I am trying to write a weird function, so bear with me here. This function should take a list L as a parameter and have a sum variable. If L is not a list, it should return nil. Otherwise, it should iterate through each element of the list and do the following:
If the element is a number and less than zero, it should subtract 1 from the sum.
If the element is a number and greater than zero, it should add 1 to the sum.
if the element is 0 or not a number, then it should add 0 to the sum.
Here's the code I have, but it returns 0 regardless of arguments passed in:
(defun sigsum (L)
(let ((sum 0)) ;;variable sum
(if (not (listp L)) ;;if L is not a list
nil ;;return nil
(dotimes (i (length L)) ;;otherwise loop through L
(if (numberp (elt L i)) ;;if each element is a number
(if (< (elt L i) 0) ;;if is less than 0 subtract 1 from sum
(- sum 1)
(if (> (elt L i) 0) ;;if greater than 0 add 1 to sum
(+ sum 1))
(+ sum 0)) ;;else add 0 to sum
(+ sum 0))) ;;not a number so add 0 to sum
)
sum) ;;return sum
)
As always, any help is greatly appreciated.
Other answers have described the problems in your code, but it might be beneficial to look at other ways of solving the problem. This is a pretty typical case for a reduction with a key function (see reduce). You can sum up the elements in list with (reduce '+ list). However, you don't want to just sum up the elements, some of which might not be numbers, you want to map each element to a number (-1, 0, or 1), and add those up. That means you need a key function. First, let's define the function that takes an element to either -1, 0, or 1:
(defun to-number (x)
(cond
((and (numberp x) (< x 0)) -1)
((and (numberp x) (> x 0)) 1)
((or (not (numberp x)) (zerop x)) 0)))
Then your sum function needs to return nil if its argument is not a list, or (reduce '+ … :key 'to-number) if its argument is a list:
(defun sum (thing)
(if (not (listp thing))
nil
(reduce '+ thing :key 'to-number)))
Conceptually, this approach is the same as applying the addition operator to the result of (mapcar 'to-number list), but reduce is generally preferred, because there can be a maximum number of arguments a function can be called with, so (apply '+ (mapcar …)) breaks if (mapcar …) returns a list that's longer than that. The other problem is that mapcar will allocate a whole new list to hold the intermediate values (the results of to-number), which is an unnecessary space usage.
(- sum 1) does not update any variables. Since you don't use the result it goes away. It's like this is all programming languages. sum + 1 in an Algol language (like C) does not update sum.
If you want to update a variable you can use setf:
(setf sum (+ sum 1))
(setf sum (- sum 1))
Or you can use incf and decf that are macroes that expand to the same as the above expressions.
There are many other ways you can do it in CL. You can use reduce
(defun mysignum (list)
(if (listp list)
(reduce (lambda (e acc)
(+ acc
(cond ((or (not (numberp e)) (zerop e)) 0)
((< e 0) -1)
(t 1))))
list)
nil))
You can use loop:
(defun mysignum (list)
(if (listp list)
(loop :for e :in list
:when (and (numberp e) (< e 0))
:summing -1
:end
:when (and (numberp e) (> e 0))
:summing +1
:end)
nil))
The expression (+ sum 1) returns one more than sum. It does not change sum. You can get where you want using incf or decf, which modify places.
You could also use loop:
(loop :for element :in list
:count (and (numberp element) (plusp element)) :into plus
:count (and (numberp element) (minusp element)) :into minus
:finally (return (- plus minus)))
Signum is also useful, but it has float contagion (if any of the numbers are floats, the result will be a float as well).
(loop :for element :in list
:when (numberp element)
:sum (signum element))
(defun discrete (lis)
(cond
((and (listp lis) (not (equal lis nil)))
(let ((sum 0))
(loop for item in lis do
(cond ((or (not (numberp item)) (equal 0 item)) t)
((and (numberp item) (> item 1)) (setf sum (+ 1 sum)))
((and (numberp item) (< item 1)) (setf sum (- sum 1)))))
sum))))
Usage:(discrete '(-1 2 3 0 2 2 -1 2 34 0 -1))=> 3
(discrete '(-4 a b))=> -1
(discrete '()) => NIL
(discrete '(a s d f))=> 0
Related
Working on this program that's supposed to take a vector of integers as input and return the one with the longest integer. Example (vector 20 738 5942 125) and would return 4 as its the longest one. I'm pretty sure I have most of this done the only issue I have is in the conditional as I have to call an outside function (count-integers), this is what I have so far:
(require while)
(define (empty-VINT? low high) (> low high))
(define (count-integers n)
(cond [(< n 10) 1]
(else(+ 1 (count-integers [/ n 10])))))
(define (count-digits V)
(local [
(define x (void))
(define accum (void))
(define largest 0)]
(begin
(set! x (vector-length V))
(set! accum 0)
(while (< accum (vector-length V))
(cond [(empty-VINT? x accum) accum]
[(> (count-integers (vector-ref V accum) largest)
(add1 x) accum(vector-ref V accum))]
[else add1 accum])))))
Right now when its run, I get this message: cond: expected a clause with a question and an answer, but found a clause with only one part. Any suggestions would be great, thanks
First of all, it's not clear what do you want to return. 4 isn't the longest integer (that's 5942), 4 is a maximal digit count among integers in given vector.
Secondly, your code isn't idiomatic and without your comment, it's very hard to say what's going on. Programming in functional languages requies functional way of thinking. Forget about while, set!, void, local and nested define and instead spend some time learning about apply, map, filter and foldl.
I would solve this problem like this:
(define (digits number)
(string-length (number->string number)))
(define (max-digit-count vec)
(apply max (map digits (vector->list vec))))
(max-digit-count (vector 20 738 5942 125))
=> 4
From comments:
Design and implement a function to find the number of digits in the longest integer in a (vectorof integer) ...
use ... while loops
So a plan (design) might be:
count-digits: integer -> natural
max-digit-count: (vectorof integer) -> natural
..something while something max count-digits something ???
Implementing count-digits seems straightforward (but
integers can be negative, and in Racket (integer? 123.000) is true).
#lang racket
(define (count-digits int) ;; Integer -> Natural
;; produce count of digits in int
(string-length (number->string (abs (exact-truncate int)))))
As #Gwang-Jin Kim mentions, while could be defined:
(define-syntax-rule (while condition body ...)
;; From: https://stackoverflow.com/questions/10968212/while-loop-macro-in-drracket
(let loop ()
(when condition
body ...
(loop))))
and then one could use it:
(define (max-digit-count vec) ;; VectorOfInteger -> Natural
;; produce maximum of digit counts of vec elements
(define vx 0)
(define acc 0)
(while (< vx (vector-length vec))
(set! acc (max accum (count-digits (vector-ref vec vx))))
(set! vx (add1 vx)))
acc)
(max-digit-count (vector 20 -738.00 5942 125)) ;=> 4
One of the problems with while is that it can't produce a value (where would it come
from if the condition is false on entry?)
If one "enhanced" while a bit:
(define-syntax-rule (while< x-id limit a-id a-init update)
;; "while loop" incrementing x-id from 0 to limit-1, updating a-id
(let loop ([x-id 0] [a-id a-init])
(if (< x-id limit)
(loop (add1 x-id) update)
a-id)))
max-digit-count could be neater:
(define (max-digit-count vec) ;; VectorOfInteger -> Natural
;; produce maximum of digit counts of vec elements
(while< vx (vector-length vec)
acc 0 (max acc (count-digits (vector-ref vec vx)))))
#MartinPuda's answer is quite good.
I would have defined:
(define (digits n (acc 0))
(if (< n 1)
acc
(digits (/ n 10) (+ acc 1))))
(define (max-digits lst)
(digits (car (sort lst >))))
To apply it:
(max-digits (vector->list (vector 20 738 5942 125)))
Why you should not use while
Using while would force you to mutate variable values. It is much more "natural" for lisp languages to follow the functional style (recursive functions instead of while loops or other loops) rather than the imperative style with mutation of variables.
That is why while is not in the lisp languages.
But if you want to use it:
(define-syntax-rule (while condition body ...)
;; From: https://stackoverflow.com/questions/10968212/while-loop-macro-in-drracket
(let loop ()
(when condition
body ...
(loop))))
(define (digits n (acc 0))
(cond ((< n 1) acc)
(else (digits (/ n 10) (+ acc 1)))))
(define (max-digits lst)
(let ((max-digit 0))
(while (not (null? lst))
(let ((digit (digits (car lst))))
(when (< max-digit digit)
(set! max-digit digit))
(set! lst (cdr lst))))
max-digit))
Then you can try:
> (max-digits (vector->list v))
4
> (max-digits '(1111 123456 2345 34))
6
Prefer let over define
Why? Because if you use let, you can control the scope of the to-be-mutated variable very precisely. You can define in your definition, from where on your variable canNOT have any effect on your code (since its scope ended at some point). While with define you don't have this fine-grained control. (Or this control is implicit not explicite like with let). You could delete/unbind the variable explicitely but that is rarely done in real life.
Therefore, in Lisp, for variable declarations use whenever possible let, especially whenever you deal with mutated variables.
All imperative = declarations should be in Lisp languages let expressions!
You can use function arguments instead of let-definitions, because they are anyway implemented using lets
Just you save syntactically some lines - and the fewer lines you occupy the cleaner the code.
#lang racket
(define (digits n)
(string-length (number->string n)))
(define (max-digit a b)
(if (< (digits a) (digits b)) b a))
(define (max-digits lst (res ""))
(while (not (null? lst))
(set! res (max-digit res (car lst)))
(set! lst (cdr lst)))
(digits res))
I am trying to write a function that take a list and iterates over each element in the list if the number is even I want that number to be added to the previous number in the list.
I was thinking an accumulator will count up from 0 with each iteration giving a position for each element in the list.
If the number in the list is even I want to add that number to the previous number in the list.
Hence why I am trying to use the accumulator as an index for list-ref. I don't know how to write it to get the accumulator value for the previous iteration (+ i (list-ref a-list(- acc 1)))?
(define loopl (lambda (l)
(for/fold
([acc 0])([i l])
(cond
[(even? i)(+ i (list-ref (- acc 1) l))]
enter image description here
The question is not quite clear about the value to be returned by this function:
this answer assumes that it is a total of even elements together with their previous elements.
The function is developed using the
HtDF (How to Design Functions)
design method with a BSL (Beginning Student language) in DrRacket.
Start with a stub, incorporating signature and purpose, and a minimal "check-expect" example:
(Note: layout differs slightly from HtDF conventions)
(define (sum-evens-with-prev xs) ;; (Listof Integer) -> Integer ; *stub define* ;; *signature*
;; produce total of each even x with its previous element ; *purpose statement*
0) ; *stub body* (a valid result)
(check-expect (sum-evens-with-prev '()) 0) ; *minimal example*
This can be run in DrRacket:
Welcome to DrRacket, version 8.4 [cs].
Language: Beginning Student with List Abbreviations
The test passed!
>
The next steps in HtDF are template and inventory. For a function with one list
argument the "natural recursion" list template is likely to be appropriate;
(define (fn xs) ;; (Listof X) -> Y ; *template*
(cond ;
[(empty? xs) ... ] #|base case|# ;; Y ;
[else (... #|something|# ;; X Y -> Y ;
(first xs) (fn (rest xs))) ])) ;
With this template the function and the next tests become:
(define (sum-evens-with-prev xs) ;; (Listof Number) -> Number
;; produce total of each even x with its previous element (prev of first is 0)
(cond
[(empty? xs) 0 ] #|base case: from minimal example|#
[else (error "with arguments: " #|something: ?|#
(first xs) (sum-evens-with-prev (rest xs))) ]))
(check-expect (sum-evens-with-prev '(1)) 0)
(check-expect (sum-evens-with-prev '(2)) 2)
These tests fail, but the error messages and purpose statement suggest what is required:
the (... #|something|# from the template has to choose whether to add (first xs):
(define (sum-evens-with-prev xs) ;; (Listof Integer) -> Integer
;; produce total of each even x with its previous element (prev of first is 0)
(cond
[(empty? xs) 0 ]
[else
(if (even? (first xs))
(+ (first xs) (sum-evens-with-prev (rest xs)))
(sum-evens-with-prev (rest xs))) ]))
Now all 3 tests pass! Time for more check-expects (note: careful introduction of
check-expects is a way of clarifying ones understanding of the requirements, and
points one to the code to be added):
(check-expect (sum-evens-with-prev '(1 1)) 0)
(check-expect (sum-evens-with-prev '(1 2)) 3)
Ran 5 tests.
1 of the 5 tests failed.
Check failures:
Actual value 2 differs from 3, the expected value.
sum-evens-with-prev needs the prev value to include in the even? case:
make it available by introducing it as an argument (renaming the function), add
the appropriate arguments to the recursive calls, the function now just calls
sum-evens-and-prev:
(define (sum-evens-and-prev xs prev) ;; (Listof Integer) Integer -> Integer
;; produce total of each even x and prev
(cond
[(empty? xs) 0 ]
[else
(if (even? (first xs))
(+ prev (first xs) (sum-evens-and-prev (rest xs) (first xs)))
(sum-evens-and-prev (rest xs) (first xs))) ]))
(define (sum-evens-with-prev xs) ;; (Listof Integer) -> Integer
;; produce total of each even x with its previous element (prev of first is 0)
(sum-evens-and-prev xs 0))
(just add some more tests, and all is well :)
(check-expect (sum-evens-with-prev '(0 2)) 2)
(check-expect (sum-evens-with-prev '(2 1)) 2)
(check-expect (sum-evens-with-prev '(1 3)) 0)
(check-expect (sum-evens-with-prev '(2 2)) 6)
(check-expect (sum-evens-with-prev '(1 2 3 4)) 10)
(check-expect (sum-evens-with-prev '(1 2 3 3 5 6 6)) 26)
Welcome to DrRacket, version 8.4 [cs].
Language: Beginning Student with List Abbreviations.
All 11 tests passed!
>
The (for/fold) form requires a (values) clause, and it is in that which you would put the conditional form.
Assuming you want only the new list as the return value, you would also want a #:result clause following the iteration variables.
(define loopl
(lambda (l)
(for/fold
([index 0]
[acc '()]
#:result acc)
([i l])
(values [+ index 1]
[append acc
(if (and (> index 0)
(even? i))
(list (+ i (list-ref l (- index 1))))
(list i))]))))
This should give the correct answer.
You almost never want to repeatedly call list-ref in a loop: that makes for horrible performance. Remember that (list-ref l i) takes time proportional to i: in your case you're going to be calling list-ref with the index being, potentially 0, 1, ..., and that going to result in quadratic performance, which is bad.
Instead there's a neat trick if you want to iterate over elements of a list offset by a fixed amount (here 1): iterate over two lists: the list itself and its tail.
In addition you need to check that the first element of the list is not even (because there is no previous element in that case, so this is an error).
Finally I'm not entirely sure what you wanted to return from the function: I'm assuming it's the sum.
(define (accum-even-previouses l)
(unless (not (even? (first l)))
;; if the first elt is even this is an error
(error 'accum-even-previouses "even first elt"))
(for/fold ([accum 0])
([this (in-list (rest l))]
[previous (in-list l)])
(+ accum (if (even? this)
(+ this previous)
0))))
I have a project in scheme in which I need to implement an infinite sequence of numbers. I can't use any scheme-built-in complex functions, and I just do not know how to make my sequence infinite without program crashing in infinite loop. I don't have to really output it, but I need to be able to use it.
(seq n) ;;output: n,n+1,n+2,n+3.... to infinity (seq 5) ->5,6,7,8,9...
Right now I did a sequence until n+7, but I need this to infinity:
(define (seq n)
(define (asc-order LIST counter)
(cond ((= counter (+ n 7)) LIST)
(else (asc-order (append LIST (cons (+ counter 1) '()))
(+ counter 1)))))
(asc-order '() (- n 1))
)
IO example (It works, but I need it infinite sequence):
>(define s (seq 3))
>(car s)
3
You can represent an infinite sequence as a function that produces one element at a time. The user (consumer) can then call the function each a new element of the sequence is needed.
An example:
(define (f x) (* x x))
(define seq
(let ()
(define n 0) ; current index
(lambda () ; the function that is to be called repeatedly
(define a (f n)) ; compute the new element
(set! n (+ n 1)) ; compute new index
a))) ; return the new element
(seq) ; compute element 0
(seq) ; compute element 1
(seq) ; ...
(seq)
(seq)
(seq)
This evaluates to:
0
1
4
9
16
25
In order to write (sequence->list s n) which computes the first n elements of the sequence s, make a loop that calls s in total n times - and collect the results in a list.
The key is to delay evaluation of the list by wrapping a procedure around it.
Here's the simplest implementation I can think of.
It's only "lazy" in the tail.
(define (seq n)
(cons n (lambda () (seq (+ n 1)))))
(define (seq-car s)
(car s))
(define (seq-cdr s)
((cdr s)))
Example use:
; Get the 'n' first elements of 's'.
(define (seq-take n s)
(if (<= n 0)
'()
(cons (seq-car s) (seq-take (- n 1) (seq-cdr s)))))
> (define s (seq 10))
> s
'(10 . #<procedure>)
> (seq-take 5 s)
'(10 11 12 13 14)
Here is another solution using delayed evaluation:
(use-modules (ice-9 receive))
(define (seq f)
(let loop ((n 0))
(lambda ()
(values (f n) (loop (1+ n))))))
(define squares (seq (lambda (x) (* x x))))
(receive (square next) (squares)
(pk square) ;; => 0
(receive (square next) (next)
(pk square) ;; => 1
(receive (square next) (next)
(pk square) ;; => 4
(receive (square next) (next)
(pk square))))) ;; => 9
I'm trying to create a function that takes in user input, x, and displays all of the numbers from 1 up to x. Not quite sure where to go from here:
(define (iota x)
(if (>= x 1)
(display
;; Takes a number n
;; and returns a list with 1..n
;; in order.
(define (make-list n)
(let loop ((n n) (accumulator '()))
(if (zero? n)
accumulator
(loop (- n 1) (cons n accumulator)))))
;; prints data and adds
;; a newline
(define (println x)
(display x)
(newline))
;; prints 1..n by applying println
;; for-each element over the list 1..n
(define (iota n)
(for-each println (make-list n)))
Basically you want to use recursion. So think of it was counting up. As you count up either you've added enough numbers and you are done building your list or you need to add the current number to your list and go on to the next number.
Look at the following code:
(define (range-2 next-num max-num cur-list)
;;if we've added enough numbers return
(if (> next-num max-num)
cur-list
;; otherwise add the current number to the list and go onto the next one
(range-2
(+ 1 next-num)
max-num
(append cur-list (list next-num))
)))
But to get the function you wanted you need to start off with the constants you specified (ie 1), so lets create a convenience function for calling range with the starter values you need to set up the recursion, sometimes called priming the recursion:
(define (range X)
(range-2 1 X `() ))
You could do it without the second function using lambda, but this is pretty common style from what I've seen.
Once you've constructed a list of the numbers you need you just display it using
(display (range 10))
If you're using Racket you're in luck, you can use iterations and comprehensions, for instance in-range:
(define (iota x)
(for ([i (in-range 1 (add1 x))])
(printf "~a " i)))
Or for a more standard solution using explicit recursion - this should work in most interpreters:
(define (iota x)
(cond ((positive? x)
(iota (- x 1))
(display x)
(display " "))))
Either way, it works as expected:
(iota 10)
=> 1 2 3 4 5 6 7 8 9 10
(define iota2
(lambda (y)
(let loop ((n 1))
(if (<= n y)
(cons n (loop (+ n 1)))
'()))))
(define (iota x)
(if (>= x 1)
(begin
(iota (- x 1))
(display x)
(newline))))
Trying to learn lisp, want to delete every nth. I only managed to delete the first (nth) element
(defun delete-nth (n list)
(if (zerop n)
(cdr list)
(let ((cons (nthcdr (1- n) list)))
(if cons
(setf (cdr cons) (cddr cons))
cons))))
I'd like to delete the next nth and so on
Also I tried this:
(defun remove-nth (list n)
(remove-if (constantly t) list :start n :end (+ 1 n)))
No idea how to start again
What I was thinking was concatenating, but I have no idea of how to keep track of my position.
Counting from 1 (changing to 0 is trivial):
(defun remove-every-nth (n list)
(loop for element in list
for index from 1
unless (zerop (rem index n))
collect element))
Also: Please indent your code correctly.
An alternative way to do the same thing:
(defun remove-all-nth (list period)
(remove-if
(let ((iterator 0))
(lambda (x)
(declare (ignore x))
(= 0 (mod (incf iterator) period)))) list))
(remove-all-nth '(1 2 3 4 5 6 7 8 9 0) 3)
; (1 2 4 5 7 8 0)
Perhaps a more academic recursive solution here:
(defun delete-nth (n list)
(labels ((rec (i list)
(cond ((null list) nil)
((= i 1) (rec n (cdr list)))
(t (cons (car list) (rec (1- i) (cdr list)))))))
(rec n list)))
But in real life I'd use the loop option above.