I'm learning Common Lisp and I've been trying to write a function to reverse a list. This is the function but every time I try to run it I get "Segmentation Fault (Core Dumped)"
(defun reverseList(x)
(if (cdr x)
(cons (reverseList x) (car x))
(car x)))
(reverseList '(1 2 3))
I think this will work:
(define reverseList(lambda(x)(
if (null? (cdr x))
(cons (car x) '())
(append (reverseList (cdr x)) (cons (car x) '()))
)
)
)
Related
I'm trying to teach myself Racket. I'm currently trying to write a function to help understand nested lists. The function takes a nested list and a procedure and applies the procedure to each element to produce a new list. An example:
(map-tree even? '(1 2 3 4)) => '(#f #t #f #t)
Here's what I've got so far:
(define (map-tree proc tree)
(map-tree-aux tree proc '() ))
(define (map-tree-aux tree proc lst)
(if (null? tree)
lst
(if (list? tree)
(if (null? (cdr tree))
(if (number? (car tree))
(map-tree-aux (car tree) proc (append-end (proc (car tree)) lst))
(map-tree-aux (car tree) proc lst))
(if (number? (car tree))
(map-tree-aux (cdr tree) proc (append-end (proc (car tree)) (map-tree-aux (car tree) proc lst)))
(map-tree-aux (cdr tree) proc lst)))
lst)))
(define (append-end elem lst)
(append lst (list elem)))
While this works with the original example I supplied, a more complex example comes out incorrectly:
(map-tree even? '(1 (2 (3 (4))))) should be '(#f (#t (#f (#t)))), but is currently (#f #t #f #t).
I know it's just a matter is "listing" somewhere, but I'm having an issue finding out how to do it.
My first thought was to apply the list procedure to the lst if the tree is null and (car tree) is not a number, but I get the opposite of what I want (the resultant list is nested in the opposite direction). I'd really appreciate your help.
Thanks!
When iterating over list of lists, the general idea for the cases to check is:
if list is empty (null? lst), do something ...
if the first item in list is atomic (not (pair? (car lst))), do something else ...
if the first item in list is a list itself (pair? (car lst)), else ...
Choosing the right construct is also important, ie. instead of nesting if statements, using cond or match etc. is preferred.
Also try and avoid using non-constant time procedures (such as append) in your recursive steps to improve efficiency.
With these in mind, one approach to create the function in question is by simply using cons to build a new list while preserving the structure of the old, as follows:
(define (my-map pred lst)
(cond
((null? lst) '())
((not (pair? (car lst)))
(cons (pred (car lst))
(my-map pred (cdr lst))))
(else
(cons (my-map pred (car lst))
(my-map pred (cdr lst))))))
You can write the same function using match instead of cond:
(define (my-map pred lst)
(match lst
['() '()]
[(cons (? pair?) b)
(cons (my-map pred (car lst))
(my-map pred (cdr lst)))]
[(cons a b)
(cons (pred (car lst))
(my-map pred (cdr lst)))]))
You can also build a tail-recursive function that does this:
(define (my-map pred lst)
(let loop ((lst lst)
(acc '()))
(cond
((null? lst)
(reverse acc))
((not (pair? (car lst)))
(loop (cdr lst) (cons (pred (car lst)) acc)))
(else
(loop (cdr lst) (cons (loop (car lst) '()) acc))))))
Notice that (reverse acc) is returned in the base case because the list being built in the accumulator acc is in reverse order from the original list lst. To avoid this, we can modify this function to accumulate a continuation instead:
(define (my-map pred lst)
(let loop ((lst lst)
(acc identity))
(cond
((null? lst)
(acc '()))
((not (pair? (car lst)))
(loop (cdr lst) (lambda (r)
(acc (cons (pred (car lst)) r)))))
(else
(loop (cdr lst)
(lambda (r)
(acc (cons (loop (car lst) identity) r))))))))
For all cases, you will have:
(my-map even? '(1 2 3 4 5 7))
=> '(#f #t #f #t #f #f)
(my-map even? '(1 (2 (3 (4 (5 (7)))))))
=> '(#f (#t (#f (#t (#f (#f))))))
Hello why do i get *** - EVAL/APPLY: too many arguments given to F on function call with nested lists parameter. I cannot figure it out, since I passed a simple nested list.
(defun f (L)
(cond
((NULL l) nil)
((listp (car L))
(append (F(car L))) (F(cdr L) (car (F (car L)))))
(T (list(car L)))
)
)
(setq A '((1) 2 3))
(f A)
This better formatting should make it easy to spot the error:
(defun f (l)
(cond ((null l) nil)
((listp (car l))
(append (f (car l)))
(f (cdr l)
(car (f (car l)))))
(t (list (car l)))))
If that does not help, use SBCL to compile the function. It will give you a very clear error message.
I am learning Lisp and I had to write a function whose return value was a list containing the odd integers (if any) from the given input. In code I have this:
(defun f3 (a)
(cond
((null a) nil )
((and (numberp (car a)) (oddp (car a))) (cons (car a) (f3 (cdr a))))
(T (f3 (cdr a)))
) ; end cond
)
I originally wanted to use the append function, but I kept getting errors.
It was recommended to me to use cons function. When I did this my function started working (code is above). I originally had this:
(defun f3 (a)
(cond
((null a) ())
((and (numberp (car a)) (oddp (car a))) (append (f3 (cdr a)) (car a))))
(T (append () (f3 (cdr a))))
)
)
but kept getting errors. For example, if I called (f3 '(1 2 3)) it would say "error 3 is not type LIST". So, my questions are why does cons work here and why did append not work? How does cons work? Thanks in advance.
append wants list arguments, and (car a) is not a list. Instead of (car a) you'd need (list (car a)). In other words, (append (f3 (cdr a)) (list (car a))).
That will basically work, but you'll get the result in reverse order. So that should be (append (list (car a)) (f3 (cdr a))).
Also note that your (append () (f3 (cdr a))) is equivalent to just (f3 (cdr a)).
The resulting changes in your original would be:
(defun f3 (a)
(cond
((null a) ())
((and (numberp (car a)) (oddp (car a)))
(append (list (car a)) (f3 (cdr a)))))
(T (f3 (cdr a)))))
But, you wouldn't normally use append to prepend a single element to a list. It would more naturally be done using cons. So
(append (list (car a)) (f3 (cdr a)))
Is more appropriately done by:
(cons (car a) (f3 (cdr a)))
Which finally takes you right to the working version you showed.
While something like mbratch's answer will help you in learning about list manipulation (and so is probably a more useful answer for you at this point in your study), it's also important to learn about the standard library of the language that you're using. In this case, you're trying to filter out everything except odd numbers. Using remove-if-not, that's just:
(defun keep-odd-numbers (list)
(remove-if-not (lambda (x)
(and (numberp x) (oddp x)))
list))
CL-USER> (keep-odd-numbers '(1 a 2 b 3 c 4 d 5 e))
;=> (1 3 5)
While this isn't a fix to your actual problem, which #mbratch provided, here's the way I would implement something like this using the LOOP macro (another part of the standard library):
(defun keep-odd-numbers (list)
(loop for x in list collecting x when (and (numberp x) (oddp x))))
I ran over an example of a problem which should determine the list of all non-numeric atoms at any level in a non-linear list.
(Defun Lis(L)
(Cond
((Null L) Nil)
((Not (Numberp (Car L))) (Cons (Car L) (Lis (Cdr L))))
((Atom (Car L)) (Lis (Cdr L)))
(T (Append (Lis (Car L)) (Lis (Cdr L))))
))
I took an example, (Lis '(1 A ((B) 6) (2 (C 3)) D 4)) which should return (A B C D)
Now I don't understand how can the list be created when the 3rd element of the list is evaluated ((B) 6).It will enter on the 2nd branch and do the cons?But that isn't constructing the new list with ((B) 6)?When will it enter on the last branch? I'm a little confused of how this algorithm works,can somebody make it clear for me?
The code works fine if you "invert" the 2 middle tests:
(defun lis(L)
(cond
((null L) nil)
((numberp (car L)) (lis (cdr L)))
((atom (car L)) (cons (car L) (lis (cdr L))))
(t (append (lis (car L)) (lis (cdr L))))))
because (not (numberp (car L))) is also true for lists so in the initial version the code never recurses down into a sublist.
I would write it as:
(defun tree-keep-if (predicate tree)
"Returns the list of all non-numeric atoms at any level in a cons tree."
(mapcan (lambda (item)
(cond ((consp item) (tree-keep-if predicate item))
((funcall predicate item) (list item))
((atom item) nil)))
tree))
Using it:
CL-USER > (tree-keep-if (complement #'numberp) '(1 A ((B) 6) (2 (C 3)) D 4))
(A B C D)
A more sophisticated version might remove the recursion to not be limited by stack size.
since yesterday I've been trying to program a special case statement for scheme that would do the following:
(define (sort x)
(cond ((and (list? x) x) => (lambda (l)
(sort-list l)))
((and (pair? x) x) => (lambda (p)
(if (> (car p) (cdr p))
(cons (cdr p) (car p))
p)))
(else "here")))
instead of using all the and's and cond's statement, I would have:
(define (sort x)
(scase ((list? x) => (lambda (l)
(sort-list l)))
((pair? x) => (lambda (p)
(if (> (car p) (cdr p))
(cons (cdr p) (car p))
p)))
(else "here")))
What I could do so far, was this:
(define (sort x)
(scase (list? x) (lambda (l)
(sort-list l)))
(scase (pair? x) (lambda (p)
(if (> (car p) (cdr p))
(cons (cdr p) (car p))
p))))
with this code:
(define-syntax scase
(syntax-rules ()
((if condition body ...)
(if condition
(begin
body ...)))))
What I wanted to do now, is just allow the scase statement to have multiple arguments like this:
(scase ((list? (cons 2 1)) 'here)
((list? '(2 1)) 'working))
but I can't seem to figure out how I can do that. Maybe you guys could give me a little help?
Thanks in advance ;)
If this is an exercise in learning how to use syntax-rules, then disregard this answer.
I see a way to simplify your code that you are starting with.
(define (sort x)
(cond ((list? x)
(sort-list x))
((pair? x)
(if (> (car x) (cdr x))
(cons (cdr x) (car x))
x)))
(else "here")))
Since all the (and (list? x) x) => (lambda l ... does is see if x is a list, and then bind l to x, (since #f is not a list, and '() is not false, at least in Racket), you can just skip all that and just use x. You do not need to use => in case, and in this case it doesn't help. => is useful if you want to do an test that returns something useful if successful, or #f otherwise.
Now, if you want to use a macro, then you're going to need to clarify what you want it to do a bit better. I think that case already does what you want. Your existing macro is just if, so I'm not sure how to extend it.
I found the solution for my question, here it goes:
(define-syntax cases
(syntax-rules ()
((_ (e0 e1 e2 ...)) (if e0 (begin e1 e2 ...)))
((_ (e0 e1 e2 ...) c1 c2 ...)
(if e0 (begin e1 e2 ...) (cases c1 c2 ...)))))
Thank you all anyway :)
Here's a solution :
#lang racket
(require mzlib/defmacro)
(define-syntax scase
(syntax-rules (else)
((_ (else body1)) body1)
((_ (condition1 body1) (condition2 body2) ...)
(if condition1
body1
(scase (condition2 body2) ...)))))
(define (sort1 x)
((scase ((list? x) (lambda (l)
(sort l <)))
((pair? x) (lambda (p)
(if (> (car p) (cdr p))
(cons (cdr p) (car p))
p)))
(else (lambda (e) "here")))
x))
It works in DrRacket. I made three changes to your solution. First, i renamed your sort procedure to sort1 since sort is inbuilt in scheme ( I have used it inside sort1). Second, I have changed the sort1 itself so that the input given will be passed to the procedure returned by scase and you will directly get the sorted result. Third, I have modified the scase syntax extension, so that it will accept the else condition.
>(sort1 (list 3 1 2))
'(1 2 3)
> (sort1 (cons 2 1))
'(1 . 2)
> (sort1 'here)
"here"
I suggest you read "The Scheme Programming Language" by Kent Dybvig. There is an entire chapter on syntactic extensions.