(7 8 9))
would return #f.
My solution is
\version "2.19.6"
validate =
#(define-scheme-function (parser location lst) (list?)
(if (memv #f (map (lambda sig
(and (list? (car sig))
(= 2 (length (car sig))))) lst))
(display "invalid")
(display "valid"))
(newline))
{
\validate #'((1 2)(3 4))
\validate #'((5 6)(7 8 9))
}
The "framework" doesn't matter, it's just a compilable example. My
question is only about the "if" expression.
What it does is:
- go through the elements of lst
- produce a list of boolean values,
- #t if we have a two element list,
- #f if not
- check if this intermediate list contains at least one #f
Somehow this looks clumsy to me, and I'd like to know (and learn) if
there's a better solution for this.
TIA
Urs
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