Say, we would like to use a function that requires a predicate, but for some reason we're interested in other features of the function (like :start and :end parameters), so we need to supply a predicate, that always returns true.
Obviously, it's not a problem at all:
CL-USER> (defparameter *list* '(0 1 2 3 4 5))
*LIST*
CL-USER> (remove-if (lambda (x) t) *list* :start 1 :end 3)
(0 3 4 5)
Works, but not beautifully at all. And we may get ugly message that variable x is not used. Since I'm in LISP for beauty, I'm curious if there is a 'always t' predicate?
We can define it:
(defun tp (&rest rest)
(declare (ignore rest))
t)
..but may be it exists?
You're looking for the function constantly that takes an argument and returns a function that always returns that value. The predicate you need, then is (constantly t). Thus:
CL-USER> (remove-if (constantly t) '(0 1 2 3 4 5) :start 1 :end 3)
(0 3 4 5)
The notes on constantly show that you were absolutely on the right track with your proposed implementation. (You did even better, though, by adding the (declare (ignore …)).)
Notes:
constantly could be defined by:
(defun constantly (object)
#'(lambda (&rest arguments) object))
After writing this, it occurred to me that this might be a duplicate. I didn't find a proper duplicate, found a similar, more specific, question that was looking to remove a single element at a position, Is there a common lisp macro for popping the nth element from a list?, in which Rainer Joswig's answer includes:
Removing the nth element of a list:
(defun remove-nth (list n)
(remove-if (constantly t) list :start n :end (1+ n)))
This is really just a generalization of that approach, since you're working with arbitrary sequence boundaries. Thus we could have (making the boundary argument analogous to subseq's):
(defun remove-subseq (sequence &optional (start 0) end)
(remove-if (constantly t) sequence :start start :end end))
(defun remove-nth (sequence n)
(remove-subseq sequence n (1+ n)))
CL-USER> (remove-subseq '(0 1 2 3 4 5) 1 3)
(0 3 4 5)
CL-USER> (remove-nth '(0 1 2 3 4 5) 3)
(0 1 2 4 5)
Related
i would like to visit all the cons cells in a list and perform some action on them (including such things as setcar). is there an idiomatic way of doing this?
i can, i think, do something like this
(progn
(setq a (list 1 2 3 4 5 6))
(setq okay a)
(while okay
(if (eq (car okay) 3)
(setcar okay 22))
(setq okay (cdr okay))))
(where the if expression is my "application logic", say.)
but, if there's a terser way of doing this, i'd be interested in hearing about it.
If you want to mutate the cars of the list, then in recent emacsen the likely think you want is cl-mapl, which maps a function over successive tails of the list. This is essentially Common Lisp's mapl function: CL has
maplist which maps a function over tails and returns a new list of the values of the function, so (maplist (lambda (e) e) '(1 2 3)) is ((1 2 3) (2 3) (3));
mapl which is like maplist but returns the original list.
elisp (courtesy of some now-standard library) now has both cl-mapl and cl-maplist.
So:
> (let ((y (list 1 2 3 4 5 6 7)))
(cl-mapl (lambda (tail)
(rplaca tail 0))
y)
y)
(0 0 0 0 0 0 0)
or
> (let ((y (list 1 2 3 4 5 6 7)))
(cl-mapl (lambda (tail)
(rplaca tail (if (cdr tail) (cadr tail) 'fish)))
y)
y)
(2 3 4 5 6 7 fish)
(In neither of these cases do you need to make sure that y is returned: I just did it to make it clear that y is being destructively modified by this.)
(setq a (mapcar (lambda (x) (if (equal x 3) 22 x)) a))
That sets the value of variable a to the result of changing any members of a that are 3 to 22.
Choose the equality operator you want, equal, eql, or =, depending on whether you know that either all list members are numbers (use =) or you know that they are either numbers or you want to test object equality otherwise, (use eql), or you don't know what they might be (use equal).
You haven't indicated any need to do list-structure modification (setcar). It appears that all you care about is for a to be a list as I described.
I am struggling to find the right approach to solve the following function
(FOO #'– '(1 2 3 4 5))
=> ((–1 2 3 4 5) (1 –2 3 4 5) (1 2 –3 4 5) (1 2 3 –4 5) (1 2 3 4 –5))
The first Parameter to the foo function is supposed to be a function "-" that has to be applied to each element returning a list of list as shown above. I am not sure as to what approach I can take to create this function. I thought of recursion but not sure how I will preserve the list in each call and what kind of base criteria would I have. Any help would be appreciated. I cannot use loops as this is functional programming.
It's a pity you cannot use loop because this could be elegantly solved like so:
(defun foo (fctn lst)
(loop
for n from 0 below (length lst) ; outer
collect (loop
for elt in lst ; inner
for i from 0
collect (if (= i n) (funcall fctn elt) elt))))
So we've got an outer loop that increments n from 0 to (length lst) excluded, and an inner loop that will copy verbatim the list except for element n where fctn is applied:
CL-USER> (foo #'- '(1 2 3 4 5))
((-1 2 3 4 5) (1 -2 3 4 5) (1 2 -3 4 5) (1 2 3 -4 5) (1 2 3 4 -5))
Replacing loop by recursion means creating local functions by using labels that replace the inner and the outer loop, for example:
(defun foo (fctn lst)
(let ((len (length lst)))
(labels
((inner (lst n &optional (i 0))
(unless (= i len)
(cons (if (= i n) (funcall fctn (car lst)) (car lst))
(inner (cdr lst) n (1+ i)))))
(outer (&optional (i 0))
(unless (= i len)
(cons (inner lst i) (outer (1+ i))))))
(outer))))
Part of the implementation strategy that you choose here will depend on whether you want to support structure sharing or not. Some of the answers have provided solutions where you get completely new lists, which may be what you want. If you want to actually share some of the common structure, you can do that too, with a solution like this. (Note: I'm using first/rest/list* in preference to car/car/cons, since we're working with lists, not arbitrary trees.)
(defun foo (operation list)
(labels ((foo% (left right result)
(if (endp right)
(nreverse result)
(let* ((x (first right))
(ox (funcall operation x)))
(foo% (list* x left)
(rest right)
(list* (revappend left
(list* ox (rest right)))
result))))))
(foo% '() list '())))
The idea is to walk down list once, keeping track of the left side (in reverse) and the right side as we've gone through them, so we get as left and right:
() (1 2 3 4)
(1) (2 3 4)
(2 1) (3 4)
(3 2 1) (4)
(4 3 2 1) ()
At each step but the last, we take the the first element from the right side, apply the operation, and create a new list use revappend with the left, the result of the operation, and the rest of right. The results from all those operations are accumulated in result (in reverse order). At the end, we simply return result, reversed. We can check that this has the right result, along with observing the structure sharing:
CL-USER> (foo '- '(1 2 3 4 5))
((-1 2 3 4 5) (1 -2 3 4 5) (1 2 -3 4 5) (1 2 3 -4 5) (1 2 3 4 -5))
By setting *print-circle* to true, we can see the structure sharing:
CL-USER> (setf *print-circle* t)
T
CL-USER> (let ((l '(1 2 3 4 5)))
(list l (foo '- l)))
((1 . #1=(2 . #2=(3 . #3=(4 . #4=(5))))) ; input L
((-1 . #1#)
(1 -2 . #2#)
(1 2 -3 . #3#)
(1 2 3 -4 . #4#)
(1 2 3 4 -5)))
Each list in the output shares as much structure with the original input list as possible.
I find it easier, conceptually, to write some of these kind of functions recursively, using labels, but Common Lisp doesn't guarantee tail call optimization, so it's worth writing this iteratively, too. Here's one way that could be done:
(defun phoo (operation list)
(do ((left '())
(right list)
(result '()))
((endp right)
(nreverse result))
(let* ((x (pop right))
(ox (funcall operation x)))
(push (revappend left (list* ox right)) result)
(push x left))))
The base case of a recursion can be determined by asking yourself "When do I want to stop?".
As an example, when I want to compute the sum of an integer and all positive integers below it, I can do this recusively with a base case determined by answering "When do I want to stop?" with "When the value I might add in is zero.":
(defun sumdown (val)
(if (zerop val)
0
(+ (sumdown (1- val)) val)))
With regard to 'preserve the list in each call', rather than trying to preserve anything I would just build up a result as you go along. Using the 'sumdown' example, this can be done in various ways that are all fundamentally the same approach.
The approach is to have an auxiliary function with a result argument that lets you build up a result as you recurse, and a function that is intended for the user to call, which calls the auxiliary function:
(defun sumdown1-aux (val result)
(if (zerop val)
result
(sumdown1-aux (1- val) (+ val result))))
(defun sumdown1 (val)
(sumdown1-aux val 0))
You can combine the auxiliary function and the function intended to be called by the user by using optional arguments:
(defun sumdown2 (val &optional (result 0))
(if (zerop val)
result
(sumdown2 (1- val) (+ val result))))
You can hide the fact that an auxiliary function is being used by locally binding it within the function the user would call:
(defun sumdown3 (val)
(labels ((sumdown3-aux (val result)
(if (zerop val)
result
(sumdown3-aux (1- val) (+ val result)))))
(sumdown3-aux val 0)))
A recursive solution to your problem can be implemented by answering the question "When do I want to stop when I want to operate on every element of a list?" to determine the base case, and building up a result list-of-lists (instead of adding as in the example) as you recurse. Breaking the problem into smaller pieces will help - "Make a copy of the original list with the nth element replaced by the result of calling the function on that element" can be considered a subproblem, so you might want to write a function that does that first, then use that function to write a function that solves the whole problem. It will be easier if you are allowed to use functions like mapcar and substitute or substitute-if, but if you are not, then you can write equivalents yourself out of what you are allowed to use.
In Arc there's a macro called in
> (let x 1
(in x 4 5 6))
nil
> (let x 1
(in x 1 5 6))
t
that checks if its first parameter is equal to any of the rest. I want a version of this that takes a parameter plus a list (semantically identical to Python's in), so I wrote:
(assign weak-tens* '(11 12))
(mac in? (elt lst)
(cons 'in (cons elt lst)))
(def transform-deck (deck)
(map [if (in? _ weak-tens*) #\T _] deck))
output:
arc> (load "main.arc")
*** redefining in?
map: contract violation
expected: list?
given: '(_ . weak-tens*)
argument position: 2nd
other arguments...:
#<procedure:ac-niltree>
To answer your immediate question, you can use mem, as follows:
arc> (let x 3
(mem x '(1 2 3 4)))
(3 4)
Also, there's no reason this should be a macro. It doesn't do anything that requires a macro.
But let's look at why the macro doesn't work:
arc> (macex1 '(in? 1 '(1 2 3)))
(in 1 quote (1 2 3))
Ah, we're putting the value "quote".
Here's how we want the code to expand:
(in? 1 '(1 2 3))
should expand to:
(in 1 1 2 3)
But as mentioned, we don't even want this to be a macro in the first place. Ignoring mem, it could be written as follows:
(def in? (elt lst)
(if (no lst)
nil
(is elt ;;if we've found the element
(car lst))
t
(in? elt (cdr lst)))) ;;otherwise recurse
I'm reading/working through Practical Common Lisp. I'm on the chapter about building a test framework in Lisp.
I have the function "test-+" implemented as below, and it works:
(defun test-+ ()
(check
(= (+ 1 2) 3)
(= (+ 5 6) 11)
(= (+ -1 -6) -7)))
Remember, I said, it works, which is why what follows is so baffling....
Here is some code that "test-+" refers to:
(defmacro check (&body forms)
`(combine-results
,#(loop for f in forms collect `(report-result ,f ',f))))
(defmacro combine-results (&body forms)
(with-gensyms (result)
`(let ((,result t))
,#(loop for f in forms collect `(unless ,f (setf ,result nil)))
,result)))
(defmacro with-gensyms ((&rest names) &body body)
`(let ,(loop for n in names collect `(,n (gensym)))
,#body))
(defun report-result (value form)
(format t "~:[FAIL~;pass~] ... ~a~%" value form)
value)
Now, what I've been doing is using Slime to macro-expand these, step by step (using ctrl-c RET, which is mapped to macroexpand-1).
So, the "check" call of "test-+" expands to this:
(COMBINE-RESULTS
(REPORT-RESULT (= (+ 1 2) 3) '(= (+ 1 2) 3))
(REPORT-RESULT (= (+ 5 6) 11) '(= (+ 5 6) 11))
(REPORT-RESULT (= (+ -1 -6) -7) '(= (+ -1 -6) -7)))
And then that macro-expands to this:
(LET ((#:G2867 T))
(UNLESS (REPORT-RESULT (= (+ 1 2) 3) '(= (+ 1 2) 3)) (SETF #:G2867 NIL))
(UNLESS (REPORT-RESULT (= (+ 5 6) 11) '(= (+ 5 6) 11)) (SETF #:G2867 NIL))
(UNLESS (REPORT-RESULT (= (+ -1 -6) -7) '(= (+ -1 -6) -7))
(SETF #:G2867 NIL))
#:G2867)
And it is THAT code, directly above this sentence, which doesn't work. If I paste that into the REPL, I get the following error (I'm using Clozure Common Lisp):
Unbound variable: #:G2867 [Condition of type UNBOUND-VARIABLE]
Now, if I take that same code, replace the gensym with a variable name such as "x", it works just fine.
So, how can we explain the following surprises:
The "test-+" macro, which calls all of this, works fine.
The macro-expansion of the "combine-results" macro does not run.
If I remove the gensym from the macro-expansion of "combine-results", it
does work.
The only thing I can speculate is that you cannot use code the contains literal usages of gensyms. If so, why not, and how does one work around that? And if that is not the explanation, what is?
Thanks.
GENSYM creates uninterned symbols. When the macro runs normally, this isn't a problem, because the same uninterned symbol is being substituted throughout the expression.
But when you copy and paste the expression into the REPL, this doesn't happen. #: tells the reader to return an uninterned symbol. As a result, each occurrence of #:G2867 is a different symbol, and you get the unbound variable warning.
If you do (setq *print-circle* t) before doing the MACROEXPAND it will use #n= and #n# notation to link the identical symbols together.
The code, after being printed and read back, is no longer the same code. In particular, the two instances of #:G2867 in the printed representation would be read back as two separated symbols (albeit sharing the same name), while they should be the same in the original internal representation.
Try setting *PRINT-CIRCLE* to T to preserve the identity in the printed representation of the macro-expanded code.
I'm pretty fresh to the Common Lisp scene and I can't seem to find an quick way to get the nth element from a list and remove it from said list at the same time. I've done it, but it ain't pretty, what I'd really like is something like "pop" but took a second parameter:
(setf x '(a b c d))
(setf y (popnth 2 x))
; x is '(a b d)
; y is 'c
I'm pretty sure that "popnth" would have to be a macro, in case the parameter was 0 and it had to behave like "pop".
EDIT: Here's my crap first version:
(defmacro popnth (n lst)
(let ((tempvar (gensym)))
`(if (eql ,n 0)
(pop ,lst)
(let ((,tempvar (nth ,n ,lst)))
(setf (cdr (nthcdr ,(- n 1) ,lst)) (nthcdr ,(+ n 1) ,lst))
,tempvar))))
Something like this:
Removing the nth element of a list:
(defun remove-nth (list n)
(remove-if (constantly t) list :start n :end (1+ n)))
constantly returns a function, that always returns its argument.
As a macro that accepts a place, using define-modify-macro:
(define-modify-macro remove-nth-f (n) remove-nth "Remove the nth element")
POP-NTH
(defmacro pop-nth (list n)
(let ((n-var (gensym)))
`(let ((,n-var ,n))
(prog1 (nth ,n-var ,list)
(remove-nth-f ,list ,n-var)))))
Example:
CL-USER 26 > (defparameter *list* (list 1 2 3 4))
*LIST*
CL-USER 27 > (pop-nth *list* 0)
1
CL-USER 28 > *list*
(2 3 4)
CL-USER 29 > (pop-nth *list* 2)
4
CL-USER 30 > *list*
(2 3)
Yes, Lisp has a macro for popping the N-th element of a list: it is called pop.
$ clisp -q
[1]> (defvar list (list 0 1 2 3 4 5))
LIST
[2]> (pop (cdddr list))
3
[3]> list
(0 1 2 4 5)
[4]>
pop works with any form that denotes a place.
The problem is that, unlike cddr, nthcdr isn't an accessor; a form like (nthcdr 3 list) does not denote a place; it works only as a function call.
Writing a specialized form of pop is not the best answer; rather, we can achieve a more general fix by writing a clone of nthcdr which behaves like a place accessor. Then the pop macro will work, and so will every other macro that works with places like setf and rotatef.
;; our clone of nthcdr called cdnth
(defun cdnth (idx list)
(nthcdr idx list))
;; support for (cdnth <idx> <list>) as an assignable place
(define-setf-expander cdnth (idx list &environment env)
(multiple-value-bind (dummies vals newval setter getter)
(get-setf-expansion list env)
(let ((store (gensym))
(idx-temp (gensym)))
(values dummies
vals
`(,store)
`(let ((,idx-temp ,idx))
(progn
(if (zerop ,idx-temp)
(progn (setf ,getter ,store))
(progn (rplacd (nthcdr (1- ,idx-temp) ,getter) ,store)))
,store))
`(nthcdr ,idx ,getter)))))
Test:
$ clisp -q -i cdnth.lisp
;; Loading file cdnth.lisp ...
;; Loaded file cdnth.lisp
[1]> (defvar list (list 0 1 2 3 4 5))
LIST
[2]> (pop (cdnth 2 list))
2
[3]> list
(0 1 3 4 5)
[4]> (pop (cdnth 0 list))
0
[5]> list
(1 3 4 5)
[6]> (pop (cdnth 3 list))
5
[7]> list
(1 3 4)
[8]> (pop (cdnth 1 list))
3
[9]> list
(1 4)
[10]> (pop (cdnth 1 list))
4
[11]> list
(1)
[12]> (pop (cdnth 0 list))
1
[13]> list
NIL
[14]>
A possible improvement to the implementation is to analyze the idx form and optimize away the generated code that implements the run-time check on the value of idx. That is to say, if idx is a constant expression, there is no need to emit the code which tests whether idx is zero. The appropriate code variant can just be emitted. Not only that, but for small values of idx, the code can emit special variants based on the "cadavers": cddr, cdddr, rather than the general nthcdr. However, some of these optimizations might be done by the Lisp compiler and thus redundant.
I came up with a solution that is a little more efficient than my first attempt:
(defmacro popnth (n lst)
(let ((t1 (gensym))(t2 (gensym)))
`(if (eql ,n 0)
(pop ,lst)
(let* ((,t1 (nthcdr (- ,n 1) ,lst))
(,t2 (car (cdr ,t1))))
(setf (cdr ,t1) (cddr ,t1))
,t2))))
Here is it in action:
[2]> (defparameter *list* '(a b c d e f g))
*LIST*
[3]> (popnth 3 *list*)
D
[4]> *list*
(A B C E F G)
[5]> (popnth 0 *list*)
A
[6]> *list*
(B C E F G)
I have same suspicion as #6502...If I remember right...Neither push nor pop can be defined as modify-macros, the former because the place is not its first argument, and the latter because its return value is not the modified object.
Definition of define-modify-macro
An expression of the form (define-modify-macro m (p1 ... pn) f) defines a new macro m, such that a call of the form (m place a1 ... an) will cause place to be set to (f val a1 ... an), where val represents the value of place. The parameters may also include rest and optional parameters. The string, if present, becomes the documentation of the new macro.
I have this popnth works just fine:
(defun nthpop (index lst)
(pop (nthcdr (1- index) lst)))
> *list*
(1 2 3 4 5)
> (nthpop 2 *list*)
2
> *list*
(1 3 4 5)