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Re: [Scheme-reports] Pattern matching and list comprehensions



Hello.


Wed, 19 Mar 2014 10:54:26 +0100 от Panicz Maciej Godek <godek.maciek@x>:
Hi!

2014-03-19 7:06 GMT+01:00 Denis Trapeznikoff <denin@x>:
> Mon, 17 Mar 2014 23:02:52 +0100 от Panicz Maciej Godek
> <godek.maciek@x>:
>
> > I think that there's a nice tradition among the Scheme programmers to
> > use the name "compose" to refer to a composition function. It is much
> > more descriptive than "dollar-asterisk" ("multiply dollars"? "a
> > millionare marries a star"? "jackpot"?), and hence more
> > reader-friendly, and unless you are doing some domain-specific
> > research, your programs usually won't get much longer because of that.
>
> Yes (and I did so, too, at start), but `compose' is too long a word (compare
> it to Haskell's `.').

Haskell is a very nice language, but -- compared to Scheme -- it
conveys a lot of information in its complex syntax. Correct me if I'm
wrong, but I don't think that it would allow you to define the "."
operator if it wasn't already in the language.
Actually, there is no problem in that:

import Prelude hiding ((.))

main = (putStrLn . concat) ["Hell", "o!"]

(.) :: (b -> c) -> (a -> b) -> (a -> c)
f . g = (\x -> f (g x))

Haskell is a functional language after all! :)
Interestingly, when I was exploring Scheme for the first time, I have
also been introducing such operators as

(define ($_ f . x)
  (lambda y (apply f (append x y))))

(define (_$ f . y)
  (lambda x (apply f (append x y))))
((append) is not needed here.)
It's not only that -- it is also relevant how arguments are applied to
the predicate. To be honest, I have no intuition of whether such
generalization is good. Can you come up with any real-world usage
example? (Like: multiple argument map allows you to easily define the
code that computes the dot product)
Sorry, I don't have production examples right before my eyes now, so anything I come up with would be somewhat artificial.
Let's say you have two function graphs in two buffers and want to compute their relation at all points where the denominator is not zero:

(define nom-list)
(define denom-list)
(define ratio-list (call-with-values (lambda () (filter (lambda (a b) (not (zero? b))) nom-list denom-list)) (lambda (l1 l2) (map / l1 l2))))

(I intentionally have not used any more procedures of prelude.scm in this example.)

Pozdrawiam
M.


С уважением,

denin@x
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