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Generalizing append_map

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Merciadri Luca

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Apr 6, 2012, 8:32:38 AM4/6/12
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Hi,

I've got

==
(define append_map
(lambda (f l)
(if (null? l) '() (append (f (car l)) (append_map f (cdr l))))))
==

which is equivalent to

==
(define append_map
(lambda (f l) (apply append (map f l))))
==

I now want to generalize this to an arbitrary number of arguments. I
know this can be done using (lambda v) at the place of (lambda (v))
(this is called `generalized lambda form'). How could I use this in
this context (e.g. one function f and an arbitrary number of lists)?

Thanks.
- --
Merciadri Luca
See http://www.student.montefiore.ulg.ac.be/~merciadri/
- --

If all else fails, try the obvious.
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Jussi Piitulainen

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Apr 6, 2012, 11:43:51 AM4/6/12
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Merciadri Luca writes:

> Hi,
>
> I've got
>
> ==
> (define append_map
> (lambda (f l)
> (if (null? l) '() (append (f (car l)) (append_map f (cdr l))))))
> ==
>
> which is equivalent to
>
> ==
> (define append_map
> (lambda (f l) (apply append (map f l))))
> ==
>
> I now want to generalize this to an arbitrary number of arguments. I
> know this can be done using (lambda v) at the place of (lambda (v))
> (this is called `generalized lambda form'). How could I use this in
> this context (e.g. one function f and an arbitrary number of lists)?

With care. I suggest writing append-map-1 (what you have above), then
append-map-2 (two argument lists), then append-map-3 (three argument
lists), and by then you probably see the pattern which you can
generalize to append-map-n (any positive number of argument lists). It
worked for me.

At the moment I don't see how zero argument lists would also fit the
pattern or make sense. Mind stops working.

It's useful to know that (apply proc arg1 arg2 argses) is the same as
(apply proc (cons arg1 (cons arg2 argses))).

namekuseijin

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Apr 11, 2012, 8:04:38 PM4/11/12
to
Em sexta-feira, 6 de abril de 2012 09h32min38s UTC-3, Merciadri Luca escreveu:
> -----BEGIN PGP SIGNED MESSAGE-----
> Hash: SHA1
>
> Hi,
>
> I've got
>
> ==
> (define append_map
> (lambda (f l)
> (if (null? l) '() (append (f (car l)) (append_map f (cdr l))))))
> ==
>
> which is equivalent to
>
> ==
> (define append_map
> (lambda (f l) (apply append (map f l))))
> ==
>
> I now want to generalize this to an arbitrary number of arguments. I
> know this can be done using (lambda v) at the place of (lambda (v))
> (this is called `generalized lambda form'). How could I use this in
> this context (e.g. one function f and an arbitrary number of lists)?

whatever you do, change that C identifier to lisp style... :)

Merciadri Luca

unread,
Apr 22, 2012, 2:01:49 PM4/22/12
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Thanks for the help.

- --
Merciadri Luca
See http://www.student.montefiore.ulg.ac.be/~merciadri/
- --

If it's too good to be true, then it probably is.
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