Re: Scheme question: convert a range
Hello David N. and Andrew,
Great, all your suggestions are of some use to me.
Thanks!
JM
> Le 17 nov. 2015 à 14:19, Andrew Bernard a écrit :
>
> Hi Jacques,
>
> You could base a solution on this approach:
>
> c'1 -\markup {
> \column {
> \column {
> #(string-append
> "commllen = "
> (string-concatenate (map number->string '(1 2
> }
> }
> }
>
>
> [Pardon the formatting - it’s late and my mailer won’t behave.]
>
> Andrew
>
>
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Re: Scheme question: convert a range
Hi Jacques,
You could base a solution on this approach:
c'1 -\markup {
\column {
\column {
#(string-append
"commllen = "
(string-concatenate (map number->string '(1 2
}
}
}
[Pardon the formatting - it’s late and my mailer won’t behave.]
Andrew
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Re: Scheme question: convert a range
Jacques,
On Tue, Nov 17, 2015 at 6:32 AM, Jacques Menu wrote:
> Message says that #(string-append… is not a markup.
>
> > Le 17 nov. 2015 à 13:31, Jacques Menu a écrit :
> >
> > Hello folks,
> >
> > I’ve tried to integrate such a pure Scheme function:
> >
> >
> > guile> (define (function arg)
> > (if (and (integer? (car arg)) (integer? (cdr arg)))
> > (iota (1+ (interval-length arg)) (car arg) 1)
> > )
> > )
> > guile> (function '(3 . 7))
> > (3 4 5 6 7)
> >
> >
> > as part of a markup, but to no avail. Here is one of my attempts:
> >
> >
> > \version "2.19.30"
> >
> > #(define (function arg)
> > (if (and (integer? (car arg)) (integer? (cdr arg)))
> > (iota (1+ (interval-length arg)) (car arg) 1)
> > )
> > )
> >
> > {
> > c'1 -\markup {
> >\column {
> > \column {
> >#(string-append
> > "commllen = " (list->sting (map #'number->string (function '(3
> . 7
> > )
> > }
> >}
> > }
> > }
> >
> >
> > Thanks for your help!
> >
> > JM
>
Why not:
{
c'1 -\markup {
\column {
\column {
#(string-append
"commllen = "
(format #f "~a" (function '(3 . 7)))
)
}
}
}
}
Regarding your approach:
First, it's "list->string" and that deals with a list of character data,
not strings.
You could do something like:
{
c'1 -\markup {
\column {
\column {
#(string-append
"commllen = "
(apply string-append (map number->string (function '(3 . 7
)
}
}
}
}
But I'm guessing that's not the output format you want.
Hope this helps,
DN
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Re: Scheme question: convert a range
Message says that #(string-append… is not a markup.
> Le 17 nov. 2015 à 13:31, Jacques Menu a écrit :
>
> Hello folks,
>
> I’ve tried to integrate such a pure Scheme function:
>
>
> guile> (define (function arg)
> (if (and (integer? (car arg)) (integer? (cdr arg)))
> (iota (1+ (interval-length arg)) (car arg) 1)
> )
> )
> guile> (function '(3 . 7))
> (3 4 5 6 7)
>
>
> as part of a markup, but to no avail. Here is one of my attempts:
>
>
> \version "2.19.30"
>
> #(define (function arg)
> (if (and (integer? (car arg)) (integer? (cdr arg)))
> (iota (1+ (interval-length arg)) (car arg) 1)
> )
> )
>
> {
> c'1 -\markup {
>\column {
> \column {
>#(string-append
> "commllen = " (list->sting (map #'number->string (function '(3 .
> 7
> )
> }
>}
> }
> }
>
>
> Thanks for your help!
>
> JM
>
>> Le 17 nov. 2015 à 02:10, Andrew Bernard a écrit :
>>
>> Hi Simon,
>>
>> Fellow listers have posted many answers while I was cooking up this one. All
>> good!
>>
>>
>> (use-modules (srfi srfi-1))
>>
>> (define (range r)
>> (let ((start (car r))
>> (end (cdr r)))
>> (iota (+ (- end start) 1) start 1)))
>>
>> That’s pure Scheme of course. Thomas Morley’s answer is more in the
>> vernacular of lilypond. But I just wanted to point out the list functions
>> you would expect to exist but don’t in R5RS are in SRFI-1. But lilypond
>> looks after all that for you.
>>
>>
>> The purely recursive solutions are nice, because lists are intrinsically
>> recursively defined, but Functional programmers tend avoid doing that and
>> prefer to hide the recursion behind generalised functions such as iota. A
>> lot easier on the brain and the maintainer. Just a matter of FP style. There
>> are many views!
>>
>> Andrew
>>
>>
>>
>>
>>
>>
>>
>>
>> ___
>> lilypond-user mailing list
>> lilypond-user@gnu.org
>> https://lists.gnu.org/mailman/listinfo/lilypond-user
>
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Re: Scheme question: convert a range
Hello folks,
I’ve tried to integrate such a pure Scheme function:
guile> (define (function arg)
(if (and (integer? (car arg)) (integer? (cdr arg)))
(iota (1+ (interval-length arg)) (car arg) 1)
)
)
guile> (function '(3 . 7))
(3 4 5 6 7)
as part of a markup, but to no avail. Here is one of my attempts:
\version "2.19.30"
#(define (function arg)
(if (and (integer? (car arg)) (integer? (cdr arg)))
(iota (1+ (interval-length arg)) (car arg) 1)
)
)
{
c'1 -\markup {
\column {
\column {
#(string-append
"commllen = " (list->sting (map #'number->string (function '(3 . 7
)
}
}
}
}
Thanks for your help!
JM
> Le 17 nov. 2015 à 02:10, Andrew Bernard a écrit :
>
> Hi Simon,
>
> Fellow listers have posted many answers while I was cooking up this one. All
> good!
>
>
> (use-modules (srfi srfi-1))
>
> (define (range r)
> (let ((start (car r))
> (end (cdr r)))
> (iota (+ (- end start) 1) start 1)))
>
> That’s pure Scheme of course. Thomas Morley’s answer is more in the
> vernacular of lilypond. But I just wanted to point out the list functions you
> would expect to exist but don’t in R5RS are in SRFI-1. But lilypond looks
> after all that for you.
>
>
> The purely recursive solutions are nice, because lists are intrinsically
> recursively defined, but Functional programmers tend avoid doing that and
> prefer to hide the recursion behind generalised functions such as iota. A lot
> easier on the brain and the maintainer. Just a matter of FP style. There are
> many views!
>
> Andrew
>
>
>
>
>
>
>
>
> ___
> lilypond-user mailing list
> lilypond-user@gnu.org
> https://lists.gnu.org/mailman/listinfo/lilypond-user
___
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Re: Scheme question: convert a range
Hi Simon, Fellow listers have posted many answers while I was cooking up this one. All good! (use-modules (srfi srfi-1)) (define (range r) (let ((start (car r)) (end (cdr r))) (iota (+ (- end start) 1) start 1))) That’s pure Scheme of course. Thomas Morley’s answer is more in the vernacular of lilypond. But I just wanted to point out the list functions you would expect to exist but don’t in R5RS are in SRFI-1. But lilypond looks after all that for you. The purely recursive solutions are nice, because lists are intrinsically recursively defined, but Functional programmers tend avoid doing that and prefer to hide the recursion behind generalised functions such as iota. A lot easier on the brain and the maintainer. Just a matter of FP style. There are many views! Andrew ___ lilypond-user mailing list lilypond-user@gnu.org https://lists.gnu.org/mailman/listinfo/lilypond-user
Re: Scheme question: convert a range
On Mon, Nov 16, 2015 at 3:22 PM, Simon Albrecht wrote: > On 16.11.2015 22:20, Thomas Morley wrote: > >> (define (foo pair) >>(if (and (integer? (car pair)) (integer? (cdr pair))) >>(iota (1+ (interval-length pair)) (car pair) 1)) >>#f) >> >> (foo '(3 . 7)) >> --> (3 4 5 6 7) >> > > An equally good solution. > Thank you, Simon > > Interesting that all these responses show up in my inbox hours after I sent mine, though these were sent earlier! ___ lilypond-user mailing list lilypond-user@gnu.org https://lists.gnu.org/mailman/listinfo/lilypond-user
Re: Scheme question: convert a range
On Mon, Nov 16, 2015 at 5:33 PM, David Nalesnik wrote: > >> If not, something like this would work: >> > (for all cases) ___ lilypond-user mailing list lilypond-user@gnu.org https://lists.gnu.org/mailman/listinfo/lilypond-user
Re: Scheme question: convert a range
On Mon, Nov 16, 2015 at 5:24 PM, David Nalesnik wrote: > Hi Simon, > > On Sun, Nov 15, 2015 at 12:53 PM, Simon Albrecht > wrote: > >> Hello, >> >> The subject certainly seems cryptic – it’s difficult to summarize, but an >> example will make it clear immediately. >> I want to write a scheme procedure, which takes a pair like #'(3 . 7) and >> returns a list with all the numbers in the range: #'(3 4 5 6 7) >> How is this done most easily? >> >> > Well, if the numbers are ascending you could do something like: > > (define (expand-interval arg) >(iota (- (1+ (cdr arg)) (car arg)) (car arg))) > > (display (expand-interval '(3 . 7))) > > If not, something like this would work: > (define (expand-interval arg) (let ((step (if (> (car arg) (cdr arg)) -1 1))) (iota (1+ (abs (- (cdr arg) (car arg (car arg) step))) --David ___ lilypond-user mailing list lilypond-user@gnu.org https://lists.gnu.org/mailman/listinfo/lilypond-user
Re: Scheme question: convert a range
Hi Simon, On Sun, Nov 15, 2015 at 12:53 PM, Simon Albrecht wrote: > Hello, > > The subject certainly seems cryptic – it’s difficult to summarize, but an > example will make it clear immediately. > I want to write a scheme procedure, which takes a pair like #'(3 . 7) and > returns a list with all the numbers in the range: #'(3 4 5 6 7) > How is this done most easily? > > Well, if the numbers are ascending you could do something like: (define (expand-interval arg) (iota (- (1+ (cdr arg)) (car arg)) (car arg))) (display (expand-interval '(3 . 7))) If not, something like this would work: (define (expand-interval arg) (let ((dir (if (> (car arg) (cdr arg)) -1 1))) (let loop ((elt (car arg)) (result '())) (if (= elt (+ dir (cdr arg))) (reverse result) (loop (+ dir elt) (cons elt result)) HTH, David ___ lilypond-user mailing list lilypond-user@gnu.org https://lists.gnu.org/mailman/listinfo/lilypond-user
Re: Scheme question: convert a range
2015-11-15 19:53 GMT+01:00 Simon Albrecht : > Hello, > > The subject certainly seems cryptic – it’s difficult to summarize, but an > example will make it clear immediately. > I want to write a scheme procedure, which takes a pair like #'(3 . 7) and > returns a list with all the numbers in the range: #'(3 4 5 6 7) > How is this done most easily? > > TIA, Simon Hi Simon, (define (foo pair) (if (and (integer? (car pair)) (integer? (cdr pair))) (iota (1+ (interval-length pair)) (car pair) 1)) #f) (foo '(3 . 7)) --> (3 4 5 6 7) HTH, Harm ___ lilypond-user mailing list lilypond-user@gnu.org https://lists.gnu.org/mailman/listinfo/lilypond-user
Re: Scheme question: convert a range
Simon Albrecht writes: > The subject certainly seems cryptic – it’s difficult to summarize, but > an example will make it clear immediately. > I want to write a scheme procedure, which takes a pair like #'(3 . 7) > and returns a list with all the numbers in the range: #'(3 4 5 6 7) > How is this done most easily? You mean something like this? #+BEGIN_SRC scheme :results output (define pair (cons 3 7)) (define (range first last) (if (>= first (+ last 1)) '() (cons first (range (+ first 1) last (display (range (car pair) (cdr pair))) #+END_SRC #+RESULTS: : (3 4 5 6 7) HTH Patrick ___ lilypond-user mailing list lilypond-user@gnu.org https://lists.gnu.org/mailman/listinfo/lilypond-user
Re: Scheme question: convert a range
On 16.11.2015 23:30, pls wrote: Simon Albrecht writes: The subject certainly seems cryptic – it’s difficult to summarize, but an example will make it clear immediately. I want to write a scheme procedure, which takes a pair like #'(3 . 7) and returns a list with all the numbers in the range: #'(3 4 5 6 7) How is this done most easily? You mean something like this? #+BEGIN_SRC scheme :results output (define pair (cons 3 7)) (define (range first last) (if (>= first (+ last 1)) '() (cons first (range (+ first 1) last (display (range (car pair) (cdr pair))) #+END_SRC #+RESULTS: : (3 4 5 6 7) Yeah, that’s the recursion way to do it – and without any SRFI. Thank you, Simon ___ lilypond-user mailing list lilypond-user@gnu.org https://lists.gnu.org/mailman/listinfo/lilypond-user
Re: Scheme question: convert a range
On 16.11.2015 22:20, Thomas Morley wrote: (define (foo pair) (if (and (integer? (car pair)) (integer? (cdr pair))) (iota (1+ (interval-length pair)) (car pair) 1)) #f) (foo '(3 . 7)) --> (3 4 5 6 7) An equally good solution. Thank you, Simon ___ lilypond-user mailing list lilypond-user@gnu.org https://lists.gnu.org/mailman/listinfo/lilypond-user
Re: Scheme question: convert a range
On 16.11.2015 21:59, David Kastrup wrote: Simon Albrecht writes: Hello, The subject certainly seems cryptic – it’s difficult to summarize, but an example will make it clear immediately. I want to write a scheme procedure, which takes a pair like #'(3 . 7) and returns a list with all the numbers in the range: #'(3 4 5 6 7) How is this done most easily? (let ((x '(3 . 7))) (iota (- (cdr x) (car x) -1) (car x))) Nice! Thanks. Yours, Simon ___ lilypond-user mailing list lilypond-user@gnu.org https://lists.gnu.org/mailman/listinfo/lilypond-user
Re: Scheme question: convert a range
(iota 7 3) Am 15.11.2015 um 19:53 schrieb Simon Albrecht: > Hello, > > The subject certainly seems cryptic – it’s difficult to summarize, but > an example will make it clear immediately. > I want to write a scheme procedure, which takes a pair like #'(3 . 7) > and returns a list with all the numbers in the range: #'(3 4 5 6 7) > How is this done most easily? > > TIA, Simon > > ___ > lilypond-user mailing list > lilypond-user@gnu.org > https://lists.gnu.org/mailman/listinfo/lilypond-user ___ lilypond-user mailing list lilypond-user@gnu.org https://lists.gnu.org/mailman/listinfo/lilypond-user
Re: Scheme question: convert a range
Simon Albrecht writes: > Hello, > > The subject certainly seems cryptic – it’s difficult to summarize, but > an example will make it clear immediately. > I want to write a scheme procedure, which takes a pair like #'(3 . 7) > and returns a list with all the numbers in the range: #'(3 4 5 6 7) > How is this done most easily? (let ((x '(3 . 7))) (iota (- (cdr x) (car x) -1) (car x))) -- David Kastrup ___ lilypond-user mailing list lilypond-user@gnu.org https://lists.gnu.org/mailman/listinfo/lilypond-user
Scheme question: convert a range
Hello, The subject certainly seems cryptic – it’s difficult to summarize, but an example will make it clear immediately. I want to write a scheme procedure, which takes a pair like #'(3 . 7) and returns a list with all the numbers in the range: #'(3 4 5 6 7) How is this done most easily? TIA, Simon ___ lilypond-user mailing list lilypond-user@gnu.org https://lists.gnu.org/mailman/listinfo/lilypond-user
