Total lisp beginner here.
I'm wondering how to convert a character to symbol. Simply what I want is convert #\a to a
Here what I have done so far:
(defun convert(char)
(if(eq char #\a)
(setq char 'a))
char)
This one is works actually but I don't want to add 26 conditions(letters in alphabet) and make a long-dumb code.
Also I'm wondering is there any functions in common lisp that converts character list to symbol list like: (#\h #\e #\l #\l #\o) to (h e l l o) ? I have found intern and make-symbol related to that but they require string as parameter.
CL-USER 230 > (intern (string #\Q))
Q
NIL
CL-USER 231 > (intern (string #\q))
\q
NIL
Btw., your code has a bunch of improvements necessary:
(defun convert(char) ;
(if(eq char #\a) ; use EQL instead of EQ for characters
; indentation is wrong
(setq char 'a)) ; indentation is wrong
char) ; indentation is wrong
Better write it as:
(defun convert (char)
(if (eql char #\a)
'a
char))
or
(defun convert (char)
(case char
(#\a 'a)
(#\b 'b)
(otherwise char)))
As mentioned above, the 'real' solution is:
(defun convert (char)
(intern (string char)))
(defun converter (c)
(if (characterp c)
(make-symbol (string c))))
You could use make-symbol and convert the symbol to a string.
(setq my-sym (converter #\a))
Answering my own question after a while:
Also I'm wondering is there any functions in common lisp that converts character list to symbol list like: (#\h #\e #\l #\l #\o) to (h e l l o) ?
I can convert single character to symbol with this convert function:
(defun convert (c)
(if (characterp c)
(intern (string c))))
Since one of main ideas of LISP is "processing a list" I can use one of map functions to apply one operation to every single element of the list.
(mapcar #'convert '(#\h #\e #\l #\l #\o))
Here mapcar function will result the list of symbols:
(|h| |e| |l| |l| |o|)
Related
I have a list of pairs ((3. #\K) (5 . #\J)) ... and I would like to create a function in scheme that returns the list as this: ("3K", "5J")...
I've beeing trying but I cannot make it.
This is what I have.
; The deckCards will contain the list of pairs;
; The real Deck will contain the empty list.
(define (deck->strings deckCards realDeck)
(let lenOfItem ([n (my-lenght deckCards)])
(if (= 1 n)
(list (card->string (first deckCards)))
(append realDeck (deck->strings (cdr deckCards) realDeck))))
)
I did try doing with cond but for some reason it doesnt return the list and it seems impossible to append the list to the realDeack before calling itself recursively.
I think I found an approach and it worked. Not sure if it good to use it. However, this prints all the strings from top to boottom in a new line... Will this matter? I think its because I have 48 elements.
(map (lambda (i) (card->string i))
clubs)
Total newbie to Scheme here.
I've been stuck on a scheme problem for sometime now. I don't understand how to code this right. I've looked every where on this site and others, and I just can't get this to work.
the problem: define a function Square that squares its parameters. If the parameter is not a number, print the message "invalid_input".
Here is what I have tried:
(define (square x) (cond
((number? x) (* x x))
(else (display "invalid_input\n"))
)
)
I've also tried this:
(define (square x) (cond
((number? x) (* x x))
((not (number? x)) (display "invalid_input\n"))
)
)
And this:
(define (square x) (if
(number? x) (* x x) (display "invalid_input\n")
)
)
None of these have worked when I call square like this (square h).
I keep getting this error on Linux
scheme#(guile-user)> (square h)
;;; <stdin>:44:0: warning: possibly unbound variable `h'
<unnamed port>:44:0: In procedure #<procedure 7fcc7d0a0b60 at <current input>:44:0 ()>:
<unnamed port>:44:0: In procedure module-lookup: Unbound variable: h
Entering a new prompt. Type `,bt' for a backtrace or `,q' to continue.
Shouldn't it print "invalid_input" since 'h' is not a number?
help me out here please. thank you
Functions always evaluate their arguments in Scheme. With (square h), h is an unbound variable; when h is evaluated, an error is issued since h is not bound. You can see that the OP definitions work as intended by trying (square 'h) or (square "h"). In the first case 'h is a symbol, and in the second "h" is a string; these are not numbers, so "invalid_input" is printed.
OP should work on following lispy conventions when formatting Scheme code; it is very bad style in lisps to scatter closing parentheses over multiple lines. Here is the same code in a more readable style, typical of lisps:
(define (square x)
(cond ((number? x) (* x x))
(else
(display "invalid_input\n"))))
Your code works fine. But you have do define h in order to use it.
$ guile
GNU Guile 2.0.13
Copyright (C) 1995-2016 Free Software Foundation, Inc.
Guile comes with ABSOLUTELY NO WARRANTY; for details type `,show w'.
This program is free software, and you are welcome to redistribute it
under certain conditions; type `,show c' for details.
Enter `,help' for help.
Your function is correct.
scheme#(guile-user)> (define (square x)
(if (number? x)
(* x x)
(display "invalid_input\n")))
And works as expected:
scheme#(guile-user)> (square 5)
$1 = 25
But if you pass an undefined value, you will get an error:
scheme#(guile-user)> (square h)
;;; <stdin>:6:0: warning: possibly unbound variable `h'
<unnamed port>:6:0: In procedure #<procedure 5595e9575c20 at <current input>:6:0 ()>:
<unnamed port>:6:0: In procedure module-lookup: Unbound variable: h
Entering a new prompt. Type `,bt' for a backtrace or `,q' to continue.
scheme#(guile-user) [1]> ,q
You have to make sure, that h is defined.
scheme#(guile-user)> (let ((h 5)) (square h))
$2 = 25
This is the same as in every other language. Try it in C:
int square (int x) {
return x * x;
}
int main (int argc, char** argv)
{
printf ("%d", square (h));
}
I want to define the Thue-Morse Sequence (or the fair-sharing sequence) in terms of an initial element, 0, and the rule defining the next section of the list in terms of the entire list up until this point. i.e.
fair 0 = [0]
--fair 1 = [0,1]
--fair 2 = [0,1,1,0]
--fair 3 = [0,1,1,0,1,0,0,1]
fair n = fair (n - 1) ++ map (1-) (fair (n - 1))
This works fine to generate the list up to any predefined length, but it seems ineffective to not just define the entire list at once, and use take if I need a predefined amount.
My first attempt at defining the entire list was fair = 0 : map (1-) fair but of course, this populates the list as it goes, so it doesn't ever (need to) reenter the list (and returns [0,1,0,1,0,1...]). What I want is some way to define the list so that when it reaches a not-yet-defined element in the list, it defines the next 'chunk' by reentering the list only until that point, (rather than the computation 'chasing' the new values as they're produced), so the steps in computing the list would be akin to this procedure:
begin with initial list, [0]
map (1-) over the existing list, producing [1]
append this to the existing list, producing [0,1]
map (1-) over the existing list, producing [1,0]
append this to the existing list, producing [0,1,1,0]
map (1-) over the existing list, producing [1,0,0,1]
append this to the existing list, producing [0,1,1,0,1,0,0,1]
The Wikipedia article I linked above has a helpful gif to illustrate this process.
As I presume you can see, this would continue indefinitely as new elements are needed. However, I can't for the life of me find a way to successfully encode this in a recursive function.
I have tried
reenter f xs = reenter f (xs ++ map f xs)
fair = reenter (1-) [0]
But while the logic seems correct, it hangs without producing anything, probably due to the immediate recursive call (though I thought haskell's lazy evaluation might take care of that, despite it being a rather complex case).
As you noted, you can't do the recursive call immediately - you first need to return the next result, and then recursively call, as in your last try:
Prelude> reenter prev_list = inverted_prev_list ++ reenter (prev_list ++ inverted_prev_list) where inverted_prev_list = map (1-) prev_list
Prelude> f = [0] ++ reenter [0]
Prelude> take 20 f
[0,1,1,0,1,0,0,1,1,0,0,1,0,1,1,0,1,0,0,1]
Following is code in Racket, another functional programming language, using the steps listed in the question.
(define (f n)
(define (invert s) ; sub-function to invert the numbers
(list->string
(for/list ((i (string->list s)))
(if (equal? i #\0) #\1 #\0))))
(let loop ((c 1)
(s "0")) ; starting string is "0"
(if (> c n)
s
(loop (add1 c)
(string-append s (invert s))))))
Testing:
(f 1)
(f 2)
(f 3)
(f 4)
(f 5)
Output:
"01"
"0110"
"01101001"
"0110100110010110"
"01101001100101101001011001101001"
For infinite series:
(define (f)
(define (invert s)
(list->string
(for/list ((i (string->list s)))
(if (equal? i #\0) #\1 #\0))))
(let loop ((s "0"))
(define ss (string-append s (invert s)))
(println ss)
(loop ss)))
To run:
(f)
This may give some ideas regarding a Haskell solution to this problem.
I'm looking for an equivalent of replace-regexp-in-string that just uses literal strings, no regular expressions.
(replace-regexp-in-string "." "bar" "foo.buzz") => "barbarbarbarbarbarbarbar"
But I want
(replace-in-string "." "bar" "foo.buzz") => "foobarbuzz"
I tried various replace-* functions but can't figure it out.
Edit
In return for the elaborate answers I decided to benchmark them (yea, I know all benchmarks are wrong, but it's still interesting).
The output of benchmark-run is (time, # garbage collections, GC time):
(benchmark-run 10000
(replace-regexp-in-string "." "bar" "foo.buzz"))
=> (0.5530160000000001 7 0.4121459999999999)
(benchmark-run 10000
(haxe-replace-string "." "bar" "foo.buzz"))
=> (5.301392 68 3.851943000000009)
(benchmark-run 10000
(replace-string-in-string "." "bar" "foo.buzz"))
=> (1.429293 5 0.29774799999999857)
replace-regexp-in-string with a quoted regexp wins. Temporary buffers do remarkably well.
Edit 2
Now with compilation! Had to do 10x more iteration:
(benchmark-run 100000
(haxe-replace-string "." "bar" "foo.buzz"))
=> (0.8736970000000001 14 0.47306700000000035)
(benchmark-run 100000
(replace-in-string "." "bar" "foo.buzz"))
=> (1.25983 29 0.9721819999999983)
(benchmark-run 100000
(replace-string-in-string "." "bar" "foo.buzz"))
=> (11.877136 86 3.1208540000000013)
haxe-replace-string is looking good
Try this:
(defun replace-in-string (what with in)
(replace-regexp-in-string (regexp-quote what) with in nil 'literal))
s.el string manipulation library has s-replace function:
(s-replace "." "bar" "foo.buzz") ;; => "foobarbuzz"
I recommend installing s.el from Emacs package manager, if you work with strings in your Elisp.
Emacs 28.1 (still in development at time of writing) provides this as standard:
** New function 'string-replace'.
This function works along the line of 'replace-regexp-in-string', but
matching on strings instead of regexps, and does not change the global
match state.
(string-replace FROMSTRING TOSTRING INSTRING)
Replace FROMSTRING with TOSTRING in INSTRING each time it occurs.
(string-replace ".*" "BAR" "foo.*bar.*baz")
⇒ "fooBARbarBARbaz"
I'd not hope for this to be faster:
(defun haxe-replace-string (string string-a string-b)
"Because there's no function in eLisp to do this."
(loop for i from 0 upto
(- (length string) (length string-a))
for c = (aref string i)
with alen = (length string-a)
with result = nil
with last = 0
do (loop for j from i below (+ i alen)
do (unless
(char-equal
(aref string-a (- j i))
(aref string j))
(return))
finally
(setq result
(cons (substring string last (- j alen)) result)
i (1- j) last j))
finally
(return
(if result
(mapconcat
#'identity
(reverse (cons (substring string last) result)) string-b)
string))))
Becasue replace-regexp-in-string is a native function, but you never know... Anyways, I wrote this some time ago for some reason, so, if you fill like comparing the performance - you are welcome to try :)
Another idea, using temporary buffer:
(defun replace-string-in-string (what with in)
(with-temp-buffer
(insert in)
(beginning-of-buffer)
(while (search-forward what nil t)
(replace-match with nil t))
(buffer-string)))
s-replace is fine if you are ready to require it, but say you want to use a replace in string feature early in the load process and don't yet have s.el loaded or don't need all of it. Well, here is the definition of s-replace from s.el. As you can see, it has no dependencies so you can use it without requiring the rest of s.el:
(defun s-replace (old new s)
"Replaces OLD with NEW in S."
(declare (pure t) (side-effect-free t))
(replace-regexp-in-string (regexp-quote old) new s t t))
I'm taking a list name as input with a single quote ('), but after doing a few operations, I want to actually evaluate it instead of treat it as an atom.
So for example, just for simplicity sake, I have the following list:
(setf LT '(A B C))
I have a function called SEP. To run the function, I must run it as (SEP 'LT). So as you can see, LISP will interpret LT as an atom instead of evaluate it as a list, which is not what I want.
So essentially, I want (SEP 'LT) to really become (SEP '(A B C)) somehow.
The input format can't be changed. Any help would be appreciated. Thanks!
If LT is a top-level variable, defined with defvar, then you can get its value with symbol-value as such:
* (symbol-value 'lt)
(A B C)
* (defun sep (name)
(assert (symbolp name))
(let ((value (symbol-value name)))
...