This question already has answers here:
Count occurrence of element in a list in Scheme?
(4 answers)
Closed 8 years ago.
I want to make a function that occurs how many times an element occurs in a list. For example in the list: '(a b c b b c c a) I want it to return a nested list with:
'((a 2) (b 3) (c 3))
I know the function will look sort of like this:
(define collect-similar
(lambda (elm ls)
(cond
[(null? ls) '()]
[(equal? elm (car ls))]
I know that I need to continue checking through the list until it arrives back at the base case of the null list and I can check the rest of the list by using cadr. But I'm not too sure on how to get the value and how to make it return the nested list.
The next function I'm trying to write finds the most common element in the list. For example running the function on the list '(a a a a a b c) will simply return a. I know I can make use of the collect-similar function and find which number is the highest.
This has been asked before, just adapt one of #ChrisJester-Young's bagify implementations. For example:
(define (collect-similar lst) ; a slightly modified `bagify`
(hash->list
(foldl (lambda (key ht)
(hash-update ht key add1 0))
'#hash()
lst)))
(collect-similar '(a b c b b c c a))
=> '((a . 2) (b . 3) (c . 3))
With collect-similar in place, it's simple to find the most common element:
(define (most-common lst)
(let loop ((alst (collect-similar lst)) ; use previous procedure
(maxv '(#f . -inf.0)))
(cond ((null? alst) (car maxv))
((> (cdar alst) (cdr maxv))
(loop (cdr alst) (car alst)))
(else
(loop (cdr alst) maxv)))))
(most-common '(a a a a a b c))
=> 'a
Related
I want to be able to make my list go one step to the right. Here's my example code that makes it go one step to the left.
(define (rotate-L lst) (append (cdr lst) (list (car lst))))
(rotate-L '(a b c))
(b c a)
I'm having trouble understanding why when I make it backward, i get an error
(define (rotate-L lst) (append (car lst) (list (cdr lst))))
(rotate-L '(a b c))
SchemeError: argument 0 of append has wrong type (string)
Current Eval Stack:
0: (rotate-L (quote (a b c)))
(define rotr
(lambda (l)
(if (null? l)
l
((lambda (s) (s s l cons))
(lambda (s l c)
(if (null? (cdr l))
(c (car l) '())
(s s (cdr l)
(lambda (f r)
(c f (cons (car l) r))))))))))
(define rotl
(lambda (l0)
(if (null? l0)
l0
((lambda (s) (s s (cdr l0) (lambda (r) r)))
(lambda (s l c)
(if (null? l)
(c (list (car l0)))
(s s (cdr l)
(lambda (r)
(c (cons (car l) r))))))))))
Here is a test:
(rotr '())
(rotr '(a))
(rotr '(a b))
(rotr '(a b c))
(rotl '())
(rotl '(a))
(rotl '(a b))
(rotl '(a b c))
whose output is so:
1 ]=> (rotr '())
;Value: ()
1 ]=> (rotr '(a))
;Value: (a)
1 ]=> (rotr '(a b))
;Value: (b a)
1 ]=> (rotr '(a b c))
;Value: (c a b)
1 ]=> (rotl '())
;Value: ()
1 ]=> (rotl '(a))
;Value: (a)
1 ]=> (rotl '(a b))
;Value: (b a)
1 ]=> (rotl '(a b c))
;Value: (b c a)
It's worth remembering what lists look like as structures made of conses:
(1 2 3 4) looks like this, for instance
So pretty obviously (car '(1 2 3 4)) is a number, not a list, and so (append (car '(1 2 3 4)) ...) is not going to work, because append wants its arguments to be lists. But even if you fix this to be correct, you're not doing what you think:
(define (not-rotating lst)
(append (list (car lst)) (cdr lst)))
This simply takes the first element of the list, makes a single-element list from it, and appends the rest of the list to it. Well, you could write this more easily:
(define (not-rotating lst)
(cons (car lst) (cdr list)))
and it should be clear that this is doing nothing at all useful.
To rotate the list 'right' you need to:
find the last element of the list;
construct a list which is all of the original list except for that element;
and glue the last element onto the start of that list.
The only way to find the last element of the list is to walk along the list until you get to it. And you also need to make a copy of all the elements of the list but the last one, which also means walking down the list. It is natural to do these two operations at the same time:
walk down the list, building up a copy of it;
when you get to the last element, attach it to the start of the copy.
Here is a terrible way of doing this:
(define (rotate-r/terrible lst)
(let rrl ((tail lst)
(building '()))
(if (null? (cdr tail))
;; we've got to the end
(cons (car tail) building)
(rrl (cdr tail)
(append building (list (car tail)))))))
This is terrible because at each step it appends the current element to the end of the list it is building. That's terrible for not one but two reasons:
appending two lists takes time proportional to the length of the first list;
appending two lists requires a complete copy of the first list to be made.
That makes this both quadratic in the length of the list, which is terrible, and also very 'consy' – it allocates lots of ephemeral storage. But it does work.
So: there is, not surprisingly a better way of doing this: a way which is both linear and makes no more than two copies of the list. I'm not going to give it here, but it's surprisingly close to the above terrible solution. Two hints:
what is the natural way to accumulate elements into a list? is it at the start, or at the end of a list?
how do you turn the list you accumulated like that into the list you want?
For my programming languages class I'm supposed to write a function in Scheme to reverse a list without using the pre-made reverse function. So far what I got was
(define (reverseList lst)
(COND
((NULL? lst) '())
(ELSE (CONS (reverseList(CDR lst)) (CAR lst)))
))
The problem I'm having is that if I input a list, lets say (a b c) it gives me (((() . c) . b) . a).
How am I supposed to get a clean list without multiple sets of parenthesis and the .'s?
The problem with your implementation is that cons isn't receiving a list as its second parameter, so the answer you're building isn't a proper list, remember: a proper list is constructed by consing an element with a list, and the last list is empty.
One possible workaround for this is to use a helper function that builds the answer in an accumulator parameter, consing the elements in reverse - incidentally, this solution is tail recursive:
(define (reverse lst)
(reverse-helper lst '()))
(define (reverse-helper lst acc)
(if (null? lst)
acc
(reverse-helper (cdr lst) (cons (car lst) acc))))
(reverse '(1 2 3 4 5))
=> '(5 4 3 2 1)
You are half way there. The order of the elements in your result is correct, only the structure needs fixing.
What you want is to perform this transformation:
(((() . c) . b) . a) ; input
--------------------
(((() . c) . b) . a) () ; trans-
((() . c) . b) (a) ; for-
(() . c) (b a) ; mation
() (c b a) ; steps
--------------------
(c b a) ; result
This is easy to code. The car and cdr of the interim value are immediately available to us. At each step, the next interim-result is constructed by (cons (cdr interim-value) interim-result), and interim-result starts up as an empty list, because this is what we construct here - a list:
(define (transform-rev input)
(let step ( (interim-value input) ; initial set-up of
(interim-result '() ) ) ; the two loop variables
(if (null? interim-value)
interim-result ; return it in the end, or else
(step (car interim-value) ; go on with the next interim value
(cons ; and the next interim result
(... what goes here? ...)
interim-result )))))
interim-result serves as an accumulator. This is what's known as "accumulator technique". step represents a loop's step coded with "named-let" syntax.
So overall reverse is
(define (my-reverse lst)
(transform-rev
(reverseList lst)))
Can you tweak transform-rev so that it is able to accept the original list as an input, and thus skip the reverseList call? You only need to change the data-access parts, i.e. how you get the next interim value, and what you add into the interim result.
(define (my-reverse L)
(fold cons '() L)) ;;left fold
Step through the list and keep appending the car of the list to the recursive call.
(define (reverseList lst)
(COND
((NULL? lst) '())
(ELSE (APPEND (reverseList(CDR lst)) (LIST (CAR lst))))
))
Instead of using cons, try append
(define (reverseList lst)
(if (null? lst)
'()
(append (reverseList (cdr lst)) (list (car lst)) )
)
)
a sample run would be:
1]=> (reverseList '(a b c 1 2 + -))
>>> (- + 2 1 c b a)
car will give you just one symbol but cdr a list
Always make sure that you provide append with two lists.
If you don't give two lists to the cons it will give you dotted pair (a . b) rather than a list.
See Pairs and Lists for more information.
I want to write a function that takes two list arguments and returns the longer list of the two inputs. If the two lists are equal in length, the function returns #t, and if one of the arguments is not a list, the function should return #f.
Sample runs:
(longer-list '(1 2 3 4) '(a b c d e)) returns (a b c d e)
(longer-list '(d e f) '(4 5 6)) returns #t (or true)
(longer-list '(g h i) 3) returns #f (or false)
How can I do this?
Seems to be you need to do a case analysis. You need to check if either argument are not a list and return #f there, then if it's not you need to get the lengths of the two lists to check if they are of equal length or if the one list is maller than the other. Something like this perhaps?
(define (longest lst1 lst2)
(if <??> ; check if one of the argument is not a list
#f
(let ((len1 <??>) (len2 <??>))
(cond ((= <??> <??>) #t) ; same length
((< <??> <??>) <??>) ; lst1 shorter than lst2
(else <??>>))))) ; lst2 shorter than lst1
It seems like an assignment so I let you fill in the blanks.
(define (longer-list a b)
(and (list? a)
(list? b)
(let ll ((aa a) (bb b))
(cond
((and (null? aa) (null? bb)) #t)
((null? aa) b)
((null? bb) a)
(else (ll (cdr aa) (cdr bb)))))))
This question already has answers here:
"application: not a procedure" in binary arithmetic procedures
(1 answer)
Getting every nth atom using scheme does not pick up the last atom
(3 answers)
Closed 8 years ago.
Hi I'm writing something that grabs every third element after the first element. However I can't test the logic due to ";The object (a b c d e f g) is not applicable." The code is below, atom? checks if it's a list, listful is defined as an empty list.
(DEFINE (threes var)
(if(atom? var)
((newline) (DISPLAY "Bad input") ))
(APPEND(listful (CAR var)))
(if(> 3 (length var))
(threes (cdddr(listful)))
(listful))
)
Can anyone give me some tips? Here is how I'm calling the method in the Scheme environment.
>(threes (list1))
>(threes '(A B C D E F))
An if can only have one expression as the consequent, and one as the alternative. If you need more than one you have to use a begin to enclose the sequence of expressions - surrounding the expressions with () won't work, and that's causing the error reported, because Scheme interprets () as function application. This would be the correct syntax:
(define (threes var)
(if (atom? var)
(begin
(newline)
(display "Bad input")))
(append (listful (car var)))
(if (> 3 (length var))
(begin
(threes (cdddr (listful)))
(listful))))
… However, that's unlikely to work. What you want to do has been asked a couple of times in the last days, in particular here is my own previous answer:
(define (threes lst)
(cond ((or (null? lst) (null? (cdr lst))) lst)
((null? (cdr (cdr lst))) (list (car lst)))
(else (cons (car lst)
(threes (cdr (cdr (cdr lst))))))))
For example:
(threes '(a b c d e f g h i j k l m n o p))
=> '(a d g j m p)
See the original question for other ways to solve the problem, with detailed explanations.
This is the function that removes the last element of the list.
(define (remove-last ll)
(if (null? (cdr ll))
'()
(cons (car ll) (remove-last (cdr ll)))))
So from my understanding if we cons a list (eg. a b c with an empty list, i.e. '(), we should get
a b c. However, testing in interaction windows (DrScheme), the result was:
If (cons '() '(a b c))
(() a b c)
If (cons '(a b c) '())
((a b c))
I'm like what the heck :(!
Then I came back to my problem, remove all elements which have adjacent duplicate. For example,
(a b a a c c) would be (a b).
(define (remove-dup lst)
(cond ((null? lst) '())
((null? (cdr lst)) (car lst))
((equal? (car lst) (car (cdr lst))) (remove-dup (cdr (cdr lst))))
(else (cons (car lst) (car (cdr lst))))
)
)
It was not correct, however I realize the answer have a . between a b. How could this happen?
`(a . b)`
There was only one call to cons in my code above, I couldn't see which part could generate this .. Any idea?
Thanks,
cons build pairs, not lists. Lisp interpreters uses a 'dot' to visually separate the elements in the pair. So (cons 1 2) will print (1 . 2). car and cdr respectively return the first and second elements of a pair. Lists are built on top of pairs. If the cdr of a pair points to another pair, that sequence is treated as a list. The cdr of the last pair will point to a special object called null (represented by '()) and this tells the interpreter that it has reached the end of the list. For example, the list '(a b c) is constructed by evaluating the following expression:
> (cons 'a (cons 'b (cons 'c '())))
(a b c)
The list procedure provides a shortcut for creating lists:
> (list 'a 'b 'c)
(a b c)
The expression (cons '(a b c) '()) creates a pair whose first element is a list.
Your remove-dup procedure is creating a pair at the else clause. Instead, it should create a list by recursively calling remove-dup and putting the result as the second element of the pair. I have cleaned up the procedure a bit:
(define (remove-dup lst)
(if (>= (length lst) 2)
(if (eq? (car lst) (cadr lst))
(cons (car lst) (remove-dup (cddr lst)))
(cons (car lst) (remove-dup (cdr lst))))
lst))
Tests:
> (remove-dup '(a b c))
(a b c)
> (remove-dup '(a a b c))
(a b c)
> (remove-dup '(a a b b c c))
(a b c)
Also see section 2.2 (Hierarchical Data and the Closure Property) in SICP.
For completeness, here is a version of remove-dup that removes all identical adjacent elements:
(define (remove-dup lst)
(if (>= (length lst) 2)
(let loop ((f (car lst)) (r (cdr lst)))
(cond ((and (not (null? r))(eq? f (car r)))
(loop f (cdr r)))
(else
(cons (car lst) (remove-dup r)))))
lst))
Here in pseudocode:
class Pair {
Object left,
Object right}.
function cons(Object left, Object right) {return new Pair(left, right)};
So,
1. cons('A,'B) => Pair('A,'B)
2. cons('A,NIL) => Pair('A,NIL)
3. cons(NIL,'A) => Pair(NIL,'A)
4. cons('A,cons('B,NIL)) => Pair('A, Pair('B,NIL))
5. cons(cons('A 'B),NIL)) => Pair(Pair('A,'B),NIL)
Let's see lefts and rights in all cases:
1. 'A and 'B are atoms, and whole Pair is not a list, so (const 'a 'b) gives (a . b) in scheme
2. NIL is an empty list and 'A is an atom, (cons 'a '()) gives list (a)
3. NIL and 'A as above, but as left is list(!), (cons '() 'a) gives pair (() . a)
4. Easy case, we have proper list here (a b).
5. Proper list, head is pair (a . b), tail is empty.
Hope, you got the idea.
Regarding your function. You working on LIST but construct PAIRS.
Lists are pairs (of pairs), but not all pairs are lists! To be list pair have to have NIL as tail.
(a b) pair & list
(a . b) pair not list
Despite cons, your function has errors, it just don't work on '(a b a a c c d). As this is not related to your question, I will not post fix for this here.