David Christiansen's answer was very detailed and thorough, but I wanted to
add one more. If I understand it, your goal is to take an arbitrary
(possibly nested) list of strings and concatenate all the strings. This
will do that:
(define lst '("foo" ("bar") ( ("baz") "jaz") ))
(string-append (flatten lst)) ;; returns "foobarbazjaz"
'flatten' takes any S-expression (loosely, any list) and turns it into a
single one-level list. For example, all of the following produce '(a b c d
e)
(flatten '(a b c d e))
(flatten '((a b) (c d) e))
(flatten '( (((a (b (c d) e)))) ))
(flatten '(a b (c ((d e))) ))
[Note that I used symbols instead of strings so as to avoid typing lots of
" characters, but same thing.]
(string-append (flatten lst)) will fail if there is anything in lst that is
not a string (e.g. a symbol, number, etc) To fix that you could do the
following:
(define (to-str x)
(cond ((number? x) number->string)
((symbol? x) symbol->string)
(...other clauses as desired...)
(else identity)))
This is a function that takes one item, determines what type it is, and
then returns a function that will convert that type to a string.
'identity' is a function that takes any one item and returns that item
unchanged -- in this case it's being used as 'string->string'.
Now that you have to-str you can then take the function that it returns and
use it. For example, try doing the following in the Racket REPL:
(define converter (to-str "string_foo")) ;; converter is now the
function 'identity'
(converter "string_foo") ;; returns "string_foo"
(define converter (to-str 'symbol_foo)) ;; converter is now the function
'symbol->string'
(converter 'symbol_foo) ;; returns "symbol_foo" [which is a string]
(define converter (to-str 7)) ;; converter is now the function
'number->string'
(converter 7) ;; returns "7" [which is a string]
There are more types <https://docs.racket-lang.org/guide/datatypes.html>
than I've included, but this should make the point.
At this point you can handle lists that contain non-string items:
(define lst '("foo" (bar "baz" 7) "jaz"))
(string-append (flatten lst)) ;; throws an exception because 'bar isn't a
string
(string-append (map to-str (flatten lst))) ;; returns "foobarbaz7jaz"
Hope this helps,
Dave
On Thu, Oct 13, 2016 at 8:18 AM, Gregor Kopp <gregorkop...@gmail.com> wrote:
> Hi!
>
> I'm very new to Racket and the lisp family generally, and very slowly
starting to get an idea how everything works together.
>
> Now my very first question ;)
>
> If I have a list of string's, how can i recursivly iterate over that list
and compose a single string out of the elements of that list?
>
> I prepared a simple example, so hopefully you can understand what I'm
talking about:
>
> (define mylist '("hello" "darkness" "my" "old" "fiend"))
>
> (for-each (lambda (elem)
> (displayln elem)
> )
> mylist)
>
> It prints out every element of mylist as I expect in a new line.
> But what if I want to collect all strings into one single string, so that
I can get "hellodarknessmyoldfriend"?
>
> I think I can use string-append, right?
> But how could I do that? Is there a idiomatic way to do that in Racket?
>
> Thank you very much for every input.
>
> Gregor
>
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