So far i have made two functions to apply a filter to a list. I am trying to get the functions to return only the elements that the operation applies to for instance all odd numbers. I am only returning a boolean of whether each number is odd.
(define (accumulate op initial sequence)
(if (null? sequence)
initial
(op (car sequence)
(accumulate op initial (cdr sequence)))))
(define (my-filter predicate sequence)
(accumulate (lambda (element rest)(cons (predicate element) rest)) '() sequence))
(my-filter odd? (list 1 2 3 4 5 6 7))
You just have to use the predicate in a conditional expression. If the predicate evaluates to #t then add the current element to the answer, otherwise skip it:
(define (my-filter predicate sequence)
(accumulate (lambda (element rest)
(if (predicate element)
(cons element rest)
rest))
'()
sequence))
For example:
(my-filter odd? (list 1 2 3 4 5 6 7))
=> '(1 3 5 7)
Related
I often struggle writing iterative functions in scheme: it makes writing recursive procedures much simpler. Here is an example of trying to square items in a list using an iterative procedure:
(define square (lambda (x) (* x x)))
(define (square-list items)
(define result nil) ; set result
(define (iter items-remaining)
(if (null? items-remaining)
result
(set! result (cons (car items-remaining) (iter (cdr items-remaining))))))
(iter items))
(square-list '(1 2 3 4 5))
; (4 9 16 25)
My main question about this is:
Is there a way to do this procedure without having to first define the result before the inner procedure? I was trying to make the iterative procedure have the function prototype of define (iter items-remaining answer) but was having a hard time implementing it that way.
And if not, why isn't that possible?
The posted code does not work; but even when fixed up so that it does work, this would not be an idiomatic Scheme solution.
To make the posted code work:
nil must be replaced with '(), since Scheme does not represent the empty list with nil
square must be called on the car of items-remaining
set! should modify result by adding squared numbers to it, not by trying to add the result of a recursive call. This won't work at all here because set! returns unspecified values; but even if it did work, this would not be tail-recursive (i.e., this would not be an iterative process)
The value of result must be returned, and it will have to be reversed first since result is really an accumulator
Here is a fixed-up version:
(define (square-list-0 items)
(define result '()) ; set result
(define (iter items-remaining)
(cond ((null? items-remaining)
result)
(else
(set! result (cons (square (car items-remaining))
result))
(iter (cdr items-remaining)))))
(iter items)
(reverse result))
A better solution would not use mutation, and would not need (define result '()):
(define (square-list-1 xs)
(define (iter xs acc)
(if (null? xs)
(reverse acc)
(iter (cdr xs) (cons (square (car xs)) acc))))
(iter xs '()))
Here an accumulator, acc, is added to the lambda list for the iter procedure. As results are calculated, they are consed onto acc, which means that at the end of this process the first number in acc is based on the last number in xs. So, the accumulator is reversed before it is returned.
Another way to do this, and probably a more idiomatic solution, is to use a named let:
(define (square-list-2 xs)
(let iter ((xs xs)
(acc '()))
(if (null? xs)
(reverse acc)
(iter (cdr xs) (cons (square (car xs)) acc)))))
This is a bit more concise, and it lets you bind arguments to their parameters right at the beginning of the definition of the iter procedure.
All three of the above solutions define iterative processes, and all three give the same results:
> (square-list-0 '(1 2 3 4 5))
(1 4 9 16 25)
> (square-list-1 '(1 2 3 4 5))
(1 4 9 16 25)
> (square-list-2 '(1 2 3 4 5))
(1 4 9 16 25)
Of course, you could just use map:
> (map square '(1 2 3 4 5))
(1 4 9 16 25)
Write a scheme function that returns a list containing all elements of a given
list that satisfy a given predicate eg. (lambda (x) (< x 5)) '(3 9 5 8 2 4 7) => '(3 2 4)
Any hints how to begin about this ?
Basically you are creating filter.
(define (filter included? lst)
...)
You need to check if the argument has elements and if the first element is to be included or not. Including would involve cons-ing the first element to the recursing with the cdr with the same predicate while not including would not include consing but the exact same thing as the tail in a conse situation.
(filter odd? '(3 4))
; ==> (cons 3 (filter odd? '(4)))
(filter odd? '(4))
; ==> (filter odd? '())
(filter odd? '())
; ==> '()
Putting them together shows that (filter odd? '(3 4)) should produce the result of (cons 3 '())
Here is a skeleton of what I would have done:
(define (filter included? lst)
(cond ((null? lst) <??>)
((included? (car lst)) <??>) ; since it is not null? it has to have at least one element
(else <??>))) ; neither null? nor included? => skip
For the built-in function foldr, I know the function blueprint is the following:
(foldr combine base alist)
combine is supposed to take in two parameters:
an item that foldr consumes
the result of applying foldr to the rest of alist
I cannot seem to understand how to put point #2 in parameter form ever. How did you do it?
combine is not a built-in function. I would have to code it myself based on the requirements.
Think of second parameter as the accumulated value so far. For example, if we are adding the elements, then acc is the sum of all the previous eles and we need to add the current element:
(foldr (lambda (ele acc) (+ ele acc))
0 ; we're adding numbers, so the base is 0
'(1 2 3 4 5))
=> 15
Another example - if we're copying the list, then acc contains the previous eles in the list (starting from the last one and going back from there) and we have to cons the current element at the head :
(foldr (lambda (ele acc) (cons ele acc))
'() ; we're creating a list, so the base is an empty list
'(1 2 3 4 5))
=> '(1 2 3 4 5)
The exact nature of acc depends on the problem to be solved, but you should be able get the idea from the previous examples.
Think of it as the result computed so far and that foldr iterates from end to beginning while a foldl iterates from beginning to end. It's easier to see if you look at a simple implementation of it:
(define (foldr1 f init lst)
(let r ((lst lst))
(if (null? lst)
init
(cons (f (car lst)) (r (cdr lst))))))
(foldr1 combine base '(1 2 3)) ; ==
(combine 1 (combine 2 (combine 3 base)))
(define (foldl1 f init lst)
(let r ((lst lst) (acc init))
(if (null? lst)
acc
(r (cdr lst) (f (car lst))))))
(foldl1 combine base '(1 2 3)) ; ==
(combine 3 (combine 2 (combine 1 base)))
Also note that the order or the arguments change in some implementations. Racket and SRFI-1 always have the accumulator as the last argument, but in R6RS the argument order changes for fold-left (but not fold-right):
#!r6rs
(import (rnrs))
;; swap argument order
(fold-left (lambda (acc e) (cons e acc)) '() '(1 2 3))
; ==> (3 2 1)
Currently trying to write a filter function that takes a list of procedures and a list of numbers, deletes the procedures that does not return true on every element of the list of numbers.
What I have done is the following, currently taking a single procedure and runs through the list to create a list stating whether the procedure is true or false for each element.
Using foldr, or map if required (non recursive)
(define (my-filter ps xs)
(foldr (λ (x y)
(cons (ps x) y)) '() xs))
How do I delete the procedure if it has at lease one #f?
i.e.
currently
> (my-filter even? '(1 2 3 4))
(#f #t #f #t)
> (my-filter even? (2 4))
(#t #t)
want
> (my-filter (list even?) '(1 2 3 4))
(list)
> (my-filter (list even?) '(2 4))
(list even?)
You can solve this by using Racket's built-in andmap procedure, which tests if a condition is true for all elements in a list - like this:
(define (my-filter ps xs)
(foldr (lambda (f acc)
(if (andmap f xs) ; if `f` is true for all elements in `xs`
(cons f acc) ; then add `f` to the accumulated output
acc)) ; otherwise leave the accumulator alone
'() ; initially the accumulator is an empty list
ps)) ; iterate over the list of functions
Notice that we do not "delete" functions from the output list, instead at each step we decide whether or not we should add them to the output list.
The advantage of foldr is that it preserves the same order as the input list, if that's not a requirement, foldl is more efficient because internally it uses tail recursion. Either way, it works as expected:
(my-filter (list even?) '(1 2 3 4))
=> '()
(my-filter (list even?) '(2 4))
=> '(#<procedure:even?>)
Start with some wishful thinking. Say we have a know of knowing if all xs return #t for any given f
(define (my-filter fs xs)
(foldr (λ (f y)
(if (every? f xs)
(cons f y)
y))
'()
fs))
Now let's define every?
(define (every? f xs)
(cond [(null? xs) #t]
[(f (car xs)) (every? f (cdr xs))]
[else #f]))
Let's check it out for (list even?)
(my-filter (list even?) '(1 2 3 4)) ; ⇒ '()
(my-filter (list even?) '(2 4)) ; ⇒ '(#<procedure:even?>)
Let's add another test in the mix
(define (gt3? x) (> x 3))
(my-filter (list even? gt3?) '(2 4)) ; ⇒ '(#<procedure:even?>)
(my-filter (list even? gt3?) '(4 6)) ; ⇒ '(#<procedure:even?> #<procedure:gt3?>)
Cool !
If you want to see "pretty" procedure names instead of the #<procedure:...> stuff, you can map object-name over the resulting array
(map object-name (my-filter (list even? gt3?) '(4 6))) ; ⇒ '(even? gt3?)
Even though foldl will give you a reversed output, I still think it would be better to use foldl in this case. Just my 2 cents.
You are given a list of strings.
Generate a procedure such that applying this procedure to such a list
would result in a list of the lengths of each of the strings in the
input.
Use map, filter, or fold-right.
(lengths (list "This" "is" "not" "fun")) => (4 2 3 3)
(define lengths (lambda (lst) your_code_here))
I got stuck in the following code and I do not understand how can I use filter.
(define lengths
(lambda (lst)
(if (null? lst)
nil
(fold-right list (string-length (car lst)) (cdr lst)))))
This seems like a work for map, you just have to pass the right procedure as a parameter:
(define (lengths lst)
(map string-length lst))
As you should know, map applies a procedure to each of the elements in the input list, returning a new list collecting the results. If we're interested in building a list with the lengths of strings, then we call string-length on each element. The procedure pretty much writes itself!
A word of advice: read the documentation of the procedures you're being asked to use, the code you're writing is overly complicated. This was clearly not a job for filter, although fold-right could have been used, too. Just remember: let the higher-order procedure take care of the iteration, you don't have to do it explicitly:
(define (lengths lst)
(fold-right (lambda (x a)
(cons (string-length x) a))
'()
lst))
This looks like homework so I'll only give you pointers:
map takes a procedure and applies to to every element of a list. Thus
(define (is-strings lst)
(map string? lst))
(is-strings '("hello" 5 sym "89")) ; (#t #f #f #t)
(define (add-two lst)
(map (lambda (x) (+ x 2)) lst))
(add-two '(3 4 5 6)) ; ==> (5 6 7 8)
filter takes procedure that acts as a predicate. If #f the element is omitted, else the element is in the resulting list.
(define (filter-strings lst)
(filter string? lst))
(filter-strings '(3 5 "hey" test "you")) ; ==> ("hey" "you")
fold-right takes an initial value and a procedure that takes an accumulated value and a element and supposed to generate a new value:
(fold-right + 0 '(3 4 5 6)) ; ==> 18, since its (+ 3 (+ 4 (+ 5 (+ 6 0))))
(fold-right cons '() '(a b c d)) ; ==> (a b c d) since its (cons a (cons b (cons c (cons d '()))))
(fold-right - 0 '(1 2 3)) ; ==> -2 since its (- 1 (- 2 (- 3 0)))
(fold-right (lambda (e1 acc) (if (<= acc e1) acc e1)) +Inf.0 '(7 6 2 3)) ; ==> 2
fold-right has a left handed brother that is iterative and faster, though for list processing it would reverse the order after processing..
(fold-left (lambda (acc e1) (cons e1 acc)) '() '(1 2 3 4)) ; ==> (4 3 2 1)