I am at the beginning of the road. I tried to understand this cons part for 2-3 hours, but I have fallen into trouble. I couldn't understand this cons part. How does it combine 'a d' according to result?
(define remv
(lambda (x ls)
(cond
[(null? ls) '()]
[(eqv? (car ls) x) (remv x (cdr ls))]
[else (cons (car ls) (remv x (cdr ls)))])))
When I put a input like this:
> (remv 'b '(a b b d))
I get (a d).
But I couldn't understand how it became 'a d'.
To remove all occurrences of x from a list ls, you have three cases to consider:
ls is empty, in which case the result is also empty,
The first element of ls is equal to x, in which case you just drop it and return the rest of the list after you have removed all xes from it,
The first element of ls is different from x, in which case you keep that element and stick it at the front of a list which has no xes in it.
Related
I found this Scheme code on the internet, it outputs all subsets given a list, can someone explain how it works?
(define (subsets s) (if (null? s)
(list ())
(let ((rest (subsets (cdr s))))(append rest (map (lambda (x) (cons (car s) x)) rest)))))
(subsets '(a b c))
You really need to format it:
(define (subsets s)
(if (null? s)
(list ())
(let ((rest (subsets (cdr s))))
(append rest
(map (lambda (x)
(cons (car s) x))
rest)))))
What it does is return (()) for the argument '()
For a one element argument eg. '(b) if binds rest to the subsets of every element execept the first.. Which is no elements so rest is (()) and then it returns a list consisting of all the elements in rest as well as each list element of rest with b added to front. Thus (() (b)).
For a two element argument eg. '(a b) it binds rest to the subsets of every element except the first. Which is (b) so we know from above that it is (() (b)) and we know it will use those as well as every element with a added to the beginning: (() (b) (a) (a b))
I could go on but I'm guessing you get the picture?
I'm totally new to Scheme and I am trying to implement my own map function. I've tried to find it online, however all the questions I encountered were about some complex versions of map function (such as mapping functions that take two lists as an input).
The best answer I've managed to find is here: (For-each and map in Scheme). Here is the code from this question:
(define (map func lst)
(let recur ((rest lst))
(if (null? rest)
'()
(cons (func (car rest)) (recur (cdr rest))))))
It doesn't solve my problem though because of the usage of an obscure function recur. It doesn't make sense to me.
My code looks like this:
(define (mymap f L)
(cond ((null? L) '())
(f (car L))
(else (mymap (f (cdr L))))))
I do understand the logic behind the functional approach when programming in this language, however I've been having great difficulties with coding it.
The first code snippet you posted is indeed one way to implement the map function. It uses a named let. See my comment on an URL on how it works. It basically is an abstraction over a recursive function. If you were to write a function that prints all numbers from 10 to 0 you could write it liks this
(define (printer x)
(display x)
(if (> x 0)
(printer (- x 1))))
and then call it:
(printer 10)
But, since its just a loop you could write it using a named let:
(let loop ((x 10))
(display x)
(if (> x 0)
(loop (- x 1))))
This named let is, as Alexis King pointed out, syntactic sugar for a lambda that is immediately called. The above construct is equivalent to the snippet shown below.
(letrec ((loop (lambda (x)
(display x)
(if (> x 0)
(loop (- x 1))))))
(loop 10))
In spite of being a letrec it's not really special. It allows for the expression (the lambda, in this case) to call itself. This way you can do recursion. More on letrec and let here.
Now for the map function you wrote, you are almost there. There is an issue with your two last cases. If the list is not empty you want to take the first element, apply your function to it and then apply the function to the rest of the list. I think you misunderstand what you actually have written down. Ill elaborate.
Recall that a conditional clause is formed like this:
(cond (test1? consequence)
(test2? consequence2)
(else elsebody))
You have any number of tests with an obligatory consequence. Your evaluator will execute test1? and if that evaluated to #t it will execute the consequence as the result of the entire conditional. If test1? and test2? fail it will execute elsebody.
Sidenote
Everything in Scheme is truthy except for #f (false). For example:
(if (lambda (x) x)
1
2)
This if test will evaluate to 1 because the if test will check if (lambda (x) x) is truthy, which it is. It is a lambda. Truthy values are values that will evaluate to true in an expression where truth values are expected (e.g., if and cond).
Now for your cond. The first case of your cond will test if L is null. If that is evaluated to #t, you return the empty list. That is indeed correct. Mapping something over the empty list is just the empty list.
The second case ((f (car L))) literally states "if f is true, then return the car of L".
The else case states "otherwise, return the result mymap on the rest of my list L".
What I think you really want to do is use an if test. If the list is empty, return the empty list. If it is not empty, apply the function to the first element of the list. Map the function over the rest of the list, and then add the result of applying the function the first element of the list to that result.
(define (mymap f L)
(cond ((null? L) '())
(f (car L))
(else (mymap (f (cdr L))))))
So what you want might look look this:
(define (mymap f L)
(cond ((null? L) '())
(else
(cons (f (car L))
(mymap f (cdr L))))))
Using an if:
(define (mymap f L)
(if (null? L) '()
(cons (f (car L))
(mymap f (cdr L)))))
Since you are new to Scheme this function will do just fine. Try and understand it. However, there are better and faster ways to implement this kind of functions. Read this page to understand things like accumulator functions and tail recursion. I will not go in to detail about everything here since its 1) not the question and 2) might be information overload.
If you're taking on implementing your own list procedures, you should probably make sure they're using a proper tail call, when possible
(define (map f xs)
(define (loop xs ys)
(if (empty? xs)
ys
(loop (cdr xs) (cons (f (car xs)) ys))))
(loop (reverse xs) empty))
(map (λ (x) (* x 10)) '(1 2 3 4 5))
; => '(10 20 30 40 50)
Or you can make this a little sweeter with the named let expression, as seen in your original code. This one, however, uses a proper tail call
(define (map f xs)
(let loop ([xs (reverse xs)] [ys empty])
(if (empty? xs)
ys
(loop (cdr xs) (cons (f (car xs)) ys)))))
(map (λ (x) (* x 10)) '(1 2 3 4 5))
; => '(10 20 30 40 50)
I wish to replace the first occurrence of a symbol within pairs. For example:
take
(define n '((a . b) . (a . d)))
and i define a method context to replace the first instance (left most) of X with '()
replacing a should give me:
((() . b) a . d)
however i am stuck as my method replaces ALL instances and i am not sure how to add a check for this.
my code is as follows:
(define (context s sym)
(cond ((null? s) #f)
((atom? s)
(if (equal? s sym) '() s ))
(else (cons (context (car s) sym)
(context (cdr s) sym)))))
which gives : ((() . b) () . d)
any help? Thank you
The quickest way is to use a flag indicating whether the replacement has already been done, something along the lines of:
(define (context sxp sym)
(define done #f)
(let loop ((sxp sxp))
(cond (done sxp)
((pair? sxp) (cons (loop (car sxp)) (loop (cdr sxp))))
((eq? sym sxp) (set! done #t) '())
(else sxp))))
It's not very elegant to use set!, but the alternative would be to have the procedure return 2 values, and the resulting let-values code would be even worse in terms of readability IMO.
Also note that I didn't use atom? because it's not defined in standard Scheme; the usual way is to successively test null? then pair?, and handle the atom case in the else clause.
This is a bit more general (you can replace things other than symbols, and you can customize the test, and you can specify any particular number of instances to replace, not just one), and may be a little bit more complicated at first glance than what you're looking for, but here's a solution that works by internally using a continuation-passing style helper function. The main function, subst-n takes a new element, and old element, a tree, a test, and a count. It replaces the first count occurrences of new (as compared with test) with old (or all, if count is not a non-negative integer).
(define (subst-n new old tree test count)
(let substs ((tree tree)
(count count)
(k (lambda (tree count) tree)))
(cond
;; If count is a number and zero, we've replaced enough
;; and can just "return" this tree unchanged.
((and (number? count) (zero? count))
(k tree count))
;; If the tree is the old element, then "return" the new
;; element, with a decremented count (if count is a number).
((test old tree)
(k new (if (number? count) (- count 1) count)))
;; If tree is a pair, then recurse on the left side,
;; with a continuation that will recurse on the right
;; side, and then put the sides together.
((pair? tree)
(substs (car tree) count
(lambda (left count)
(substs (cdr tree) count
(lambda (right count)
(k (cons left right) count))))))
;; Otherwise, there's nothing to do but return this
;; tree with the unchanged count.
(else
(k tree count)))))
> (display (subst-n '() 'a '((a . b) . (a . d)) eq? 1))
((() . b) a . d)
> (display (subst-n '() 'a '((a . b) . (a . d)) eq? 2))
((() . b) () . d)
I'm new to Racket and trying to learn it. I'm working through some problems that I'm struggling with. Here is what the problem is asking:
Write a definition for the recursive function occur that takes a data expression a and a list s and returns the number of times that the data expression a appears in the list s.
Example:
(occur '() '(1 () 2 () () 3)) =>3
(occur 1 '(1 2 1 ((3 1)) 4 1)) => 3 (note that it only looks at whole elements in the list)
(occur '((2)) '(1 ((2)) 3)) => 1
This is what I have written so far:
(define occur
(lambda (a s)
(cond
((equal? a (first s))
(else (occur a(rest s))))))
I'm not sure how to implement the count. The next problem is similar and I have no idea how to approach that. Here is what this problem says:
(This is similar to the function above, but it looks inside the sublists as well) Write a recursive function atom-occur?, which takes two inputs, an atom a and a list s, and outputs the Boolean true if and only if a appears somewhere within s, either as one of the data expressions in s, or as one of the data expression in one of the data expression in s, or…, and so on.
Example:
(atom-occur? 'a '((x y (p q (a b) r)) z)) => #t
(atom-occur? 'm '(x (y p (1 a (b 4)) z))) => #f
Any assistance would be appreciated. Thank you.
In Racket, the standard way to solve this problem would be to use built-in procedures:
(define occur
(lambda (a s)
(count (curry equal? a) s)))
But of course, you want to implement it from scratch. Don't forget the base case (empty list), and remember to add one unit whenever a new match is found. Try this:
(define occur
(lambda (a s)
(cond
((empty? s) 0)
((equal? a (first s))
(add1 (occur a (rest s))))
(else (occur a (rest s))))))
The second problem is similar, but it uses the standard template for traversing a list of lists, where we go down on the recursion on both the first and the rest of the input list, and only test for equality when we're in an atom:
(define atom-occur?
(lambda (a s)
(cond
((empty? s) #f)
((not (pair? s))
(equal? a s))
(else (or (atom-occur? a (first s))
(atom-occur? a (rest s)))))))
I need a function that will take in a list of characters and numbers, and then return the numbers added up (ignoring the characters). This is what I have so far:
(define (adder lst)
(cond
((null? lst)
0)
((number? (car lst))
(+(adder (car lst)) (adder (cdr lst))))
((char? (car lst))
((adder(cdr lst))))
))
(display (adder '(asd12sdf)))
Running it on codepad.org just displays void. I know the code is wrong because it looks wrong, but I have no idea how to fix it... How do I have the function keep track of the first number it finds and add it to the next one it finds, while skipping all characters?
In your second cond case, there's no reason to run adder on (car lst). Just adding (car list) itself to the recursive step should work.
For the last line, don't test (char? (car lst)). Just make the last line the else clause, meaning that anything BUT a number will go to the else line.
The reason you're getting void is because your input doesn't satisfy any of the cond conditions, and you have no else, so the answer is nothing (i.e. (void)).
The last mistake is in the input you're giving it. '(asd12sdf) is literally a list with one symbol named "asd12sdf". I think you want to give it '(a s d 1 2 s d f) (a list of 6 symbols and 2 numbers) which should result in 3. Notice that there's a very important difference between the symbol 'a and the character #\a.
It looks like you have the logic down, so your problem doesn't seem to be functional languages, just Scheme's syntax.
Edit: and in the last line, you have ((adder(cdr lst))) which has one too many parens wrapped around it. That will cause Scheme to attempt to evaluate the result of adder (which is a number) as a procedure (error!).
You should observe that this function is more or less sum which can be defined simply by using fold.
(define (adder lst)
(fold + 0 lst))
What does fold do? Basically, it's defined like so:
(define (fold f initial lst)
(if (null? lst)
initial
(fold f (f (car lst) initial) (cdr lst))))
(In other words, it calls f, a function of 2 arguments, on each element of lst, using the car of the lst as the first argument, and the accumulated result as the second argument to f.)
The issue here which you need to address is that + doesn't know how to operate on non-numeric values. No problem, you've already dealt with that. What happens if it's a character instead? Well, you're not adding anything to the total value, so replace it with a 0. Therefore, your solution is as simple as:
(define (adder lst)
(fold your-new-protected-+ 0 lst))
In Common Lisp:
(reduce #'+ '(1 #\a #\b 2 1 2 #\c #\d 4)
:key (lambda (item) (if (numberp item) item 0)))
or
(loop for item in '(1 #\a #\b 2 1 2 #\c #\d 4)
when (numberp item) sum item)