Abstracting Lists in Dr. Racket - racket

Currently, I have a function item counts that is meant to count the number of elements in a string and order them into pairs if it occurs once or more.
(check-expect
(item-counts
(cons "hello"
(cons "hey"
(cons "hello" '())))
equal?)
(cons "hello" 2
(cons "hey" 1) '()))
(check-expect
(item-counts
(cons #true
(cons #false
(cons #true '())))
equal?)
(cons #true 2
(cons #false 1 '())))
(define (item-counts lop ef)
(cond
[(empty? (rest lop))
(first lop)]
[(cons? (rest lop))
(if
(ef (first lop) (item-counts (rest lop) ef))) ... ???
As shown by the tests, I wish to add a number to the existing list if one of the elements of this list occurs more than once. I used the equal? function to identify the number of times an element occurs, but I am lost on how to add the number to the existing list as well as condense the repeating elements when producing a new list. I have tried using the append function but it creates a list in a different format ( (list (list) ).

I don't know if this is self-learning, for a class, etc. I know there are standard HtDP ways to do this, recipes, etc. But I don't know these things, sorry. What I will say is that Racket is quite easy to prototype things. You can just describe your problem and don't think too much about the details. I usually like to start in the middle.
For example:
At some point in the middle of execution you will have strings yet to examine and an association list of the strings examined so far. Then your task is 1) if there are no more strings to get, we're done, otherwise 2) get the next string, add it to the table, and continue because we're not done.
(define (build-assoc from-string current-assoc)
(if (done? from-string)
current-assoc
(let ((first-string (get-first-string from-string)))
(let ((result-table (add-string first-string current-assoc))
(result-string (drop-first-string from-string)))
(build-assoc result-string
result-table)))))
Maybe you don't use this procedure in the end. But it's a picture of what you're doing, and it suggests what sort of helper procedures you will need to finish your quest. Give yourself a fighting chance and just use DrRacket to outline your problem. When you see it, then maybe it gives you some ideas how to answer the questions. DrRacket helpfully points out all your unanswered questions like "done?: unbound identifier in: done?" Ah yes, how do we know we're done, good question DrRacket. Well you see, I was given this list of strings, so...

Related

lisp: concat arbitrary number of lists

In my ongoing quest to recreate lodash in lisp as a way of getting familiar with the language I am trying to write a concat-list function that takes an initial list and an arbitrary number of additional lists and concatenates them.
I'm sure that this is just a measure of getting familiar with lisp convention, but right now my loop is just returning the second list in the argument list, which makes sense since it is the first item of other-lists.
Here's my non-working code (edit: refactored):
(defun concat-list (input-list &rest other-lists)
;; takes an arbitrary number of lists and merges them
(loop
for list in other-lists
append list into input-list
return input-list
)
)
Trying to run (concat-list '(this is list one) '(this is list two) '(this is list three)) and have it return (this is list one this is list two this is list three).
How can I spruce this up to return the final, merged list?
The signature of your function is a bit unfortunate, it becomes easier if you don't treat the first list specially.
The easy way:
(defun concat-lists (&rest lists)
(apply #'concatenate 'list lists))
A bit more lower level, using loop:
(defun concat-lists (&rest lists)
(loop :for list :in lists
:append list))
Going lower, using dolist:
(defun concat-lists (&rest lists)
(let ((result ()))
(dolist (list lists (reverse result))
(setf result (revappend list result)))))
Going even lower would maybe entail implementing revappend yourself.
It's actually good style in Lisp not to use LABELS based iteration, since a) it's basically a go-to like low-level iteration style and it's not everywhere supported. For example the ABCL implementation of Common Lisp on the JVM does not support TCO last I looked. Lisp has wonderful iteration facilities, which make the iteration intention clear:
CL-USER 217 > (defun my-append (&rest lists &aux result)
(dolist (list lists (nreverse result))
(dolist (item list)
(push item result))))
MY-APPEND
CL-USER 218 > (my-append '(1 2 3) '(4 5 6) '(7 8 9))
(1 2 3 4 5 6 7 8 9)
Some pedagogical solutions to this problem
If you just want to do this, then use append, or nconc (destructive), which are the functions which do it.
If you want to learn how do to it, then learning about loop is not how to do that, assuming you want to learn Lisp: (loop for list in ... append list) really teaches you nothing but how to write a crappy version of append using arguably the least-lispy part of CL (note I have nothing against loop & use it a lot, but if you want to learn lisp, learning loop is not how to do that).
Instead why not think about how you would write this if you did not have the tools to do it, in a Lispy way.
Well, here's how you might do that:
(defun append-lists (list &rest more-lists)
(labels ((append-loop (this more results)
(if (null this)
(if (null more)
(nreverse results)
(append-loop (first more) (rest more) results))
(append-loop (rest this) more (cons (first this) results)))))
(append-loop list more-lists '())))
There's a dirty trick here: I know that results is completely fresh so I am using nreverse to reverse it, which does so destructively. Can we write nreverse? Well, it's easy to write reverse, the non-destructive variant:
(defun reverse-nondestructively (list)
(labels ((r-loop (tail reversed)
(if (null tail)
reversed
(r-loop (rest tail) (cons (first tail) reversed)))))
(r-loop list '())))
And it turns out that a destructive reversing function is only a little harder:
(defun reverse-destructively (list)
(labels ((rd-loop (tail reversed)
(if (null tail)
reversed
(let ((rtail (rest tail)))
(setf (rest tail) reversed)
(rd-loop rtail tail)))))
(rd-loop list '())))
And you can check it works:
> (let ((l (make-list 1000 :initial-element 1)))
(time (reverse-destructively l))
(values))
Timing the evaluation of (reverse-destructively l)
User time = 0.000
System time = 0.000
Elapsed time = 0.000
Allocation = 0 bytes
0 Page faults
Why I think this is a good approach to learning Lisp
[This is a response to a couple of comments which I thought was worth adding to the answer: it is, of course, my opinion.]
I think that there are at least three different reasons for wanting to solve a particular problem in a particular language, and the approach you might want to take depends very much on what your reason is.
The first reason is because you want to get something done. In that case you want first of all to find out if it has been done already: if you want to do x and the language a built-in mechanism for doing x then use that. If x is more complicated but there is some standard or optional library which does it then use that. If there's another language you could use easily which does x then use that. Writing a program to solve the problem should be something you do only as a last resort.
The second reason is because you've fallen out of the end of the first reason, and you now find yourself needing to write a program. In that case what you want to do is use all of the tools the language provides in the best way to solve the problem, bearing in mind things like maintainability, performance and so on. In the case of CL, then if you have some problem which naturally involves looping, then, well, use loop if you want to. It doesn't matter whether loop is 'not lispy' or 'impure' or 'hacky': just do what you need to do to get the job done and make the code maintainable. If you want to print some list of objects, then by all means write (format t "~&~{~A~^, ~}~%" things).
The third reason is because you want to learn the language. Well, assuming you can program in some other language there are two approaches to doing this.
the first is to say 'I know how to do this thing (write loops, say) in languages I know – how do I do it in Lisp?', and then iterate this for all the thing you already know how to do in some other language;
the second is to say 'what is it that makes Lisp distinctive?' and try and understand those things.
These approaches result in very approaches to learning. In particular I think the first approach is often terrible: if the language you know is, say, Fortran, then you'll end up writing Fortran dressed up as Lisp. And, well, there are perfectly adequate Fortran compilers out there: why not use them? Even worse, you might completely miss important aspects of the language and end up writing horrors like
(defun sum-list (l)
(loop for i below (length l)
summing (nth i l)))
And you will end up thinking that Lisp is slow and pointless and return to the ranks of the heathen where you will spread such vile calumnies until, come the great day, the golden Lisp horde sweeps it all away. This has happened.
The second approach is to ask, well, what are the things that are interesting about Lisp? If you can program already, I think this is a much better approach to the first, because learning the interesting and distinctive features of a language first will help you understand, as quickly as possible, whether its a language you might actually want to know.
Well, there will inevitably be argument about what the interesting & distinctive features of Lisp are, but here's a possible, partial, set.
The language has a recursively-defined data structure (S expressions or sexprs) at its heart, which is used among other things to represent the source code of the language itself. This representation of the source is extremely low-commitment: there's nothing in the syntax of the language which says 'here's a block' or 'this is a conditiona' or 'this is a loop'. This low-commitment can make the language hard to read, but it has huge advantages.
Recursive processes are therefore inherently important and the language is good at expressing them. Some variants of the language take this to the extreme by noticing that iteration is simply a special case of recursion and have no iterative constructs at all (CL does not do this).
There are symbols, which are used as names for things both in the language itself and in programs written in the language (some variants take this more seriously than others: CL takes it very seriously).
There are macros. This really follows from the source code of the language being represented as sexprs and this structure having a very low commitment to what it means. Macros, in particular, are source-to-source transformations, with the source being represented as sexprs, written in the language itself: the macro language of Lisp is Lisp, without restriction. Macros allow the language itself to be seamlessly extended: solving problems in Lisp is done by designing a language in which the problem can be easily expressed and solved.
The end result of this is, I think two things:
recursion, in addition to and sometimes instead of iteration is an unusually important technique in Lisp;
in Lisp, programming means building a programming language.
So, in the answer above I've tried to give you examples of how you might think about solving problems involving a recursive data structure recursively: by defining a local function (append-loop) which then recursively calls itself to process the lists. As Rainer pointed out that's probably not a good way of solving this problem in Common Lisp as it tends to be hard to read and it also relies on the implementation to turn tail calls into iteration which is not garuanteed in CL. But, if your aim is to learn to think the way Lisp wants you to think, I think it is useful: there's a difference between code you might want to write for production use, and code you might want to read and write for pedagogical purposes: this is pedagogical code.
Indeed, it's worth looking at the other half of how Lisp might want you to think to solve problems like this: by extending the language. Let's say that you were programming in 1960, in a flavour of Lisp which has no iterative constructs other than GO TO. And let's say you wanted to process some list iteratively. Well, you might write this (this is in CL, so it is not very like programming in an ancient Lisp would be: in CL tagbody establishes a lexical environment in the body of which you can have tags – symbols – and then go will go to those tags):
(defun print-list-elements (l)
;; print the elements of a list, in order, using GO
(let* ((tail l)
(current (first tail)))
(tagbody
next
(if (null tail)
(go done)
(progn
(print current)
(setf tail (rest tail)
current (first tail))
(go next)))
done)))
And now:
> (print-list-elements '(1 2 3))
1
2
3
nil
Let's program like it's 1956!
So, well, let's say you don't like writing this sort of horror. Instead you'd like to be able to write something like this:
(defun print-list-elements (l)
;; print the elements of a list, in order, using GO
(do-list (e l)
(print e)))
Now if you were using most other languages you need to spend several weeks mucking around with the compiler to do this. But in Lisp you spend a few minutes writing this:
(defmacro do-list ((v l &optional (result-form nil)) &body forms)
;; Iterate over a list. May be buggy.
(let ((tailn (make-symbol "TAIL"))
(nextn (make-symbol "NEXT"))
(donen (make-symbol "DONE")))
`(let* ((,tailn ,l)
(,v (first ,tailn)))
(tagbody
,nextn
(if (null ,tailn)
(go ,donen)
(progn
,#forms
(setf ,tailn (rest ,tailn)
,v (first ,tailn))
(go ,nextn)))
,donen
,result-form))))
And now your language has an iteration construct which it previously did not have. (In real life this macro is called dolist).
And you can go further: given our do-list macro, let's see how we can collect things into a list:
(defun collect (thing)
;; global version: just signal an error
(declare (ignorable thing))
(error "not collecting"))
(defmacro collecting (&body forms)
;; Within the body of this macro, (collect x) will collect x into a
;; list, which is returned from the macro.
(let ((resultn (make-symbol "RESULT"))
(rtailn (make-symbol "RTAIL")))
`(let ((,resultn '())
(,rtailn nil))
(flet ((collect (thing)
(if ,rtailn
(setf (rest ,rtailn) (list thing)
,rtailn (rest ,rtailn))
(setf ,resultn (list thing)
,rtailn ,resultn))
thing))
,#forms)
,resultn)))
And now we can write the original append-lists function entirely in terms of constructs we've invented:
(defun append-lists (list &rest more-lists)
(collecting
(do-list (e list) (collect e))
(do-list (l more-lists)
(do-list (e l)
(collect e)))))
If that's not cool then nothing is.
In fact we can get even more carried away. My original answer above used labels to do iteration As Rainer has pointed out, this is not safe in CL since CL does not mandate TCO. I don't particularly care about that (I am happy to use only CL implementations which mandate TCO), but I do care about the problem that using labels this way is hard to read. Well, you can, of course, hide this in a macro:
(defmacro looping ((&rest bindings) &body forms)
;; A sort-of special-purpose named-let.
(multiple-value-bind (vars inits)
(loop for b in bindings
for var = (typecase b
(symbol b)
(cons (car b))
(t (error "~A is hopeless" b)))
for init = (etypecase b
(symbol nil)
(cons (unless (null (cddr b))
(error "malformed binding ~A" b))
(second b))
(t
(error "~A is hopeless" b)))
collect var into vars
collect init into inits
finally (return (values vars inits)))
`(labels ((next ,vars
,#forms))
(next ,#inits))))
And now:
(defun append-lists (list &rest more-lists)
(collecting
(looping ((tail list) (more more-lists))
(if (null tail)
(unless (null more)
(next (first more) (rest more)))
(progn
(collect (first tail))
(next (rest tail) more))))))
And, well, I just think it is astonishing that I get to use a programming language where you can do things like this.
Note that both collecting and looping are intentionally 'unhygenic': they introduce a binding (for collect and next respectively) which is visible to code in their bodies and which would shadow any other function definition of that name. That's fine, in fact, since that's their purpose.
This kind of iteration-as-recursion is certainly cool to think about, and as I've said I think it really helps you to think about how the language can work, which is my purpose here. Whether it leads to better code is a completely different question. Indeed there is a famous quote by Guy Steele from one of the 'lambda the ultimate ...' papers:
procedure calls may be usefully thought of as GOTO statements which also pass parameters
And that's a lovely quote, except that it cuts both ways: procedure calls, in a language which optimizes tail calls, are pretty much GOTO, and you can do almost all the horrors with them that you can do with GOTO. But GOTO is a problem, right? Well, it turns out so are procedure calls, for most of the same reasons.
So, pragmatically, even in a language (or implementation) where procedure calls do have all these nice characteristics, you end up wanting constructs which can express iteration and not recursion rather than both. So, for instance, Racket which, being a Scheme-family language, does mandate tail-call elimination, has a whole bunch of macros with names like for which do iteration.
And in Common Lisp, which does not mandate tail-call elimination but which does have GOTO, you also need to build macros to do iteration, in the spirit of my do-list above. And, of course, a bunch of people then get hopelessly carried away and the end point is a macro called loop: loop didn't exist (in its current form) in the first version of CL, and it was common at that time to simply obtain a copy of it from somewhere, and make sure it got loaded into the image. In other words, loop, with all its vast complexity, is just a macro which you can define in a CL which does not have it already.
OK, sorry, this is too long.
(loop for list in (cons '(1 2 3)
'((4 5 6) (7 8 9)))
append list)

Make a sublist from list using LiSP

I have difficulty creating a sublist: '(((3)2)1)' from the list '(3 2 1)'.
I don't know where to start, any help would be greatly appreciated.
Take a careful look at the input. You'll see that every list in the result is two elements long - except the first. Since we need to handle the first cons in one way, and all of the others another, we'll split the work between two functions, the second of which will be a loop:
(defun nest-left (list)
(if (null list)
nil
(nest-aux <?> (cdr list))))
(defun nest-aux (acc list)
(if (null list)
<?>
<?>))
I'll leave you to fill in the <?>s.

Need help to understand LISP

I am trying to write my own maximum function (with 2 elements in list at present) but getting error while executing simple function as:
(defun max_for_vararg (list)
(if (null list)
(nil))
(if (> (car list) (cdr list))
(car list)
(cdr list)))
Error as:
? (max_for_vararg '(2 4))
> Error: The value (4) is not of the expected type REAL.
> While executing: CCL::>-2, in process listener(1).
> Type :POP to abort, :R for a list of available restarts.
I appreciate if someone can help me understand it. Error seems confusing to me as similar function like below is running fine but not returning max value.
(defun max_for_vararg (list)
(if (null list)
(nil))
(if (> (car list))
(car list)
(cdr list)))
Use cadr instead of cdr. Cdr gets you the rest of the list, which is a single element list. Thus, you have to call car on that list (car (cdr list)). Since this is a common thing to want to do, they made cadr a function that evaluates out to that.
There are several errors in you code. I'll give you some pointers on how to improve it.
You try to call a function named nil.
The first if has a consequence that does (nil), thus call nil as if it is a defined function. nil in other positions is the empty list so this might be an error unless you have made a function called nil.
The first if is dead code
As long as the result of the first if does not throw you into the debugger, the second if will run. Thus when the first if is fixed it will be redundant code. You really should try to have both a consequence and an alternative even though the standard doesn't require it.
(if test-1 ; predicate
test-1-true-expression ; consequent
test-1-false-expression) ; alternative
The second if should of course be one of those expressions and not something that happens unconditional to the first.
In the updated code > needs at least two arguments to be useful.
You can think of > as a function that tests if all the arguments are in descending order. (> 4) is T since all arguments are in descending order. If you find car, cadr and caddr cryptic you may want to try the aliases first, second, third instead. eg
(> (first list) (second list)) ; is first element greater than second element?

CLISP - Reversing a simple list

I have to reverse the elements of a simple (single-dimension) list. I know there's a built-in reverse function but I can't use it for this.
Here's my attempt:
(defun LISTREVERSE (LISTR)
(cond
((< (length LISTR) 2) LISTR) ; listr is 1 atom or smaller
(t (cons (LISTREVERSE (cdr LISTR)) (car LISTR))) ; move first to the end
)
)
Output pretty close, but is wrong.
[88]> (LISTREVERSE '(0 1 2 3))
((((3) . 2) . 1) . 0)
So I tried to use append instead of cons:
(t (append (LISTREVERSE (cdr LISTR)) (car LISTR)))
But got this error:
*** - APPEND: A proper list must not end with 2
Any help?
I can give you a couple of pointers, because this looks like homework:
The base case of the recursion is when the list is empty (null), and not when there are less than two elements in the list
Consider defining a helper function with an extra parameter, an "accumulator" initialized in the empty list. For each element in the original list, cons it at the head of the accumulator. When the input list is empty, return the accumulator
As an aside note, the above solution is tail-recursive.
As a follow-up to Óscar López (and fighting the temptation to just write a different solution down):
Using both append and length makes the posted solution just about the least efficient way of reversing a list. Check out the documentation on cons and null for some better ideas on how to implement this.
Please, please indent properly.
Tail recursion really is both more efficient and reasonably simple in this case. Try it if you haven't already. labels is the form you want to use to define local recursive functions.
It may be worth your while to flip through The Little Schemer. It'll give you a better feel for recursion in general.
It's ok what you did. You only missed the composition of the result list.
Think about it: You have to append the 1-element list of the CAR to the end of the list of the reversed CDR:
(defun LISTREVERSE (LISTR)
(cons
((< (length LISTR) 2) LISTR) ; listr is 1 atom or smaller
(t (append (LISTREVERSE (cdr LISTR)) (list (car LISTR))))))
(defun listreverse (list)
(let (result)
(dolist (item list result)
(push item result))))
Don't use recursion in Common Lisp when there is a simple iterative way to reach the same goal. Common Lisp does not make any guarantees about tail recursion, and your tail recursive function invocation may not be optimized to a jump at the discretion of the compiler.
push prepends the item to the result
dolist has an optional third argument which is the value returned. It is evaluated when the loop is exited.

Get Last Element of each list

Let us say I have a list ((3 4 5) (d e f) (h i j) (5 5 5 5))
How can I get the last element of each list in such a way that the output would look like this (5 f j 5)?
Assuming this is about Common Lisp, there is a function last which returns a list containing the last item of a list. If you use this function with mapcan, which applies a given function to each element of a list and returns the concatenated results, you'll get what you want.
Note though that accessing the last element of a list is an O(N) operation, so if this isn't just homework after all, you might want to consider if you can't solve the real problem more efficiently than taking the last item of each list (maybe use another datastructure instead).
This, like most early LISPy homework problems is an exercise in thinking recursively and/or thinking in terms of induction. The way to start is to ask yourself simple questions that you can answer easily.
For example, if you had been asked to write something that gave you the first element in each list, I would thing about it this way:
Given a list of lists:
What is first-element of every list in the list '()? (easy - null)
What is first-element of every list in the list '(a)? (easy - a, or maybe an error)
What is first-element of every list in the list '((a))? (easy - (a))
What is first-element of any list in the form '(anything), where anything is a list? (easy - (first anything))
What is the first element of every list in the form '(anything morestuff)? (easy - (cons (first anything) (first-element morestuff)) )
What is first of an atom? either the atom or an error (depends on your point of view)
What is first of null? nil.
What is first of a list? (car list)
From here we can start writing code:
;; here's first, meeting questions 6-8
(define first (lambda (l)
(cond
((null? l) nil) ; Q7
((atom? l) l) ; Q6
(t (car l))))) ; Q8
;; with first we can write first-element, meeting questions 1-5
(define first-element (lambda (l)
(cond
((null? l) nil) ; Q1
((atom? l) (first l)) ; Q2
(t (cons (first (car l) (first-element (cdr l)))))))) ; Q4-5
Now this isn't your homework (intentionally). You should play with this and understand how it works. Your next goal should be to find out how this differs from your assignment and how to get there.
With respect to MAPCAR? Don't worry about it. You need to learn how to solve recursive problems first. Then you can worry about MAPCAR. What is the point of this assignment? To help you learn to think in this mode. Dang near everything in LISP/Scheme is solved by thinking this way.
The reason I went with all the questions to break it down into the parts that I'm worried about. If I'm given the task "how do I do foo on every item in a list?" I should answer the questions: How do I do handle null? How do handle an atom? How do I do handle on the first element on the list? How do I handle everything else? Once I've answered that, then I figure out how to actually do foo. How do I do foo on null? How do I do foo on an atom? How do I do foo on a list?
(defun get-last-lists (s)
(setq rt 'nil)
(loop for i from 0 to (- (length s) 1)
do (setq rt (append rt (last (nth i s)))))
(print rt))
as a beginner of lisp, i post my solution.
Write a procedure that returns the last element of a list, then learn a little about the built-in MAP (a.k.a. MAPCAR) procedure and see if any lightbulbs go off.
probably it is already solved, but I figured this out
; SELECT-FROM-INNER-LIST :: [list] -> [list]
(DEFUN SFIL (lst)
(COND ((NULL lst) NIL)
((LISTP (FIRST lst)) (APPEND (LAST (FIRST lst)) (SFIL (REST lst))))
))
Now, this works for legit list...so if you call function SFIL with correct list.... if not, it will return NIL
hopefully this will be helpful, for anyone who finds it