Define a function, call it ‘CHECK’ in scheme - function

how to write this in scheme?
Define a function, call it ‘CHECK’ that takes an integer as its first argument, a list of integers
as its second argument and returns the count of how many times the first argument appears in the second

What you are describing is a variation of what is usually called `count':
(define (check num ls) (count (lambda (val) (= val num)) ls))
(check 2 '(2 3 4 2)) ; => 2

Related

How can a list or vector in Clojure be used to store and use functions?

Here is what I tried in the REPL.
(def mstuffs [(fn [n] (* 1 n)) (fn [n] (* 1 n))])
((mstuffs first) 2)
Here is the error I got
; Execution error (IllegalArgumentException) at nrepl.middleware.interruptible-eval/evaluate$fn$fn (interruptible_eval.clj:87).
; Key must be integer
What am I doing wrong?
Vectors in clojure are functions from index to to the corresponding item e.g.
([:foo :bar :baz] 1)
=> :bar
so (mstuffs first) is trying to index the vector mstuffs with the argument first. Since first is a function not an integer you get the resulting exception. It looks like you mean to invoke the first function in the vector so should use
((first mstuffs) 2)

json-get function in lisp

In these day i'm working to a json parse in prolog and lisp.
yesterday with your help i finished the prolog project and now i need help again.
the funcion is always json-get but now in lisp.
this is the functin that i wrote:
(defun json-get (json_obj fields &optional n)
(let ((place (assoc fields json_obj :test 'string=)))
(if (null place)
n
(ns (second place) t)))
the behavior of the funtion should be the same of the prolog predicate.
for example if the input is:
CL-prompt> (defparameter x (json-parse "{\"nome\" : \"Arthur\",\"cognome\" : \"Dent\"}"))
X
CL-prompt> x
(json-obj ("nome" "Arthur") ("cognome" "Dent"))
the output should be:
CL-prompt> (json-get x "cognome")
"Dent"
insted, if the input is:
(json-get (json-parse
"{\"name\" : \"Zaphod\",
\"heads\" : [[\"Head1\"], [\"Head2\"]]}")
"heads" 1 0)
the output should be:
"Head2"
the function that i wrote is totally wrong?
P.S. for this project are forbidden functions like SET, SETQ, SETF e MULTIPLE-VALUE-SETQ and DO, DO*, DOTIMES, DOLIST e LOOP and DEFPARAMETER, DEFVAR e DEFCOSTANT inside a function
thanks guys
edit 1:
this is the description of this funcion,
a json-get function that accepts a JSON object
(represented in Common Lisp, as produced by the json_parse function) and a series of
"Fields", retrieve the corresponding object. A field represented by N (with N a number
greater than or equal to 0) represents an index of a JSON array.
edit 2 :
if i try to run json-get lisp answer me with:
Error: The variable PLACE is unbound.
You need to implement this recursively. You also need to distinguish JSON arrays (which are implemented as a list of elements prefixed with json-array) and JSON objects (which are implemented as an association list.
(defun json-get (json_obj fields)
(if (null fields) ; base case of recursion
json_obj
(let* ((cur-key (car fields))
(current (cond ((and (integerp cur-key)
(eq (car json_obj) 'json-array))
(nth (1+ cur-key) json_obj)) ; add 1 to skip over JSON-ARRAY
((and (stringp cur-key)
(eq (car json_obj) 'json-obj))
(second (assoc cur-key (cdr json_obj) :test #'string=))) ; Use CDR to skip over JSON-OBJ
(t (error "~S is not a JSON object or array or ~s is not appropriate key" json_obj cur-key)))))
(json-get current (cdr fields)))))
fields has to be a list of fields, so your second example would be:
(json-get (json-parse
"{\"name\" : \"Zaphod\",
\"heads\" : [[\"Head1\"], [\"Head2\"]]}")
'("heads" 1 0))
and the first example should be:
(json-get x '("cognome"))

Recursion in Lisp and making my own length function

I am trying to make my own length/2 function(which allows you to find the length of a list) in lisp and am having an issue.
If I were to program this in java I would create a global index variable
that = 0
Then in the method I would do
if(list.equal(null))
{
return index;
}
else
{
index++;
return functioname(tail of list) ;
}.
Obviously this is not actual java syntax but I am just trying to relay the logic I want to apply in lisp.
My main issue is that ifs in lisp only allow you to do
if test expression
then do something
else
do something else
while I am trying to do
if test expression
then do something
else
do 2x something
Is there a way I can accomplish this in lisp or is there a better way to go about this problem?
Do it recursively:
len(list):
if list is null
return 0
else
remove first_item from list
return 1 + len(list)
(define (length items)
(if (null? items)
0
(+ 1
(length (cdr items)))))
(length (list 1 2 3))
3
Or use set!:
(define count 0)
(define (length items)
(when (not (null? items))
(set! count (+ count 1))
(length (cdr items))))
(length (list 1 2 3 4 5))
count
5
Scheme has the special operator begin (the Common Lisp equivalent is progn) that lets you group expressions which are executed in sequence, and returns the last one.
(if (null? items)
0
(begin (set! index (+ 1 index))
(my-length (cdr items))
index))
If the if expression returns nil under the false condition, then you can also use when (as in Rahn's answer) which is a bit more compact.
That said, having a global variable, or any variable at all, whose value gets changed (which is the purpose of a variable) is not the Lisp/recursive way of doing things. (Global variables are not a good thing in any language.) Rahn's first example is the only way any experienced Lisp programmer would do it.
In my example above, the (set! index (+ 1 index)) and index)) lines, and the global variable index are completely unnecessary if you add the 1 to the (my-length (cdr items)). If the list is empty, the length is zero; otherwise, the length is 1 + the length of the tail of the list.

recursive definition of map in common lisp [duplicate]

This question already has answers here:
Why do we need funcall in Lisp?
(3 answers)
Closed 8 years ago.
I was expecting the Scheme approach:
(defun mmap (f xs)
(if (equal xs NIL)
'()
(cons (f (car xs)) (mmap f (cdr xs)))))
but i get a compiler warning
;Compiler warnings for ".../test.lisp" :
; In MMAP: Undefined function F
I can see that I need to make the compiler aware that f is a function. Does this have anything to do with #'?
Common Lisp is a LISP2. It means variables that are in the first element of the form is evaluated in a function namespace while everything else is evaluated from a variable namespace. Basically it means you can use the same name as a function in your arguments:
(defun test (list)
(list list list))
Here the first list is looked up in the function namespace and the other two arguments are looked up in the variable namespace. First and the rest become two different values because they are fetched from different places. If you'd do the same in Scheme which is a LISP1:
(define (test list)
(list list list))
And pass a value that is not a procedure that takes two arguments, then you'll get an error saying list is not a procedure.
A function with name x in the function namespace can be fetched from other locations by using a special form (function x). Like quote function has syntax sugar equivalent so you can write #'x. You can store functions in the variable namespace (setq fun #'+) and you can pass it to higher order functions like mapcar (CL version of Scheme map). Since first element in a form is special there is a primitive function funcall that you can use when the function is bound in the variable namespace: (funcall fun 2 3) ; ==> 5
(defun my-mapcar (function list)
(if list
(cons (funcall function (car list))
(my-mapcar function (cdr list)))
nil))
(my-mapcar #'list '(1 2 3 4)) ; ==> ((1) (2) (3) (4))
A version using loop which is the only way to be sure not to blow the stack in CL:
(defun my-mapcar (function list)
(loop :for x :in list
:collect (funcall function x)))
(my-mapcar #'list '(1 2 3 4)) ; ==> ((1) (2) (3) (4))
Of course, mapcar can take multiple list arguments and the function map is like mapcar but it works with all sequences (lists, arrays, strings)
;; zip
(mapcar #'list '(1 2 3 4) '(a b c d)) ; ==> ((1 a) (2 b) (3 c) (4 d))
;; slow way to uppercase a string
(map 'string #'char-upcase "banana") ; ==> "BANANA"
functions in CL that accepts functions as arguments also accept symbols. Eg. #'+ is a function while '+ is a symbol. What happens is that it retrieves #'+ by fetching the function represented by + in the global namespace. Thus:
(flet ((- (x) (* x x)))
(list (- 5) (funcall #'- 5) (funcall '- 5))) ; ==> (25 25 -5)
Trivia: I think the number in LISP1 and LISP2 refer to the number of namespaces rather than versions. The very first LISP interpereter had only one namespace, but dual namespace was a feature of LISP 1.5. It was he last LISP version before commercial versions took over. A LISP2 was planned but was never finished. Scheme was originally written as a interpreter in MacLisp (Which is based on LISP 1.5 and is a LISP2). Kent Pittman compared the two approaches in 1988.

Defining a Racket Function?

I'm supposed to define the function n! (N-Factorial). The thing is I don't know how to.
Here is what I have so far, can someone please help with this? I don't understand the conditionals in Racket, so an explanation would be great!
(define fact (lambda (n) (if (> n 0)) (* n < n)))
You'll have to take a good look at the documentation first, this is a very simple example but you have to understand the basics before attempting a solution, and make sure you know how to write a recursive procedure. Some comments:
(define fact
(lambda (n)
(if (> n 0)
; a conditional must have two parts:
; where is the consequent? here goes the advance of the recursion
; where is the alternative? here goes the base case of the recursion
)
(* n < n))) ; this line is outside the conditional, and it's just wrong
Notice that the last expression is incorrect, I don't know what it's supposed to do, but as it is it'll raise an error. Delete it, and concentrate on writing the body of the conditional.
The trick with scheme (or lisp) is to understand each little bit between each set of brackets as you build them up into more complex forms.
So lets start with the conditionals. if takes 3 arguments. It evaluates the first, and if that's true, if returns the second, and if the first argument is false it returns the third.
(if #t "some value" "some other value") ; => "some value"
(if #f "some value" "some other value") ; => "some other value"
(if (<= 1 0) "done" "go again") ; => "go again"
cond would work too - you can read the racket introduction to conditionals here: http://docs.racket-lang.org/guide/syntax-overview.html#%28part._.Conditionals_with_if__and__or__and_cond%29
You can define functions in two different ways. You're using the anonymous function approach, which is fine, but you don't need a lambda in this case, so the simpler syntax is:
(define (function-name arguments) result)
For example:
(define (previous-number n)
(- n 1))
(previous-number 3) ; => 2
Using a lambda like you have achieves the same thing, using different syntax (don't worry about any other differences for now):
(define previous-number* (lambda (n) (- n 1)))
(previous-number* 3) ; => 2
By the way - that '*' is just another character in that name, nothing special (see http://docs.racket-lang.org/guide/syntax-overview.html#%28part._.Identifiers%29). A '!' at the end of a function name often means that that function has side effects, but n! is a fine name for your function in this case.
So lets go back to your original question and put the function definition and conditional together. We'll use the "recurrence relation" from the wiki definition because it makes for a nice recursive function: If n is less than 1, then the factorial is 1. Otherwise, the factorial is n times the factorial of one less than n. In code, that looks like:
(define (n! n)
(if (<= n 1) ; If n is less than 1,
1 ; then the factorial is 1
(* n (n! (- n 1))))) ; Otherwise, the factorial is n times the factorial of one less than n.
That last clause is a little denser than I'd like, so lets just work though it down for n = 2:
(define n 2)
(* n (n! (- n 1)))
; =>
(* 2 (n! (- 2 1)))
(* 2 (n! 1))
(* 2 1)
2
Also, if you're using Racket, it's really easy to confirm that it's working as we expect:
(check-expect (n! 0) 1)
(check-expect (n! 1) 1)
(check-expect (n! 20) 2432902008176640000)