Gathering elements in range of Min and Max values from a list in Scheme

Question

I am new to Scheme and not currently using (Racket version), therefore I do not have such built in functions as filter and etc.

My task for now is to get elements out of my inputted list, and take out of it only those that are above my MinValue and below my MaxValue

In my "lst" variable for example I have (2 4 1 6 5 4 7 3)

In my "MinValue" I have 2

In my MaxValue I have 5

So that now after I apply something similar like

 (remove (< # minVal) lst)

It technically should give remove and left out:

(2 4 4 3)

But I am getting the error:

READ from #<INPUT CONCATENATED-STREAM #<INPUT STRING-INPUT-STREAM> #<IO TERMINAL-STREAM>>: objects printed as # in view of *PRINT-LEVEL* cannot be read back in

Main questions:

Does Scheme has a specific defined variable like # that is iterating through the list when remove/remove-if functions are used, so I can use it?

Is there is a way to get this to work in such way:

(remove (< # minVal) lst)

Thanks!

tag "homework" should be there, but I cannot create it yet

Which Scheme are you using? That looks a lot like a Lisp error message. — molbdnilo, Nov 24 '16 at 07:06

Óscar López · Accepted Answer · 2016-11-23T18:58:01.897

Almost any Scheme I know of includes a filter procedure, it's not specific to Racket, you should be able to use it (remove is not the right tool for this job).

Simply pass a lambda that receives each element in turn and you can set any condition for determining which elements go in the output list:

(filter (lambda (e) (and (>= e 2) (< e 5)))
        '(2 4 1 6 5 4 7 3))

=> '(2 4 4 3)

There is no # variable that iterates through a list and gives you each element in turn, to do that you have to explicitly traverse the list using recursion, or use one of the built-in procedures that expect a list and a lambda - the lambda's parameter represents each of the elements.

If for some reason you are not allowed to use filter, it's easy to implement it from scratch, and again notice that we don't need to use remove, and that pred gets called with each of the elements:

(define (filter pred lst)
  (cond ((null? lst) '())
        ((pred (car lst))
         (cons (car lst) (filter pred (cdr lst))))
        (else (filter pred (cdr lst)))))

1

Thanks! That explained it! – Sparrow Nov 24 '16 at 15:04

Gathering elements in range of Min and Max values from a list in Scheme

1 Answers1