(defun replace (L z y)
(let ((f (lambda (x)
(if (equal x z)
y x) z y)))
(mapcar f L)))
(replace '(3 1 5 6 3 2 3) 3 9)
(9 9 9 9 9 9 9)
Your lambda always returns y, since you have a z and a y after your if. Change it to drop those and then it works:
(defun replace (L z y)
(let ((f (lambda (x)
(if (equal x z)
y
x))))
(mapcar f L)))
(replace '(3 1 5 6 3 2 3) 3 9)
(9 1 5 6 9 2 9)
Related
new to stackoverflow and new to racket. I've been studying racket using this documentation: https://docs.racket-lang.org/reference/pairs.html
This is my understanding of map:
(map (lambda (number) (+ 1 number))'(1 2 3 4))
this assigns '(1 2 3 4) to variable number ,then map performs (+ 1 '(1 2 3 4)).
but when I see things like:
(define (matrix_addition matrix_a matrix_b)
(map (lambda (x y) (map + x y)) matrix_a matrix_b))
I get very lost. I assume we're assigning two variables x and y, then performing (map + x y),but I don't understand what or how (map + x y) works.
Another one I'm having trouble with is
(define (matrix_transpose matrix_a)
(apply map (lambda x x) matrix_a))
what does (lambda x x) exactly do?
Thank you so much for clarifying. As you can see I've been working on matrix operations as suggested by a friend of mine.
Here is one way to think about map:
(map f (list 1 2 3))
; computes
(list (f 1) (f 2) (f 3))
and
(map f (list 1 2 3) (list 11 22 33))
; computes
(list (f 1 11) (f 2 22) (f 3 33))
So your example with + becomes:
(map + (list 1 2 3) (list 11 22 33))
; computes
(list (+ 1 11) (+ 2 22) (+ 3 33))
which is (list 12 24 36).
In the beginning it with be clearer to write
(define f (lambda (x y) (+ x y)))
(map f (list 1 2 3) (list 11 22 33)))
but when you can get used to map and lambda, the shorthand
(map (lambda (x y) (+ x y)) (list 1 2 3) (list 11 22 33)))
is useful.
this assigns '(1 2 3 4) to variable number ,then map performs (+ 1 '(1 2 3 4)).
No, that's not what it does. map is a looping function, it calls the function separately for each element in the list, and returns a list of the results.
So first it binds number to 1 and performs (+ 1 number), which is (+ 1 1). Then it binds number to 2 and performs (+ 1 number), which is (+ 1 2). And so on. All the results are collected into a list, so it returns (2 3 4 5).
Getting to your matrix operation, the matrix is represented as a list of lists, so we need nested loops, which are done using nested calls to map.
(map (lambda (x y) (map + x y)) matrix_a matrix_b)
The outer map works as follows: First it binds x and y to the first elements of matrix_a and matrix_b respectively, and performs (map + x y). Then it binds x and y to the second elements of matrix_a and matrix_b, and performs (map + x y). And so on for each element of the two lists. Finally it returns a list of all these results.
The inner (map + x y) adds the corresponding elements of the two lists, returning a list of the sums. E.g. (map + '(1 2 3) '(4 5 6)) returns (5 7 9).
So all together this creates a list of lists, where each element is the sum of the corresponding elements of matrix_a and matrix_b.
Finally,
what does (lambda x x) exactly do?
It binds x to the list of all the arguments, and returns that list. So ((lambda x x) 1 2 3 4) returns the list (1 2 3 4). It's basically the inverse of apply, which spreads a list into multiple arguments to a function.
(apply (lambda x x) some-list)
returns a copy of some-list.
this assigns '(1 2 3 4) to variable number ,then map performs (+ 1 '(1 2 3 4)).
If it was that simple why the need for map. You could just do (+ 1 '(1 2 3 4)) directly. Here is an implementation of map1 which is map that can only have one list argument:
(define (map1 fn lst)
(if (empty? lst)
empty
(cons (fn (first lst))
(map1 f (rest lst)))))
And what it does:
(map1 add1 '(1 2 3))
; ==> (cons (add1 1) (cons (add1 2) (cons (add1 3) empty)))
; same as (list (add1 1) (add1 2) (add1 3))
The real map accepts any number of list arguments and then expect the element function to take as many elements that there are list arguments. eg.
(map (lambda (l n s) (list l n s)) '(a b c) '(1 2 3) '($ % *))
; ==> ((a 1 $) (b 2 %) (c 3 *))
A very cool way to do this without knowing the number of elements is unzip
(define (unzip . lst)
(apply map list lst))
(unzip '(a b c) '(1 2 3) '($ % *))
; ==> ((a 1 $) (b 2 %) (c 3 *))
So apply flattens the call to (map list '(a b c) '(1 2 3) '($ % *)) and list takes arbitrary elements so it ends up working the same way as th eprevious example, but it will also work for other dimentions:
(unzip '(a b c d) '(1 2 3 4))
; ==> ((a 1) (b 2) (c 3) (d 4))
The first argument of map is a function. This function can require one or more arguments. Followed by the function in the arguments lists are one or more lists.
map loops from first to the last element of the lists in parallel.
And feeds therefore the i-th position of each list as arguments to the function
and collects the result into a results list which it returns.
Now three short examples which would make it clear to you how map goes through the lists:
(map list '(1 2 3))
;; => '((1) (2) (3))
(map list '(1 2 3) '(a b c))
;; => '((1 a) (2 b) (3 c))
(map list '(1 2 3) '(a b c) '(A B C))
;; => '((1 a A) (2 b B) (3 c C))
Example of what function should do:
(list 3 4 6 9 7) ←→ 3x^4 + 4x^3 + 6x^2 + 9x + 7
What I have so far:
(define (poly-eval x numlist)
(compute-poly-tail x numlist 0 0))
(define (compute-poly-tail xn list n acc)
(cond
[(null? list) acc]
[else (compute-poly-tail (first list) (rest list)
(+ acc (* (first list) (expt xn n))) (+ n 1))]))
(check-expect(poly-eval 5 (list 1 0 -1)) 24)
(check-expect(poly-eval 0 (list 3 4 6 9 7)) 7)
(check-expect(poly-eval 2 (list 1 1 0 1 1 0)) 54)
Expected results:
(check-expect(poly-eval 5(list 1 0 -1)) 24)
(check-expect(poly-eval 0 (list 3 4 6 9 7))7)
(check-expect(poly-eval 2 (list 1 1 0 1 1 0)) 54)
I am getting a run-time error. Can someone spot what I am doing wrong. I don't know why I am getting these results.
There are a couple of errors in the code:
You need to process the coefficient's list in the correct order, corresponding to their position in the polynomial! you can either:
reverse the list from the beginning and process the coefficients from right to left (simpler).
Or start n in (sub1 (length numlist)) and decrease it at each iteration (that's what I did).
The order and value of the arguments when calling the recursion in compute-poly-tail is incorrect, check the procedure definition, make sure that you pass along the values in the same order as you defined them, also the first call to (first list) doesn't make any sense.
You should not name list a parameter, this will clash with the built-in procedure of the same name. I renamed it to lst.
This should fix the issues:
(define (poly-eval x numlist)
(compute-poly-tail x numlist (sub1 (length numlist)) 0))
(define (compute-poly-tail xn lst n acc)
(cond
[(null? lst) acc]
[else (compute-poly-tail xn
(rest lst)
(- n 1)
(+ acc (* (first lst) (expt xn n))))]))
It works as expected:
(poly-eval 5 (list 1 0 -1))
=> 24
(poly-eval 0 (list 3 4 6 9 7))
=> 7
(poly-eval 2 (list 1 1 0 1 1 0))
=> 54
Build power coefficient and unknown list than use map function.
; 2*3^1+4*3^0
; input is 3 and '(2 4)
; we need '(3 3) '(2 4) '(1 0)
; use map expt build '(3^1 3^0)
; use map * build '(2*3^1 4*3^0)
; use foldr + 0 sum up
(define (poly-eval x coefficient-ls)
(local ((define power-ls (reverse (build-list (length coefficient-ls) values)))
(define unknown-ls (build-list (length coefficient-ls) (λ (i) x))))
(foldr + 0 (map * coefficient-ls (map expt unknown-ls power-ls)))))
How do I turn this:
(filter (lambda (x) (and (number? x) (> x 2)))
'(1 2 3 4 5 6 7))
=> (3 4 5 6 7)
Into a define function in Scheme where I can pass in the list and conditional number?
Well, by passing the list and the number as parameters:
(define (filter-greater lst num)
(filter (lambda (x) (and (number? x) (> x num)))
lst))
(define (pascal x y)
(cond ((or (<= x 0) (<= y 0) (< x y )) 0)
((or (= 1 y) (= x y) ) 1)
(else (+ (pascal (- x 1) y) (pascal (- x 1) (- y 1))))))
This is the function I have for a recursive pascal call that should return the number given the x and y of the triangle.
1
11
121
1331
14641
If I enter pascal 0 0, it should return 1, however it returns 0;
If I enter pascal 4 2, it should return 6, but it returns 3;
Seems like my base is off but I'm not sure how I can change it without ruining the calculation for pascals algorithm. Could someone point me to the right direction
You are very close, but your conditions aren't quite doing what you think they are. You have something like an off-by-1 error, and you haven't properly split apart your cond cases.
#lang racket/base
(for ((x (in-range 1 6)))
(for ((y (in-range (add1 x))))
(printf "~a " (pascal x y)))
(newline))
0 1
0 1 1
0 1 2 1
0 1 3 3 1
0 1 4 6 4 1
I make some small changes to your conditions and get this output:
(for ((x (in-range 6)))
(for ((y (in-range (add1 x))))
(printf "~a " (pascal x y)))
(newline))
1
1 1
1 2 1
1 3 3 1
1 4 6 4 1
1 5 10 10 5 1
If this doesn't help I can edit later and put the solution in, but this smells like homework and I don't want to just post a solution.
I have a project about the game "Oware", we are supposed to write the game in the program Dr.Racket.
These are the rules for the game, they explain it pretty well, illustrating with pictures: http://www.awale.info/jugar-a-lawale/les-regles-de-lawale/?lang=en
Im kinda stuck on the first exercise, i have the method, but its not giving the numbers in the right order.
The first function we have to write is called "distribute" which should re-put x grains in the holes, giving the result in a form of a list consisting of the number of grains rest and the new numbers for the holes.
This is whats expected:
(distribute 5 '(2 3 1 5 5 2)) -> (0 (3 4 2 6 6 2))
(distribute 5 '(2 3 1)) -> (2 (3 4 2))
What I wrote:
(define distribue
(lambda (n l)
(if (or (= n 0) (null? l))
(list l n)
(cons (+ (car l) 1) (distribue (- n 1) (cdr l))))))
What it gives:
(distribue 5 '(2 3 1 5 5 2)) -> (3 4 2 6 6 (2) 0)
(distribue 5 '(2 3 1)) -> (3 4 2 () 2)
I was trying to change the list cons append, but never got the expected form of answer
How about
(define (distribue n l)
(define (iterator n p q)
(if (or (= n 0) (null? q))
(list n (append p q))
(iterator (- n 1) (append p (list (+ 1 (car q)))) (cdr q))))
(iterator n '() l))
where
(distribue 5 '(2 3 1 5 5 2))
(distribue 5 '(2 3 1))
returns
'(0 (3 4 2 6 6 2))
'(2 (3 4 2))
as required.