I'm a beginner in lisp and I need somebody to explain to me how the prog form works, step by step. What is the initial value of l1 ? Nil ?
The problem outputs T if the list has an even number of elements on the first level, nil if not.
(defun nr_par (l)
(prog ((l1 l))
ciclu
(cond
((null l1) (return T))
((null (cdr l1)) (return NIL))
((null (cddr l1)) (return T))
(T (setf l1 (cddr l1))
(go ciclu)))))
On console:
(nr_par '(1 2 3 4 5 6 7 8))
T
The program is straightforward, but not very idiomatic lisp (it is rather imperative instead of functional). Step by step goes as follows.
prog uses a series of variable bindings, in this case, l1 is assigned the value of l initially. Then, a series of statements in which a loop starts (again, not very lisp idiomatic).
This type of loops use a tag (ciclu) and a goto instruction (go), again, not recommended, but it is there. After that, the cond checks a series of cases. When the list is empty (null), you return true, in other cases, you check if the length is even or odd, and return the value in consequence.
In the case that the list is longer than one or two elements (neither of the cases is null), the l1 list is adjusted to point to the next of the next element of itself (the cddr function).
Finally, the go function turns the program back to the ciclu tag.
The program will finish when any of the cond clauses is met, returning either T or NIL.
See PROG in CLHS: L1 is var, L is init-form, so the initial value of L1 is the value of L.
As the CLHS page for prog says, it does three things: lets you have local vars and initialize them; lets you have tags as in tagbody and use go; and lets you use return as inside a block named NIL:
(defun nr_par (l)
(prog ((l1 l)) ; local binding(s)
ciclu
(if (null l1) (return T)) ; return
(if (null (cdr l1)) (return NIL))
(setf l1 (cddr l1))
(go ciclu))) ; go
(defun nr_par1 (l) ; directly equivalent
(labels ((ciclu (l1)
(if (null l1) (return-from ciclu T))
(if (null (cdr l1)) (return-from ciclu NIL))
(ciclu (cddr l1))))
(ciclu l)))
(defun nr_par2 (l) ; also equivalent
(do ((l1 l (cddr l1)))
(NIL) ; while T ...
(cond
((null l1) (return T))
((null (cdr l1)) (return NIL)))))
Function call is a glorified goto after all, isn't it?
See also Longest decreasing sequence in Lisp for an example representing several mutually-recursive functions hand-compiled into a prog with a bunch of GO statements.
Related
Is there a standard function in Common Lisp that can check against improper lists (i.e. circular and dotted lists) without signaling an error? list-length can check against circular lists (it returns nil for them), but signals type-error when given a dotted list.
Scheme's list? traverses the whole list to make sure it is not dotted or circular; Common Lisp's listp only checks that it's given nil or a cons cell.
Here's the simplest I could come up with:
(defun proper-list-p (x)
(not (null (handler-case (list-length x) (type-error () nil)))))
Since several implementations have been suggested and many unexpected problems have been found, here's a test suite for aspiring proper-list-p writers:
(defun circular (xs)
(let ((xs (copy-list xs)))
(setf (cdr (last xs)) xs)
xs))
(assert (eql t (proper-list-p '())))
(assert (eql t (proper-list-p '(1))))
(assert (eql t (proper-list-p '(1 2))))
(assert (eql t (proper-list-p '(1 2 3))))
(assert (not (proper-list-p 1)))
(assert (not (proper-list-p '(1 . 2))))
(assert (not (proper-list-p '(1 2 . 3))))
(assert (not (proper-list-p '(1 2 3 . 4))))
(assert (not (proper-list-p (circular '(1)))))
(assert (not (proper-list-p (circular '(1 2)))))
(assert (not (proper-list-p (circular '(1 2 3)))))
(assert (not (proper-list-p (list* 1 (circular '(2))))))
(assert (not (proper-list-p (list* 1 2 (circular '(3 4))))))
There is no standard function to do this, perhaps because such a function was seen as rather expensive if it was to be correct, but, really, this just seems like am omission from the language to me.
A minimal (not very performant) implementation, which does not rely on handling errors (Python people think that's a reasonable way to program, I don't, although this is a stylistic choice), is, I think
(defun proper-list-p (l)
(typecase l
(null t)
(cons
(loop for tail = l then (cdr tail)
for seen = (list tail) then (push tail seen)
do (cond ((null tail)
(return t))
((not (consp tail))
(return nil))
((member tail (rest seen))
(return nil)))))))
This takes time quadratic in the length of l, and conses proportional to the length of l. You can obviously do better using an hashtable for the occurs check, and you can use a tortoise-&-hare algorithm do avoid the occurs check (but I'm not sure what the complexity of that is off the top of my head).
I am sure there are much better functions than this in libraries. In particular Alexandria has one.
While thinking about this question, I also wrote this function:
(defun classify-list (l)
"Classify a possible list, returning four values.
The first value is a symbol which is
- NULL if the list is empty;
- LIST if the list is a proper list;
- CYCLIC-LIST if it contains a cycle;
- IMPROPER-LIST if it does not end with nil;
- NIL if it is not a list.
The second value is the total number of conses in the list (following
CDRs only). It will be 0 for an empty list or non-list.
The third value is the cons at which the cycle in the list begins, or
NIL if there is no cycle or the list isn't a list.
The fourth value is the number if conses in the cycle, or 0 if there is no cycle.
Note that you can deduce the length of the leading element of the list
by subtracting the total number of conses from the number of conses in
the cycle: you can then use NTHCDR to pull out the cycle."
;; This is written as a tail recursion, I know people don't like
;; that in CL, but I wrote it for me.
(typecase l
(null (values 'null 0 nil 0 0))
(cons
(let ((table (make-hash-table)))
(labels ((walk (tail previous-tail n)
(typecase tail
(null
(values 'list n nil 0))
(cons
(let ((m (gethash tail table nil)))
(if m
(values 'cyclic-list n tail (- n m))
(progn
(setf (gethash tail table) n)
(walk (cdr tail) tail (1+ n))))))
(t
(values 'improper-list n previous-tail 0)))))
(walk l nil 0))))
(t (values nil 0 nil 0))))
This can be used to get a bunch of information about a list: how long it is, if it is proper, if not if it's cyclic, and where the cycle is. Beware that in the cases of cyclic lists this will return circular structure as its third value. I believe that you need to use an occurs check to do this – tortoise & hare will tell you if a list is cyclic, but not where the cycle starts.
in addition, something slightly less verbose, than the accepted answer:
(defun improper-tail (ls)
(do ((x ls (cdr x))
(visited nil (cons x visited)))
((or (not (consp x)) (member x visited)) x)))
(defun proper-list-p (ls)
(null (improper-tail ls)))
or just like this:
(defun proper-list-p (ls)
(do ((x ls (cdr x))
(visited nil (cons x visited)))
((or (not (consp x)) (member x visited)) (null x))))
seen to pass all the op's test assertions
After our hopeless attempts with tailp, here, sth which uses the
sharp-representation of circular lists :) .
With regex (to detect circular sublist)
(setf *print-circle* t)
(ql:quickload :cl-ppcre)
(defun proper-listp (lst)
(or (null lst) ; either a `'()` or:
(and (consp lst) ; a cons
(not (cl-ppcre::scan "#\d+=(" (princ-to-string lst)))) ; not circular
(null (cdr (last lst)))))) ; not a dotted list
Without regex (cannot detect circular sublists)
(defun proper-listp (lst)
(or (null lst) ; either a `'()` or:
(and (consp lst) ; a cons
(not (string= "#" (subseq (princ-to-string lst) 0 1))) ; not circular
(null (cdr (last lst)))))) ; not a dotted list
(tailp l (cdr l)) is t for circular lists but nil for non-circular lists.
Credits to #tfp and #RainerJoswig who taught me this here .
So, your function would be:
(defun proper-listp (lst)
(or (null lst) ; either a `'()` or:
(and (consp lst) ; a cons
(not (tailp lst (cdr lst))) ; not circular
(null (cdr (last lst)))))) ; not a dotted list
By the way, I use proper-listp by purpose. Correct would be - by convetion proper-list-p. However, this name is already occupied in the CLISP implementation by SYSTEM::%PROPER-LIST-Pwhy the definition of the function raises a continuable error.
Conclusion of our discussion in the comment section:
The behavior of tailp for circular lists is undefined. Therefore this answer is wrong! Thank you #Lassi for figuring this out!
I'm trying to create a function that would test whether the given list is circular with a re-starting point being the beginning of the list.
Expected results:
(setq liste '(a b c))
(rplacd (cddr liste) liste)
(circular liste) => t
(circular '(a b c a b c)) => nil
As I simply want to test if any subsequent item is 'eq' to the first one, I don't want to build the whole tortoise and hare algorithm.
Here is my code :
(defun circular (liste)
(let (beginningliste (car liste)))
(labels ( (circ2 (liste)
(cond
((atom liste) nil)
((eq (car liste) beginningliste) t)
(t (circ2 (cdr liste)))
) ) ) ) )
It doesn't give the expected result but I don't understand where my error is
I'm not sure I'm using 'labels' correctly
Is there a way to do that without using 'labels'?
Edit. I guess I have answered my third question as I think I have found a simpler way. Would this work?
(defun circular (liste)
(cond
((atom liste) nil)
((eq (car liste) (cadr liste)) t)
(t (circular (rplacd liste (cddr liste))))
)
)
First, the behavior is undefined when you mutate constant data: when you quote something (here the list), the Lisp environment has the right to treat it as a constant. See also this question for why defparameter or defvar is preferred over setq. And so...
(setq list '(a b c))
(rplacd (cddr list) list)
... would be better written as:
(defparameter *list* (copy-list '(a b c)))
(setf (cdr (last *list*)) *list*)
Second, your code is badly formatted and has bad naming conventions (please use dashes to separate words); here it is with a conventional layout, with the help of emacs:
(defun circularp (list)
(let (first (car list)))
(labels ((circ2 (list)
(cond
((atom list) nil)
((eq (car list) first) t)
(t (circ2 (cdr list))))))))
With that formatting, two things should be apparent:
The let contains no body forms: you define local variables and never use them; you could as well delete the let line.
Furthermore, the let is missing one pair of parenthesis: what you wrote defines a variable name first and another one named car, bound to list. I presume you want to define first as (car list).
You define a local circ2 function but never use it. I would expect the circularp function (the -p is for "predicate", like numberp, stringp) to call (circ2 (cdr list)). I prefer renaming circ2 as visit (or recurse), because it means something.
With the above corrections, that would be:
(defun circularp (list)
(let ((first (car list)))
(labels ((visit (list)
(cond
((atom list) nil)
((eq (car list) first) t)
(t (visit (cdr list))))))
(visit (cdr list)))))
However, if your list is not circular but contains the same element multiple times (like '(a a b)), you will report it as circular, because you inspect the data it holds instead of the structure only. Don't look into the CAR here:
(defun circularp (list)
(let ((first list))
(labels ((visit (list)
(cond
((atom list) nil)
((eq list first) t)
(t (visit (cdr list))))))
(visit (cdr list)))))
Also, the inner function is tail recursive but there is no guarantee that a Common Lisp implementation automatically eliminates tail calls (you should check with your implementation; most can do it on request). That means you risk allocating as many call stack frames as you have elements in the list, which is bad. Better use a loop directly:
(defun circularp (list)
(loop
for cursor on (cdr list)
while (consp cursor)
thereis (eq cursor list)))
Last, but not least: your approach is a very common one but fails when the list is not one big circular chain of cells, but merely contains a loop somewhere. Consider for example:
CL-USER> *list*
#1=(A B C . #1#)
CL-USER> (push 10 *list*)
(10 . #1=(A B C . #1#))
CL-USER> (push 20 *list*)
(20 10 . #1=(A B C . #1#))
(see that answer where I explain what #1= and #1# mean)
The lists with numbers in front exhibit circularity but you can't just use the first cons cell as a marker, because you will be looping forever inside the sublist that is circular. This is the kind or problems the Tortoise and Hare algorithm solves (there might be other techniques, the most common being storing visited elements in a hash table).
After your last edit, here is what I would do if I wanted to check for circularity, in a recursive fashion, without labels:
(defun circularp (list &optional seen)
(and (consp list)
(or (if (member list seen) t nil)
(circularp (cdr list) (cons list seen)))))
We keep track of all the visited cons cells in seen, which is optional and initialized to NIL (you could pass another value, but that can be seen as a feature).
Then, we say that a list is circular with respect to seen if it is a cons cell which either: (i) already exists in seen, or (ii) is such that its CDR is circular with respect to (cons list seen).
The only additional trick here is to ensure the result is a boolean, and not the return value of member (which is the sublist where the element being searched for is the first element): if your environment has *PRINT-CIRCLE* set to NIL and the list is actually circular, you don't want it to try printing the result.
Instead of (if (member list seen) t nil), you could also use:
(when (member list seen))
(position list seen)
and of course (not (not (member list seen)))
I want to ask why this function doesn't work...
(defun nenum(ls)
(cond
((null ls) nil)
((listp car(ls)) (nenum (rest ls)))
((numberp car(ls)) (nenum (rest ls)))
(t (cons (car ls) (nenum (rest ls))))))
Example: (nenum '(l 1 i (b) (5) s -2 p)) --> (l i s p)
Thank you!
Looking at the predicate you have in one of your cond terms:
(listp car (ls))
Thus apply the function listp with the two arguments car and the result of calling the function ls with no arguments. car and ls both need to be free variables and listp needs to be a different function than the one defined in CLHS since it only takes one argument.
Perhaps you have though you were writing Algol? An Algol function call look like operator(operand) but not CL. CL is a LISP dialect and we have this form on our function calls:
(operand operator)
If we nest we do the same:
(operand (operand operator))
You got it right in the alternative (cons (car ls) (nenum (rest ls)))
Replace car(ls) with (car ls).
Here's a much easier way to write that function:
(defun nenum (list)
(remove-if (lambda (item)
(or (listp item)
(numberp item)))
list))
Note that NIL doesn't need its own test because listp covers it.
There's no need to write a function like this from scratch. Common Lisp already provides remove-if, and you can give it a predicate that matches numbers and non-atoms:
CL-USER> (remove-if #'(lambda (x)
(or (numberp x)
(not (atom x))))
'(l 1 i (b) (5) s -2 p))
;=> (L I S P)
Or, to make it even clearer that you're keeping non-numeric atoms, you can use remove-if-not with a predicate that checks for numeric atoms:
CL-USER> (remove-if-not #'(lambda (x)
(and (atom x)
(not (numberp x))))
'(l 1 i (b) (5) s -2 p))
;=> (L I S P)
Note that the empty list, which is often written as (), is just the symbol nil. As such, it too is a non-numeric atom. If you'd want to keep other symbols, e.g.,
CL-USER> (remove-if-not #'(lambda (x)
(and (atom x)
(not (numberp x))))
'(li (b) -1 (5) sp))
;=> (LI SP)
then you'll probably want to keep nil as well:
CL-USER> (remove-if-not #'(lambda (x)
(and (atom x)
(not (numberp x))))
'(van (b) () (5) a))
;=> (VAN NIL A)
This is the Common Lisp code:
(defun take (L)
(if (null L) nil
(cons (car L) (skip (cdr L)))))
(defun skip (L)
(if (null L) nil
(cons (car L) (take (cdr L)))))
The idea here is that, "take" will give all the odd sequence elements in the input list and "skip" will give all the even sequence elements in the input list. However, in both cases the entire list is returned.
What is the error in this code? Is this something to do with how CL handles lists, because the similar code in SML gives the desired output.
fun take(lst) =
if lst = nil then nil
else hd(lst)::skip(tl(lst))
and
skip(lst) =
if lst = nil then nil
else hd(lst)::take(tl(lst));
To expound on what Sylwester has said, your skip is wrong in both Lisp and SML. It should be
(defun take (L) ; even-indexed elements of a list L
(if (not (null L))
(cons (car L) (skip (cdr L)))))
(defun skip (L) ; odd-indexed elements of a list L
(if (not (null L))
(take (cdr L))))
and
fun take(lst) =
if lst = nil then nil
else hd(lst)::skip(tl(lst))
and
skip(lst) =
if lst = nil then nil
else take(tl(lst));
The take and skip are identical so that is no mystery. skip should just tail call instead of cons-ing. It's the consing that makes the return here.
It's worth pointing out that indexing in Common Lisp (like many other programming languages) starts with 0, so the even-indexed elements of a list are the first, the third, the fifth, and so on, since those have indices 0, 2, 4, etc. It's also worth noting that in Common Lisp, you can take the rest of the empty list and get back the empty list. (You can't do this in every Lisp, though. E.g., in Scheme it's an error to call cdr on something that's not a pair.) This means that you can implement even-elements and odd-elements rather easily. even-elementsjust returns a list of the first element, and the odd elements of the rest of the list. odd-elements returns the even-elements of the rest of the list:
(defun even-elements (list)
(if (endp list) list
(list* (first list) (odd-elements (rest list)))))
(defun odd-elements (list)
(even-elements (rest list)))
These behave in the expected fashion:
CL-USER> (even-elements '(0 1 2 3 4 5))
(0 2 4)
CL-USER> (odd-elements '(0 1 2 3 4 5))
(1 3 5)
Of course, if you note that the call to (odd-elements x) is just a call to (even-elements (rest x)), we could have implemented even-elements as follows, and had the same result:
(defun even-elements (list)
(if (endp list) list
(list* (first list) (even-elements (rest (rest list))))))
How can I remove nested parentheses recursively in Common LISP Such as
(unnest '(a b c (d e) ((f) g))) => (a b c d e f g)
(unnest '(a b)) => (a b)
(unnest '(() ((((a)))) ())) => (a)
Thanks
Here's what I'd do:
(ql:quickload "alexandria")
(alexandria:flatten list)
That works mainly because I have Quicklisp installed already.
(defun flatten (l)
(cond ((null l) nil)
((atom l) (list l))
(t (loop for a in l appending (flatten a)))))
I realize this is an old thread, but it is one of the first that comes up when I google lisp flatten. The solution I discovered is similar to those discussed above, but the formatting is slightly different. I will explain it as if you are new to lisp, as I was when I first googled this question, so it's likely that others will be too.
(defun flatten (L)
"Converts a list to single level."
(if (null L)
nil
(if (atom (first L))
(cons (first L) (flatten (rest L)))
(append (flatten (first L)) (flatten (rest L))))))
For those new to lisp, this is a brief summary.
The following line declares a function called flatten with argument L.
(defun flatten (L)
The line below checks for an empty list.
(if (null L)
The next line returns nil because cons ATOM nil declares a list with one entry (ATOM). This is the base case of the recursion and lets the function know when to stop. The line after this checks to see if the first item in the list is an atom instead of another list.
(if (atom (first L))
Then, if it is, it uses recursion to create a flattened list of this atom combined with the rest of the flattened list that the function will generate. cons combines an atom with another list.
(cons (first L) (flatten (rest L)))
If it's not an atom, then we have to flatten on it, because it is another list that may have further lists inside of it.
(append (flatten (first L)) (flatten (rest L))))))
The append function will append the first list to the start of the second list.
Also note that every time you use a function in lisp, you have to surround it with parenthesis. This confused me at first.
You could define it like this for example:
(defun unnest (x)
(labels ((rec (x acc)
(cond ((null x) acc)
((atom x) (cons x acc))
(t (rec (car x) (rec (cdr x) acc))))))
(rec x nil)))
(defun flatten (l)
(cond ((null l) nil)
((atom (car l)) (cons (car l) (flatten (cdr l))))
(t (append (flatten (car l)) (flatten (cdr l))))))
Lisp has the function remove to remove things. Here I use a version REMOVE-IF that removes every item for which a predicate is true. I test if the thing is a parenthesis and remove it if true.
If you want to remove parentheses, see this function:
(defun unnest (thing)
(read-from-string
(concatenate
'string
"("
(remove-if (lambda (c)
(member c '(#\( #\))))
(princ-to-string thing))
")")))
Note, though, as Svante mentions, one does not usually 'remove' parentheses.
Most of the answers have already mentioned a recursive solution to the Flatten problem. Using Common Lisp Object System's multiple dispatching you could solve the problem recursively by defining 3 methods for 3 possible scenarios:
(defmethod flatten ((tree null))
"Tree is empty list."
())
(defmethod flatten ((tree list))
"Tree is a list."
(append (flatten (car tree))
(flatten (cdr tree))))
(defmethod flatten (tree)
"Tree is something else (atom?)."
(list tree))
(flatten '(2 ((8) 2 (9 (d (s (((((a))))))))))) ; => (2 8 2 9 D S A)
Just leaving this here as I visited this question with the need of only flattening one level and later figure out for myself that (apply 'concatenate 'list ((1 2) (3 4) (5 6 7))) is a cleaner solution in that case.
This is a accumulator based approach. The local function %flatten keeps an accumulator of the tail (the right part of the list that's already been flattened). When the part remaining to be flattened (the left part of the list) is empty, it returns the tail. When the part to be flattened is a non-list, it returns that part prefixed onto the tail. When the part to be flattened is a list, it flattens the rest of the list (with the current tail), then uses that result as the tail for flattening the first part of the list.
(defun flatten (list)
(labels ((%flatten (list tail)
(cond
((null list) tail)
((atom list) (list* list tail))
(t (%flatten (first list)
(%flatten (rest list)
tail))))))
(%flatten list '())))
CL-USER> (flatten '((1 2) (3 4) ((5) 6) 7))
(1 2 3 4 5 6 7)
I know this question is really old but I noticed that nobody used the push/nreverse idiom, so I am uploading that here.
the function reverse-atomize takes out each "atom" and puts it into the output of the next call. At the end it produces a flattened list that is backwards, which is resolved with the nreverse function in the atomize function.
(defun reverse-atomize (tree output)
"Auxillary function for atomize"
(if (null tree)
output
(if (atom (car tree))
(reverse-atomize (cdr tree) (push (car tree) output))
(reverse-atomize (cdr tree) (nconc (reverse-atomize (car tree)
nil)
output)))))
(defun atomize (tree)
"Flattens a list into only the atoms in it"
(nreverse (reverse-atomize tree nil)))
So calling atomize '((a b) (c) d) looks like this:
(A B C D)
And if you were to call reverse-atomize with reverse-atomize '((a b) (c) d) this would occur:
(D C B A)
People like using functions like push, nreverse, and nconc because they use less RAM than their respective cons, reverse, and append functions. That being said the double recursive nature of reverse-atomize does come with it's own RAMifications.
This popular question only has recursive solutions (not counting Rainer's answer).
Let's have a loop version:
(defun flatten (tree &aux todo flat)
(check-type tree list)
(loop
(shiftf todo tree nil)
(unless todo (return flat))
(dolist (elt todo)
(if (listp elt)
(dolist (e elt)
(push e tree))
(push elt flat))))))
(defun unnest (somewhat)
(cond
((null somewhat) nil)
((atom somewhat) (list somewhat))
(t
(append (unnest (car somewhat)) (unnest (cdr somewhat))))))
I couldn't resist adding my two cents. While the CL spec does not require tail call optimization (TCO), many (most?) implementations have that feature.
So here's a tail recursive version that collects the leaf nodes of a tree into a flat list (which is one version of "removing parentheses"):
(defun flatten (tree &key (include-nil t))
(check-type tree list)
(labels ((%flatten (lst accum)
(if (null lst)
(nreverse accum)
(let ((elem (first lst)))
(if (atom elem)
(%flatten (cdr lst) (if (or elem include-nil)
(cons elem accum)
accum))
(%flatten (append elem (cdr lst)) accum))))))
(%flatten tree nil)))
It preserves null leaf nodes by default, with the option to remove them. It also preserves the left-to-right order of the tree's leaf nodes.
Note from Google lisp style guide about TCO:
You should favor iteration over recursion.
...most serious implementations (including SBCL and CCL) do implement proper tail calls, but with restrictions:
The (DECLARE (OPTIMIZE ...)) settings must favor SPEED enough and not favor DEBUG too much, for some compiler-dependent meanings of "enough" and "too much".
And this from SBCL docs:
... disabling tail-recursion optimization ... happens when the debug optimization quality is greater than 2.