I'm just starting out on lisp. I'm trying to make a product function written in Lisp. The function should take in an arbitrary parameter x, and returns the product of all numerical values contained in x. It should produce the following:
>(product 'x) -> 1
>(product '(x 5)) -> 5
>(product '((2 2)3)) -> 12
>(product '((a 3)(2 1))) -> 6
I was able to do the following:
(defun product (x)
"This function takes in an arbitrary parameter x, and returns the product of all numeric values contained within x."
(cond
((consp x) (* (car x)(product (cadr x))))
((numberp x) x)
(t 1)
)
)
This handles the cases like
(product '(2 5))-> 10
(product 'x) -> 1
But not for ones like:
>(product '(x 5)) -> 5
>(product '((2 2)3)) -> 12
I'm not sure where to go from here.
The first clause of your cond expression reads as follows:
If x is a cons cell, multiply its first element (assumed to be a number) with the result of calling product on its second element (or NIL).
Both (product '(x 5)) and (product '((2 2) 3)) fail because the first element is not a number. You should multiply by (product (car x)) instead of (car x), in order to convert your arbitrary terms into numbers.
Note also that recursing on cadr is not enough to iterate over all elements of your input list. You probably intended to use cdr.
Related
I having trouble with this function,
I would like a function which
returns a list of number inferior for a given a number.
What I did so far,
(defun inf (n l)
(cond
((not l) 0)
((>= n (car l))(cons n (inf n(cdr l))))
(cons (car l) (inf n (cdr l))) ))
But it keeps returns
(inf 12 '(12 5 3))
(12 12 10)
Instead of :
(inf 12 '(12 5 3 53 45))
(12 5 3)
What did I miss ?
First of all, the function you posted does not behave the way you claim.
The 1st invocation returns (12 12 12 . 0) (because you return 0 instead of nil in the 1st cond clause)
and the second invocation raises an exception
COND: variable CONS has no value
because you got cond syntax wrong.
Second, the question summary, question text, and the attempted
implementation, specify three different problems.
Here is a fix for your code (I replaced
car and cdr with
first and
rest for pedagogical reasons):
(defun inf (n l)
(cond
((not l) ()) ; return empty list
((>= n (first l))
(cons n (inf n (rest l))))
(t
(cons (first l) (inf n (rest l))))))
If this is, in fact, what you want, you can implement in a more
idiomatic way:
(defun inf-1 (n l)
(and l (cons (max n (first l)) (inf-1 n (rest l)))))
or even
(defun inf-2 (n l)
(mapcar (lambda (x) (max n x)) l))
If you actually want the list of numbers less than the given one, you
can use remove:
(remove 12 '(12 5 3 100) :test #'<=)
==> (5 3)
A not-so-obvious way to solve the same problem using Common Lisp's existing functions is to pass a comparison operator to REMOVE.
(remove 10 '(0 3 5 11 22 10 22 3 2) :test #'<)
The above removes all elements "equal" to 10 according to #'<, which are thus all elements u such that (< 10 u) holds. In other words, all elements strictly above 10:
(0 3 5 10 3 2)
It turns out that there is an example of this in the section linked above:
(remove 3 '(1 2 4 1 3 4 5) :test #'>) => (4 3 4 5)
Edit: since this is now the accepted answer, note also that this approach is probably hard to read, be careful when using it (add a comment, etc.).
I ran this code and I don't understand how the output is 2? As far as I'm concerned the output should always be 1?
Code:
(define (count p l)
(if (empty? l)
0
(if (p (first l))
(+l (count p (rest l)))
(count p (rest l)))))
(define (ident x)
x)
(count ident '(#true #true #false))
Your function count counts the number of elements in the list that satisfy the predicate p. Since you are using ident as the predicate, we have that (ident #true) = #true and (ident #false) = #false does not. This means that you have two values in your example list that are counted. This matches the result 2.
First, you have used the number 1 instead of the letter l for the list, so your code will not even run the way it is. After fixing that to:
(define (count p l)
(if (empty? l)
0
(if (p (first l))
(+ 1 (count p (rest l)))
(count p (rest l)))))
you should get an output of 2, as you noticed, because the procedure is essentially counting up by one whenever (p (first l)) evaluates to true. And it holds that:
> (if #true 1 0)
1
> (if #false 1 0)
0
so for the list '(#true #true #false), the procedure will count up twice, hence the result being 2.
I am basically new to Scheme, and this is my second attempt to solve a certain problem.
So what I am trying to do basically is sum each parameter we pass into the function, with an appropriate value
for example:
(sum 3 1 2 0 2) ;would return 228
Here is my code:
(define (sum one five ten twenty fifty hundred)
(+ (* 1 one)(* 5 five) (* 10 ten) (* twenty 20) (* 50 fifty) (* 100 hundred))
I think a possible solution is using the lambda function, but I could't know how to implement it.
Here is one way you can calculate the numbers by sending in only 5 arguments instead of 6:
(define sum
(lambda (L skip) ;stores yourList and the number you want to skip
(define loop
(lambda (L L2 total) ;yourList, (1,5,10...) list, running total
(cond
((null? L) (list total)) ;print the total when finish yourList
;if the next number is the skip number, just move on to next value
((= (car L2) skip) (loop L (cdr L2) total))
;otherwise, increase total and send in both lists, minus the head, to the loop
(else (loop (cdr L) (cdr L2) (+ total (* (car L) (car L2)) )))
)
)
)(loop L (list 1 5 10 20 50 100) 0) ;initial values of loop
)
)
;send in your list of numbers, plus the value place you want to skip
(sum (list 3 1 2 0 2) 20) ; ==> 228
I would be much easier, though, to just fill in all the places that you don't want with a 0. Once you have 6 arguments, the following code will work.
(define sum
(lambda (one five ten twenty fifty hundred)
(+ (* 1 one) (* 5 five) (* 10 ten) (* 20 twenty) (* 50 fifty) (* 100 hundred) )
)
)
(sum 3 1 2 0 0 2)
The OP wants to be able to pass fewer arguments than there are parameters. In this specific case, it's best to use keyword (named) arguments. Here's how you might do this (in Racket syntax):
(define (total-bills #:ones (ones 0)
#:fives (fives 0)
#:tens (tens 0)
#:twenties (twenties 0)
#:fifties (fifties 0)
#:hundreds (hundreds 0))
(+ ones (* fives 5) (* tens 10) (* twenties 20) (* fifties 50) (* hundreds 100)))
(The 0 after each of the variable names are the default values, that is, what the value would be if you don't specify it.)
Example usage:
> (total-bills #:ones 3 #:fives 1 #:tens 2 #:hundreds 2)
228
If you want to use standard scheme with a similar scheme as Chris' answer in plain old Scheme you need to plan how to identify which currency your number is in. Perhaps by sending them in pairs would be both simpler to implement and simpler to use:
#!r6rs
(import (rnrs))
(define (total-bills . args)
(fold-left (lambda (acc x)
(+ acc (apply * x)))
0
args))
(total-bills '(5 1) '(2 10) '(3 100)) ; ==> 325
(apply total-bills '((1 3) (5 1) (10 1) (20 2))) ; ==> 58
You can make a procedure that makes a specialized procedure based on currencies you pass to it:
#!r6rs
(import (rnrs))
(define (make-currency-procedure . currencies)
(lambda amounts
(fold-left (lambda (acc currency amount)
(+ acc (* currency amount)))
0
currencies
amounts)))
(define small-bills (make-currency-procedure 1 5 10 20))
(define large-bills (make-currency-procedure 10 100 1000 10000))
(small-bills 3 1 1 2) ; ==> 58
(small-bills 0 1 0 1) ; ==> 25
(large-bills 3 1 1 2) ; ==> 21130
(large-bills 0 1 0 1) ; ==> 10100
(define a1 (list 1 2 3 4))
(define a2 (list + - * /))
(define a3 (list 5 6 7 8))
(map (lambda (x y z) (y x z))
a1 a2 a3)
How do I call this lambda function directly without using map?
All it does is switching y and x, so that (1 + 5) becomes (+ 1 5)
You can write you map expression without switching the arguments:
(map (lambda (x y z) (x y z)) a2 a1 a3) ; ==> (6 -4 21 1/2)
Notice I have just switched the order of the arguments to map.
You can call a lambda by wrapping it in parenthesis and adding arguments.. eg.
((lambda (op1 proc op2) (proc op1 op2)) + 2 3) ; ==> 5
The map function is just a way of doing that with every element of the different lists. You can get the same result without using lambda if you know the length of the lists:
(list ((car a2) (car a1) (car a3))
((cadr a2) (cadr a1) (cadr a3))
((caddr a2) (caddr a1) (caddr a3))
((cadddr a2) (cadddr a1) (cadddr a3))) ; ==> (6 -4 21 1/2)
Since every element of a2 is a procedure wrapping it and arguments in parenthesis applies the procedure.
A lambda form (lambda (arg ...) body ...) gets evaluated and turns into a procedure object. When you define named procedure the same happens but the name gets bound to that procedure object. In fact. there is not difference between these 3 versions:
;; version 1 using syntactic sugar define for procedures
(define (test x) (* x x))
(test 10) ;==> 100
;; version 2 defineing a variable to a procedure
(define test (lambda (x) (* x x)))
(test 10) ;==> 100
;; version 3 using the procdure directly
((lambda (x) (* x x)) 10) ; ==> 100
As you probably know, to call a function in Scheme you write (f args), where f is the function and args is the list of arguments. This isn't any different whether f is a named function or a lambda. So to call your lambda function directly, you'd write:
;( -----------f------------ args- )
( (lambda (x y z) (y x z)) 1 + 2 )
Of course that's just a rather convoluted way of writing (+ 1 2), but there you go.
I have a function that takes 2 lists as input and I would like to make it so that everything in the first list, is multiplied with everything in the second list, and then the sum is calculated.
Here's my function so far:
(defun multiply(a b)
(if (eq a nil)
0
(progn
(* (car a) (car b))
(multiply (car a) (cdr b)))
))
Currently all I'm trying to get it to do is take the first number from the first list, and multiply one by with everything in the second list. However I get this error from when the function is re-called within the function:
(This is what I inputted, '(1 2 3) and '(4 5 6))
The value 1 is not of type list.
(MULTIPLY 1 '(5 6))
Any help would be much appreciated.
Is loop acceptable?
Case 1
If the result should be 90 as in
(+ (* 1 4) (* 1 5) (* 1 6) (* 2 4) (* 2 5) (* 2 6) (* 3 4) (* 3 5) (* 3 6))
then you could do
(defun multiply (a b)
(loop for i in a
sum (loop for j in b
sum (* i j))))
Case 2
If the result is supposed to be 32 as in
(+ (* 1 4) (* 2 5) (* 3 6))
then it would be
(defun multiply (a b)
(loop
for i in a
for j in b
sum (* i j)))
If you want to multiply two lists per element, you should just be able to use mapcar:
(mapcar #'* '(3 4 5) '(4 5 6))
As to the error in your existing function, this line should be:
...
(multiply (cdr a) (cdr b)))
...
What you describe sounds like the dot product:
(defun dot (l1 l2)
(reduce #'+ (mapcar #'* l1 l2)))
The above solution is nice because it is purely functional, but, alas, it creates an unnecessary intermediate list. Of course, a Sufficiently Smart Compiler should be able to eliminate that, but, practically speaking, you are better off using loop:
(defun dot (l1 l2)
(loop for x in l1 and y in l2 sum (* x y)))
Note also that using lists to represent mathematical vectors is not a very good idea, use vectors instead.
I would modify multiply as
(defun multiply(a b)
(if (eq a nil)
0
(+
(* (car a) (car b))
(multiply (cdr a) (cdr b)))))
When you call
(multiply '(1 2 3) '(4 5 6))
it will return the sum
32
(define (*2Ls L1 L2)
(apply map * (list L1 L2)))