I'm trying to make a code in lisp to count occurrences of atoms in a list in lisp.
The problem is the code works for all atoms except the atom (), which appears as NIL.
Example in the code:
(defun flatten (list_)
(cond ((atom list_) (list list_))
((null list_) NIL)
(t (append (flatten (car list_)) (flatten (cdr list_))) )
)
)
(defun toUniqueList (list_ out)
(cond ((null list_) NIL)
((not (member (car list_) out)) (append (list (car list_)) (toUniqueList (cdr list_) (append (list (car list_)) out)) ))
(t (toUniqueList (cdr list_) out))
)
)
(defun countOccurences (list_ x)
(cond ((null list_) 0)
((eql (car list_) x) (+ (countOccurences (cdr list_) x) 1))
(t (countOccurences (cdr list_) x))
)
)
(defun countOccurencesAll (list_)
(setq flatList (flatten list_))
(setq parsed (toUniqueList flatList '()))
(setq result '())
(dolist (x parsed)
(setq result (append result (list (list x (countOccurences flatList x)) ))))
result
)
(write (countOccurencesAll '(x y z 4.6 (a x) () (5 z x) ())))
; ((X 3) (Y 1) (Z 2) (4.6 1) (A 1) (NIL 5) (5 1))
Any idea in how to show () rather than NIL?
The expressions nil, 'nil, (), and '() all gets evaluated to nil which is displayed as nil unless it is the cdr of a pair in which it will just close the list. eg. '(() . ()) gets evaluated to (NIL . NIL) and it is displayed as (NIL). There is nothing you can do about that.
So the question is then, because ((a) (()) (c)) is really ((a . nil) . ((nil . nil) . ((c . nil) . nil))) should it count nil/() 5 times or ignore when nil in the cdr of a pair and just count it as one?
BTW using setq in countOccurencesAll on undefined bindings means your code is in the mercy of the implementation. The hyperspec does not define how it should be handled and SBCL makes warnings about how it interprets the code and other might just choose an interpretation. A better approach would be to use let to define the bindings. Using a hash and iterate over the list once would make an O(n) solution.
Related
I'm writing a macro to generate codes used by another macro in Common Lisp. But I'm new at this and have difficulty in constructing a macro that takes in a list (bar1 bar2 ... barn) and produces the following codes by a loop.
`(foo
,#bar1
,#bar2
...
,#barn)
I wonder whether this can be achieved not involving implement-dependent words such as SB-IMPL::UNQUOTE-SPLICE in sbcl.
Maybe I didn't give a clear description about my problem. In fact I want to write a macro gen-case such that
(gen-case
(simple-array simple-vector)
('(dotimes ($1 $5)
(when (and (= (aref $4 $2 $1) 1) (zerop (aref $3 $1)))
$0))
'(dolist ($1 (aref $4 $2))
(when (zerop (aref $3 $1))
$0)))
objname body)
produces something like
`(case (car (type-of ,objname))
(simple-array
,#(progn
(setf temp
'(dotimes ($1 $5)
(when (and (= (aref $4 $2 $1) 1) (zerop (aref $3 $1)))
$0)))
(code-gen body)))
(simple-vector
,#(progn
(setf temp
'(dolist ($1 (aref $4 $2))
(when (zerop (aref $3 $1))
$0)))
(code-gen body))))
In general cases, the lists taken in by gen-case may contain more than two items.
I have tried
``(case (car (type-of ,,objname))
,',#(#|Some codes that produce target codes|#))
but the target codes are inserted to the quote block and thus throw an exception in the macro who calls the macro gen-case. Moreover, I have no way to insert ,# to the target codes as a straightforward insertion will cause a "comma not inside a backquote" exception.
The codes generated are part of another macro
(defmacro DSI-Layer ((obj-name tag-name) &body body)
"Data Structure Independent Layer."
(let ((temp))
(defun code-gen (c)
(if (atom c) c
(if (eq (car c) tag-name)
(let ((args (cadr c)) (codes (code-gen (cddr c))) (flag nil))
(defun gen-code (c)
(if (atom c) c
(if (eq (car c) *arg*)
(let ((n (cadr c)))
(if (zerop n) (progn (setf flag t) codes)
(nth (1- n) args)))
(let ((h (gen-code (car c))))
(if flag
(progn
(setf flag nil)
(append h (gen-code (cdr c))))
(cons h (gen-code (cdr c))))))))
(gen-code temp))
(cons (code-gen (car c)) (code-gen (cdr c))))))
`(case (car (type-of ,obj-name))
(simple-array
,#(progn
(setf temp
'(dotimes ($1 $5)
(when (and (= (aref $4 $2 $1) 1) (zerop (aref $3 $1)))
$0)))
(code-gen body)))
(simple-vector
,#(progn
(setf temp
'(dolist ($1 (aref $4 $2))
(when (zerop (aref $3 $1))
$0)))
(code-gen body))))))
and I've set up a read-macro
(defvar *arg* (make-symbol "ARG"))
(set-macro-character #\$
#'(lambda (stream char)
(declare (ignore char))
(list *arg* (read stream t nil t))))
The intention of DSI-Layer is to add a piece of code to determine the type of input parameters. For example, the codes
(defun BFS (G v)
(let* ((n (car (array-dimensions G)))
(visited (make-array n :initial-element 0))
(queue (list v))
(vl nil))
(incf (aref visited v))
(DSI-Layer (G next-vertex)
(do nil ((null queue) nil)
(setf v (pop queue)) (push v vl)
(next-vertex (i v visited G n)
(setf queue (nconc queue (list i)))
(incf (aref visited i)))))
vl))
will be converted to
(defun BFS (G v)
(let* ((n (car (array-dimensions G)))
(visited (make-array n :initial-element 0))
(queue (list v))
(vl nil))
(incf (aref visited v))
(case (car (type-of G))
(simple-array
(do nil ((null queue) nil)
(setf v (pop queue))
(push v vl)
(dotimes (i n)
(when (and (= (aref G v i) 1) (zerop (aref visited i)))
(setf queue (nconc queue (list i)))
(incf (aref visited i))))))
(simple-vector
(do nil ((null queue) nil)
(setf v (pop queue))
(push v vl)
(dolist (i (aref G v))
(when (zerop (aref visited i))
(setf queue (nconc queue (list i)))
(incf (aref visited i)))))))))
Now I just wonder that whether the DSI-Layer can be generated from another macro gen-case by passing the type names and corresponding code templates to it or not.
By the way, I don't think the specific meaning of generated codes matters in my problem. They are just treated as data.
Don't be tempted to use internal details of backquote. If you have the lists you want to append in distinct variables, simply append them:
`(foo
,#(append b1 b2 ... bn))
If you have a list of them in some single variable (for instance if they've come from an &rest or &body argument) then do something like
`(foo
,#(loop for b in bs
appending b))
I see your problem - you need it not for a function call
but for a macro-call with case.
One cannot use dynamically macros - in a safe way.
One has to use eval but it is not safe for scoping.
#tfb as well as me answered in this question for type-case
lengthily.
previous answer (wrong for this case)
No need for a macro.
`(foo
,#bar1
,#bar2
...
,#barn)
with evaluation of its result
by pure functions would be:
(apply foo (loop for bar in '(bar1 bar2 ... barn)
nconc bar))
nconc or nconcing instead of collect fuses lists together and is very useful in loop. - Ah I see my previous answerer used append btw appending - nconc nconcing however is the "destructive" form of "append". Since the local variable bar is destructed here which we don't need outside of the loop form, using the "destructive" form is safe here - and comes with a performance advantage (less elements are copied than when using append). That is why I wired my brain always to use nconc instead of append inside a loop.
Of course, if you want to get the code construct, one could do
`(foo ,#(loop for bar in list-of-lists
nconc bar))
Try it out:
`(foo ,#(loop for bar in '((1 2 3) (a b c) (:a :b :c)) nconc bar))
;; => (FOO 1 2 3 A B C :A :B :C)
The answers of all of you inspired me, and I came up with a solution to my problem. The macro
(defmacro Layer-Generator (obj-name tag-name callback body)
(let ((temp (gensym)) (code-gen (gensym)))
`(let ((,temp))
(defun ,code-gen (c)
(if (atom c) c
(if (eq (car c) ,tag-name)
(let ((args (cadr c)) (codes (,code-gen (cddr c))) (flag nil))
(defun gen-code (c)
(if (atom c) c
(if (eq (car c) *arg*)
(let ((n (cadr c)))
(if (zerop n) (progn (setf flag t) codes)
(nth (1- n) args)))
(let ((h (gen-code (car c))))
(if flag
(progn
(setf flag nil)
(append h (gen-code (cdr c))))
(cons h (gen-code (cdr c))))))))
(gen-code ,temp))
(cons (,code-gen (car c)) (,code-gen (cdr c))))))
(list 'case `(car (type-of ,,obj-name))
,#(let ((codes nil))
(dolist (item callback)
(push
`(cons ',(car item)
(progn
(setf ,temp ,(cadr item))
(,code-gen ,body)))
codes))
(nreverse codes))))))
produces codes which are not the same as DSI-Layer but produce codes coincident with what the latter produces. Because the codes
`(case (car (type-of ,obj-name))
(tag1
,#(#|codes1|#))
(tag2
,#(#|codes2|#))
...)
are equivalent to
(list 'case `(car (type-of ,obj-name))
(cons 'tag1 (#|codes1|#))
(cons 'tag2 (#|codes2|#))
...)
And now we can use a loop to generate it just as what the Layer-Generator does.
I am new to LISP apparently... I am writing a function that takes a list and returns that list with all the duplicates removed, so (myPurge '(p a c e p c))->(a e p c)
This is my (edited)code:
(defun myPurge (L)
(if (eq L nil) ;if empty return nil
nil
(if(eq (car L)(car(cdr L)) ) ;if I find a match call function on
;rest of list
(myPurge (cdr L))
;else return that term and than call on
(progn ;rest of list
(car L)
(myPurge(cdr L)) ))))
*BUTTTTTTTT, when I call this function I get NIL!
*:
Why am i getting NIL instead of it returning car L ?
******Thank you all, this is my finished code that I ended up using( it uses another function 'myMember' I defined earlier.******
(defun myPurge (L)
(if (eq L nil)
nil
(if(myMember(car L)(cdr L))
(myPurge(cdr L))
(cons (car L)(myPurge(cdr L))))))))
****here is myMember*****it returns true if X is a member of L
(defun myMember (X L)
(if(eq L nil)
nil
(if(eq X(car L))
t
(myMember X(cdr L)) )) )
After reformatting your code looks like this:
(defun myPurge (L)
(if (eq L nil)
nil
(if (eq (car L) (car (cdr L)))
(myPurge (cdr L))
(car L)
(myPurge (cdr L)))))
As you can see, the last if is called with 4 arguments. But if only takes 3 arguments (condition, then-part, else-part).
Your code:
(defun myPurge (L)
(if (eq L nil) ;if empty return nil
nil
(if(eq (car L)(car(cdr L)) ) ;if I find a match call function on rest
;of list
(myPurge (cdr L))
;else return that term and than call on
(progn ;rest of list
(car L)
(myPurge(cdr L)) ))))
Correctly indented/formatted/named:
(defun my-purge (list)
(if (null list)
nil
(if (eql (first list) (second list))
(myPurge (rest list))
(progn
(first list)
(my-purge (rest list))))))
What looks strange? See the comment below.
(defun my-purge (list)
(if (null list)
nil
(if (eql (first list) (second list))
(myPurge (rest list))
(progn
(first list) ; <- what does this do?
(my-purge (rest list))))))
Bonus: you can get rid of the nested IFs:
(defun my-purge (list)
(cond ((null list)
nil)
((eql (first list) (second list))
(myPurge (rest list)))
(t
(first list)
(my-purge (rest list)))))
I am trying to check if a list has a mountain aspect or not in lisp.
e.g:1,5,9,6,4,3
l is my list and aux is 0-the ascending part of l or 1-the descending part of the list.
muntemain just call munte starting with aux=0,the ascending part
my error is :
Badly formed lambda: (AND (< (CAR L) (CAR (CDR L))) (EQ AUX 0))
and I can't see the problem.Can someone help please?
(defun munte (l aux)
(cond
((and (atom l) (null aux)) NIL)
((and (null l) (null aux)) NIL)
((and (atom l) (eq aux 1)) T)
((and (null l) (eq aux 1) T)
((and (< (car l) (car(cdr l))) (eq aux 0)) (munte(cdr l) 0))
((and (or (> (car l) (cadr l)) (= (car l) (cadr l))) (eq aux 0))(munte(cdr l) 1))
( and (> (car l) (cadr l)) (eq aux 1)) (munte(cdr l) 1))
(T NIL)
)
)
(defun muntemain (l)
(cond
((> (car l) (cadr l)) NIL)
((< (length l) 2) NIL)
(T (munte l 0))
)
)
Formatting
As noted by Barmar, you really need to use an editor to help you with the parenthesis. There are many tutorials for installing Emacs+Slime. Take some time to install proper tools.
Don't use EQ for numbers and characters
An implementation is permitted to make "copies" of characters and
numbers at any time. The effect is that Common Lisp makes no guarantee
that eq is true even when both its arguments are "the same thing" if
that thing is a character or number.
Factorize tests
((and (atom l) (null aux)) NIL)
((and (null l) (null aux)) NIL)
((and (atom l) (eq aux 1)) T)
((and (null l) (eq aux 1) T)
From the definition of atom, NIL is an atom, so you don't need (null L). The different cases for aux can be grouped too. The clause below is sufficient to account for all the above ones:
((atom L) (eql aux 1))
But I don't understand why aux is not a boolean in the first place if you always bind it to 0 or 1. Just use t and nil and return aux in the above clause.
Use meaningful functions
(< (car l) (car(cdr l)))
Of course, (car(cdr ..)) is known as (cadr ..), but also as second. The above test is equivalent to:
(< (first L) (second L))
And what if your list has no second element? You will compare a number against nil and signal an error (not what you want). You need more tests. In muntemain, you seem to have a special case for when length is below 2, but the test is done only if the previous returns nil, which won't happen if an error is signaled.
An iterative alternative
Here is a completely different way to attack the problem, just to give you ideas.
(lambda (list)
(loop
;; memories
for px = nil then x
for pdx = nil then dx
;; current element
for x in list
;; first and second "derivatives" (signs only)
for dx = 1 then (signum (- x px))
for ddx = 0 then (signum (- dx pdx))
;; checks
sum ddx into total
always (and (<= dx 0) (<= -1 total 0))
finally (return (= total -1))))
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))))
Need to write a union function in lisp that takes two lists as arguments and returns a list that is the union of the two with no repeats. Order should be consistent with those of the input lists
For example: if inputs are '(a b c) and '(e c d) the result should be '(a b c e d)
Here is what I have so far
(defun stable-union (x y)
(cond
((null x) y)
((null y) x))
(do ((i y (cdr i))
(lst3 x (append lst3
(cond
((listp i)
((null (member (car i) lst3)) (cons (car i) nil) nil))
(t (null (member i lst3)) (cons i nil) nil)))))
((null (cdr i)) lst3)))
My error is that there is an "illegal function object" with the segment (null (member (car i) lst3))
Advice?
You've got your parens all jumbled-up:
(defun stable-union (x y)
(cond
((null x) y)
((null y) x) ) END OF COND form - has no effect
(do ((i y (cdr i))
^^
(lst3 x (append lst3
(cond
((listp i)
( (null (member (car i) lst3))
^^ ~~~~~~~~~~~~~~~~~~~~~~~~~~~~ called as a function
(cons (car i) nil) with two arguments
nil ) )
^^
(t NEXT 3 forms have no effect
(null (member i lst3))
(cons i nil)
nil )))) )
^^
((null (cdr i)) lst3)))
Here's your code as you probably intended it to be, with corrected parenthesization and some ifs added where needed:
(defun stable-union (x y)
(cond
((null x) y)
((null y) x)
(t
(do ((i y (cdr i))
(lst3 x (append lst3
(cond
((listp i)
(if (null (member (car i) lst3))
(cons (car i) nil)
nil))
(t
(if (null (member i lst3))
(cons i nil)
nil))))))
((null (cdr i)) lst3)))))
There are still problems with this code. Your do logic is wrong, it skips the first element in y if it contains just one element. And you call append all the time whether it is needed or not. Note that calling (append lst3 nil) makes a copy of top-level cons cells in lst3, entirely superfluously.
Such long statements as you have there are usually placed in do body, not inside the update form for do's local variable.
But you can use more specialized forms of do, where appropriate. Here it is natural to use dolist. Following "wvxvw"'s lead on using hash-tables for membership testing, we write:
(defun stable-union (a b &aux (z (list nil)))
(let ((h (make-hash-table))
(p z))
(dolist (i a)
(unless (gethash i h)
(setf (cdr p) (list i) p (cdr p))
(setf (gethash i h) t)))
(dolist (i b (cdr z))
(unless (gethash i h)
(setf (cdr p) (list i) p (cdr p))
(setf (gethash i h) t)))))
using a technique which I call "head-sentinel" (z variable pre-initialized to a singleton list) allows for a great simplification of the code for the top-down list building at a cost of allocating one extra cons cell.
The error is because you're trying to execute the result of evaluating (null (member (car i) lst3)). In your cond expression, if i is a list, then it attempts to evaluate the expression
((null (member (car i) lst3)) (cons (car i) nil) nil))
And return the result. The first element in an expression should be a function, but
(null (member (car i) lst3))
Is going to return a boolean value. Hence the failure. The structure of your code needs some attention. What you've missed is that you need an inner cond, there.
Incidentally, this would be a much cleaner function if you did it recursively.
I'm a Schemer rather than a Lisper, but I had a little think about it. Here's the skeleton of a recursive implementation:
(defun stable-union (x y)
(cond
((null x) y)
((null y) x)
((listp y)
(cond
((member (car y) x) (stable-union ??? (???)))
(t (stable-union (append x (??? (???))) (cdr y)))))
((not (member y x)) (append x (list y)))
(t x)))
(Edited to correct simple tyop in second-last line, thanks to Will Ness for spotting it)
(remove-duplicates (append '(a b c) '(e c d)) :from-end t)
Because you started off with do, and because a recursive solution would be even worse, here's what you could've done:
(defun union-stable (list-a list-b)
(do ((i list-b (cdr i))
filtered back-ref)
((null i) (append list-a back-ref))
(unless (member (car i) list-a)
(if back-ref
(setf (cdr filtered) (list (car i))
filtered (cdr filtered))
(setf back-ref (list (car i))
filtered back-ref)))))
This is still quadratic time, and the behaviour is such that if the first list has duplicates, or the second list has duplicates, which are not in the first list - they will stay. I'm not sure how fair it is to call this function a "union", but you'd have to define what to do with the lists if they have duplicates before you try to unify them.
And this is what you might've done if you were interested in the result, rather than just exercising. Note that it will ensure that elements are unique, even if the elements repeat in the input lists.
(defun union-stable-hash (list-a list-b)
(loop for c = (car (if list-a list-a list-b))
with back-ref
with hash = (make-hash-table)
for key = (gethash c hash)
with result
do (unless key
(if back-ref
(setf (cdr result) (list c)
result (cdr result))
(when (or list-a list-b)
(setf back-ref (list c)
result back-ref)))
(setf (gethash c hash) t))
do (if list-a (setf list-a (cdr list-a))
(setf list-b (cdr list-b)))
do (unless (or list-a list-b)
(return back-ref))))