I'm trying to write a function that merges and then sorts a list, but now I have two different functions; one that merges it and one that sorts it. So I'm trying to write another function, that calls either functions, so it can merge and sort a list at once in that function.
This is what I have:
;; this merges the list
(define (merge l1 l2)
(cond ((null? l1) l2)
((null? l2) l1)
((< (car l1) (car l2)) (cons (car l1) (merge (cdr l1) l2)))
(else (cons (car l2) (merge l1 (cdr l2))))))
;; this sorts the list
(define sort
(lambda (lst)
(if (null? lst)
'()
(insert (car lst)
(sort (cdr lst))))))
(define insert
(lambda (elt sorted-lst)
(if (null? sorted-lst)
(list elt)
(if (<= elt (car sorted-lst))
(cons elt sorted-lst)
(cons (car sorted-lst)
(insert elt (cdr sorted-lst)))))))
You define your merge-sort like this:
(define (merge-sort l1 l2) (sort (merge l1 l2)))
Example:
> (merge-sort (list 8 3 7 4 9 2) (list 5 1 0 6 4))
(0 1 2 3 4 4 5 6 7 8 9)
Why not use one already written for you :)
Related
I've appreciate some helping hand over here. I'm trying to construct a procedure which delete duplicated elements in a list. This part is easy. But then I also want to delete duplicated elements (which may also be lists) and if it is a list the duplicated elements in that list should also be deleted, e.g (make-set (list 1 2 3 2 (list 1 3 2 4 3 4) (list 1 3 2 4 3 4))) should be '(1 3 2 (1 2 3 4)) but in our case it becomes '(1 3 2 2 3 4). Which isn't what we want. What am I doing wrong? Thanks :)
;; Checks if an element x appears in a list (set)
(define (element-of-set? x set)
(cond (( null? set) false)
((equal? x (car set)) true)
(else (element-of-set? x (cdr set)))))
;; Delete duplicated elements of a list (set)
(define make-set
(lambda (lst)
(cond ((null? lst) '())
((if (list? (car lst))
(cond ((null? (car lst))
'()
)
((element-of-set? (caar lst) (car lst)) (make-set (cdar lst))
)
(else (cons (caar lst) (make-set cadr lst))))
(cond ((element-of-set? (car lst) (cdr lst)) (make-set (cdr lst)))
(else (cons (car lst) (make-set (cdr lst))))))))))
Actually, if you want to build a function make-set that manages a generalized, untyped concept of set (that is a set that can contain either numbers or recursively other sets), the definition is quite complex. Here is my try.
;; check if x is contained in set
(define (contained? x set)
(cond ((null? set) false)
((my-equal? x (car set)) true)
(else (contained? x (cdr set)))))
;; check if all the elements of set1 are contained in set2
(define (set-contained? set1 set2)
(cond ((null? set1) true)
((null? set2) false)
(else (and (contained? (car set1) set2)
(set-contained? (cdr set1) set2)))))
;; check if set1 is equal to set2
(define (set-equal? set1 set2)
(and (= (length set1) (length set2))
(set-contained? set1 set2)))
;; check if x1 is equal to x2, when x1 and x2 can be sets or elements
(define (my-equal? x1 x2)
(cond ((list? x1) (and (list? x2) (set-equal? x1 x2)))
((list? x2) false)
(else (eq? x1 x2))))
;; add the element x to set, if not already present
(define (add-to-set x set)
(cond ((null? set) (list x))
((my-equal? x (car set)) set)
(else (cons (car set) (add-to-set x (cdr set))))))
;; make a set from a list lst
(define (make-set lst)
(cond ((null? lst) '())
((list? (car lst)) (add-to-set (make-set (car lst)) (make-set (cdr lst))))
(else (add-to-set (car lst) (make-set (cdr lst))))))
(make-set (list 1 2 3 2 (list 1 3 2 4 3 4) (list 1 3 2 4 3 4))) ; => '(1 3 (1 2 3 4) 2)
The function make-set builds the set by inserting in turn each element of the original list in a new set, so to check if the element is already present (also, if the element is a list, first it is transformed in a set). The other functions should be easy to understand, given the following convention:
If a parameter is called set, the function expects a list which has been already represented as set.
If a parameter is called x, then it is either a number or a set.
The specification of make-set is a little unclear, but maybe this works for you:
(define make-set
(lambda (lst)
(cond ((null? lst) '())
((list? (car lst)) (cons (make-set (car lst)) (make-set (cdr lst))))
((element-of-set? (car lst) (cdr lst)) (make-set (cdr lst)))
(else (cons (car lst) (make-set (cdr lst)))))))
Note that using lst is not in common use.
A nice convention is to use x as an element in a list and use xs as a list of x-elements.
I'm attempting to create a function for flattening lists in the R5RS language in scheme and am experiencing the issue where my function simply returns the input list without removing the parenthesis. I figured this was due to the extra cons, but when I remove it the output becomes the list without the elements that were in the parenthesis. Can someone point me in the right direction?
(define (denestify lst)
(cond ((null? lst)'())
((list? (car lst))(cons (denestify (cons (car (car lst))(cdr (car lst))))
(denestify (cdr lst))))
(else (cons (car lst)(denestify (cdr lst))))))
This shows how to convert Óscar López answer into one that doesn't use append and is also tail recursive:
(define (denestify-helper lst acc stk)
(cond ((null? lst)
(if (null? stk) (reverse acc)
(denestify-helper (car stk) acc (cdr stk))))
((pair? (car lst))
(denestify-helper (car lst) acc (cons (cdr lst) stk)))
(else
(denestify-helper (cdr lst) (cons (car lst) acc) stk))))
(define (denestify lst) (denestify-helper lst '() '()))
(denestify '(1 (2 (3 4 (5) (6 (7) (8)) (9))) 10))
Note how it uses the accumulator to build up the list in reverse and also a list as a stack.
Which results in
'(1 2 3 4 5 6 7 8 9 10)
as expected.
After I posted this I thought of this change:
(define (denestify-helper lst acc stk)
(cond ((null? lst)
(if (null? stk) (reverse acc)
(denestify-helper (car stk) acc (cdr stk))))
((pair? (car lst))
(denestify-helper (car lst) acc (if (null? (cdr lst))
stk
(cons (cdr lst) stk))))
(else
(denestify-helper (cdr lst) (cons (car lst) acc) stk))))
Which eliminates some useless consing by effectively doing tail-call optimization on our stack. One could go further and optimize handling of one element lists.
If you want to flatten a list of lists, then you have to use append to combine each sublist. Besides, your implementation is overly complicated, try this instead:
(define (denestify lst)
(cond ((null? lst) '())
((pair? (car lst))
(append (denestify (car lst))
(denestify (cdr lst))))
(else (cons (car lst) (denestify (cdr lst))))))
For example:
(denestify '(1 (2 (3 4 (5) (6 (7) (8)) (9))) 10))
=> '(1 2 3 4 5 6 7 8 9 10)
The original code I try to implement.. the output should be (1.4) (2.5) from my code.. I think you all know what I try to do....this is also tail recursion practice
my code
(define (myFunc lst1 lst2)
(if (or (null? lst1) (null? lst2))
'()
(list (cons (car lst1) (car lst2))
(myFunc (cdr lst1) (cdr lst2)))
))
after several of you gave me good advice about cons-pair.. so now it get's the dotted symbol in the middle.. then problem is that the improper list with empty list in the end..
when 2 input lists are like this ..... '(1 2 3 4) '(4 5 6))
my output is like this ; ((1 . 4) ((2 . 5) ((3 . 6) ())))
the empty list in the end of output shouldn't be there... so I couldn't understand about improper list , proper list....? is there are any document, I can look at?
Consider the difference between cons and list:
That is, (cons a b) creates a cell whose car is a and cdr is b.
(list a b) creates a cell whose car is a, but the cdr is a list, and the car of that list is b, while its cdr is nil.
If b is a list, the one on the left will be a list which has b as its tail, and with a added at the front of b.
The one on the right will also be a list, but one which has b as its second element, not as its tail like you want.
To fix your program, you only need to replace your list with a cons.
But your function is not tail-recursive, because it does things with the result of the recursive call.
To make it tail-recursive, a good way is usually to make a helper function which has an accumulator parameter.
I would probably write it something like this:
(define (zip-cars l1 l2)
(cons (car l1) (car l2)))
(define (zip-help l1 l2 result)
(if (or (null? l1) (null? l2))
result
(zip-help (cdr l1) (cdr l2) (cons (zip-cars l1 l2) result))))
(define (zip l1 l2)
(zip-help l1 l2 '()))
Just replace list with cons. Then your code will evaluate to `(cons (cons (cons .... (cons ... '())) and your list will be properly terminated.
(define (zip lst1 lst2)
(if (or (null? lst1) (null? lst2))
'()
(cons (cons (car lst1) (car lst2))
(zip (cdr lst1) (cdr lst2)))))
then
(zip '(1 2 3 4) '(4 5 6))
=> '((1 . 4) (2 . 5) (3 . 6))
This is not tail-recursive, though, since after returning from zip the consing still has to be done.
EDIT
An example of a tail-recursive version:
(define (zip lst1 lst2)
(let loop ((lst1 lst1) (lst2 lst2) (res '()))
(if (or (null? lst1) (null? lst2))
(reverse res)
(loop (cdr lst1)
(cdr lst2)
(cons (cons (car lst1) (car lst2)) res)))))
I have a list '(1 2 1 1 4 5) and want output list as '((1 3)(2 1)(4 1)(5 1)). I have written a small code but I am stuck with how to calculate the cardinality for each number and then put it as pair in list. Can anyone please look at my code and give some ideas?
(define set2bags
(lambda (randlist)
(cond ((null? randlist) '())
(else
(sort randlist)
(makepairs randlist)))))
(define makepairs
(lambda (inlist)
(let ((x 0)) ((newlist '()))
(cond ((zero? (car inlist)) '())
(else
(eq? (car inlist)(car (cdr inlist)))
(+ x 1)
(makepairs (cdr inlist))
(append newlist (cons (car inlist) x)))))))
Your current solution is incorrect - it doesn't even compile. Let's start again from scratch, using a named let for traversing the input list:
(define set2bags
(lambda (randlist)
(cond ((null? randlist) '())
(else (makepairs (sort randlist >))))))
(define makepairs
(lambda (inlist)
(let loop ((lst inlist)
(prv (car inlist))
(num 0)
(acc '()))
(cond ((null? lst)
(cons (list prv num) acc))
((= (car lst) prv)
(loop (cdr lst) prv (add1 num) acc))
(else
(loop (cdr lst) (car lst) 1 (cons (list prv num) acc)))))))
Now it works as expected:
(set2bags '(1 2 1 1 4 5))
=> '((1 3) (2 1) (4 1) (5 1))
The trick is keeping a counter for the cardinality (I called it num), and incrementing it as long as the same previous element (I named it prv) equals the current element. Whenever we find a different element, we add a new pair to the output list (called acc) and reset the previous element and the counter.
Your code is fairly hard to read without proper formating.
I notice a two branch cond, which is easier to read as an if.
In your else clause of set2bags, you call (sort randlist) but leave it as is. You actually want to use this in the next s-expression (makepairs (sort randlist))
So far a pretty good idea.
Now in makepairs you should have better abstraction, say let variables like-first and unlike-first. If the inlist is null, then the function should be the null list, else it's the pair with the car being the list of the car of like-first and the length of like-first and the cdr being the result of calling makepairs on the unlike-first list
(define (makepairs inlist)
(let ((like-first (filter (lambda (x) (equal? x (car inlist)) inlist))
(unlike-first (filter (lambda (x) (not (equal? x (car inlist))) inlist)))
(if (null? inlist)
'()
(cons (list (car inlist) (length like-first)) (makepairs unlike-first)))))
more effecient version
(define (makepairs inlist)
(if (null? inlist)
'()
(let loop ((firsts (list (car inlist)))
(but-firsts (cdr inlist)))
(if (or (null? but-firsts)
(not (equal? (car firsts) (car but-firsts))))
(cons (list (car firsts) (length firsts))
(makepairs but-firsts))
(loop (cons (car but-firsts) firsts) (cdr but-firsts))))))
]=> (makepairs (list 1 1 1 2 4 5))
;Value 17: ((1 3) (2 1) (4 1) (5 1))
If you have your own implementation of sort, say a mergesort you could write this right into the merge part for the best effeciency.
(define (set2bags lst)
(mergesort2bags lst <))
(define (mergesort2bags lst pred)
(let* ((halves (divide-evenly lst))
(first-half (car halves))
(other-half (cadr halves)))
(cond ((null? lst) '())
((null? (cdr lst)) (list (list (car lst) 1)))
(else
(merge-bags
(mergesort2bags first-half pred)
(mergesort2bags other-half pred)
pred)))))
(define (divide-evenly lst)
(let loop
((to-go lst)
(L1 '())
(l2 '()))
(if (null? to-go)
(list L1 L2)
(loop (cdr to-go) (cons (car to-go) L2) L1))))
(define (merge-bags L1 L2 pred)
(cond ((null? L1) L2)
((null? L2) L1)
((pred (caar L1) (caar L2))
(cons (car L1) (merge-bags (cdr L1) L2 pred)))
((equal? (caar L1) (caar L2))
(cons (list (caar L1) (+ (cadar L1) (cadar L2)))
(merge-bags (cdr L1) (cdr L2) pred)))
(else (cons (car L2) (merge-bags L1 (cdr L2) pred)))))
(mergesort2bags (list 1 2 1 1 4 5) <)
;Value 46: ((1 3) (2 1) (4 1) (5 1))
I'm thinking for very large datasets with a lot of repetition this method would pay off.
I'm using R5RS Scheme and I just want to implement a function that returns the intersection of two given lists, but I can't do that because I cannot add an element to a list. Here is my code. How can I fix it? I'm really a beginner in Scheme - this is my first work using Scheme.
thx in advance..
(define list3 '())
(define (E7 list1 list2)
(cond
((null? list1)
list3)
((member (car list1) list2) (append list3 (list (car list1))))
)
(cond
((null? list1)
list3)
((not(null? list1)) (E7 (cdr list1) list2)
)
)
)
(E7 '(4 5) '(3 4))
Here is a recursive version that does the intersection instead of the union.
(define (intersect list1 list2)
(cond ((null? list1) list1)
((member (car list1) list2) (cons (car list1) (intersect (cdr list1) list2)))
(t (intersect (cdr list1) list2))))
I think I see your problem. There are two ways to add an element to a list.
The first way would be actually adding it:
(define (intersect list1 list2)
(define newlist list2)
(do ((iter1 list1 (cdr iter1)))
(not (null? iter1))
(if (not (member (car iter1) newlist))
(set! newlist (cons (car iter1) newlist)))))
(You'll probably have to look up the definition of do if you really want to use this.)
You may notice that that's quite ugly. That's because no one actually does it this way. Instead, you have to realize that calling a function creates a new variable as well. Try this:
(define (intersect list1 list2)
(cond ((null? list1) list2)
((member (car list1) list2) (intersect (cdr list1) list2))
(else (intersect (cdr list1) (cons (car list1) list2)))))
If you're familiar with algorithms, you'll notice that the code I just wrote is quite slow, but it illustrates the point: in each case, you do a little bit of work and then call your function again. If you're having trouble seeing why this works, run this function instead on your example:
(define (intersect list1 list2)
(display list1) (display " ") (display list2) (newline)
(cond ((null? list1) list2)
((member (car list1) list2) (intersect (cdr list1) list2))
(else (intersect (cdr list1) (cons (car list1) list2)))))
Here is some simplistic elisp:
(defun is (l1 l2)
(let ((rtn))
(mapc
(lambda (e)
(if (member e l1)
(push e rtn)))
l2)
rtn))
This behaves the same as the built-in intersection for these simple tests:
(is '(1 2 5) '(1 4 10 5)) => (5 1)
(intersection '(1 2 5) '(1 4 10 5)) => (5 1)
(is '(1 4 10 5) '(1 2 5)) => (5 1)
(intersection '(1 4 10 5) '(1 2 5)) => (5 1)
You're better off using set operations from srfi-1.