Reputation: 57
how to delete an element from a list ex:- list=[1 2 3 4]
I have come up with some code.I think I got wrong somewhere.
(define delete item
(lambda (list)
(cond
((equal?item (car list)) cdr list)
(cons(car list)(delete item (cdr list))))))
Upvotes: 4
Views: 44348
Reputation: 1
Thank @ad absurdum for pointing out my problems
delete element from a list without nested lists
bad code:
(define (remove item lst)
(define (filter-lst l)
(cond
((null? l) nil)
((= item (car l)) (filter-lst (cdr l)))
(else (cons (car l) (filter-lst (cdr l)))))
)
(if (null? lst) () (filter-lst lst))
)
;;; Tests
(remove 2 '(4 3 2))
; expect (4 3)
(remove 3 nil)
; expect ()
(remove 3 '(1 3 5))
; expect (1 5)
(remove 5 '(5 5 1 4 5 4))
; expect (1 4 4)
the code above tested in Berkeley CS61A scheme interpreter
(define (remove item lst)
(cond
((null? lst) '())
((equal? item (car lst)) (remove item (cdr lst)))
(else (cons (car lst) (remove item (cdr lst)))))
)
(remove 2 '(1 3 2))
; expect (1 3)
(remove 3 '())
; expect ()
(remove 3 '(1 3 5))
; expect (1 5)
(remove 5 '(5 3 5 5 1 4 5 4))
; expect (3 1 4 4)
the code above tested in chez scheme interpreter
Upvotes: -1
Reputation: 8564
This code seems to work just fine, but only deletes an element that should be in the list:
(define (delete element lst)
(let loop ([temp lst])
(if (= element (car temp)) (cdr temp)
(cons (car temp) (loop (cdr temp))))))
Upvotes: 0
Reputation: 177
1) if consider the input list may be a simple list, or you just want to delete the item in the top-level of a nested list for example:
delete 2 from (1 2 3 4) will return (1 2 3)
delete 2 from (1 2 3 (2 3) 3 2 4) will return (1 3 (2 3) 3 4)
as we can see the 2nd example above, it just delete the item in the top-level of the nested list, within the inner list, we doesn't change it.
this code should be:
(define (deleteitem list1 item)
( cond
((null? list1) ’())
((equal? (car list1) item) (deleteItem (cdr list1) item))
(else (cons (car list1) (deleteitem (cdr list1) item)))
))
2) if consider the input list may be a nested list
for example:
input list: (1 2 3 (3 2 (2 4 (2 5 6) 2 5 6) 2 4) 2 3 (2 3 4))
and delete the element 2 in the input list
the output list should be: (1 3 (3 (3 (5 6) 5 6) 4) 3 (3 4))
and the code should be:
(define (delete2 list1 item)
( cond
((null? list1) '())
((pair? (car list1)) (con (delete2 (car list1) item) (delete2 (cdr list1) item)))
((equal? (car list1) item) (delete2 (cdr list1) item))
(else (cons (car list1) (delete2 (cdr list1) item)))
))
Upvotes: 1
Reputation: 313
Shido Takafumi wrote a tutorial about Scheme, Yet Another Scheme Tutorial. In chapter 7, exercise 1, the 3rd problem.
A function that takes a list (ls) and an object (x) as arguments and returns a list removing x from ls.
The author gave the solution code bottom of the page.
; 3
(define (remove x ls)
(if (null? ls)
'()
(let ((h (car ls)))
((if (eqv? x h)
(lambda (y) y)
(lambda (y) (cons h y)))
(remove x (cdr ls))))))
The code maybe difficult to comprehend for beginner. It's same as the code below.
(define (rm x ls)
(if (null? ls)
'()
(if (eqv? x (car ls))
(rm x (cdr ls))
(cons (car ls)
(rm x (cdr ls))))))
This can delete the same elements in list. :D
Upvotes: 5
Reputation: 1
(define (deleteItem(list item))
(cond
((eq ? item (car(list)))cdr(list))
(cons(car(list)(deleteItem(cdr list)))
)
)
Upvotes: -4
Reputation: 1049
Your code is almost correct.
The item
also should be a parameter, so the function may begin with like this:
(define delete
(lambda (item list)
...
Also, your code needs paren around the cdr list
and else
in the last clause.
Then, the code may be like this:
(define delete
(lambda (item list)
(cond
((equal? item (car list)) (cdr list))
(else (cons (car list) (delete item (cdr list)))))))
Upvotes: 13