racket

问题 What I would like to do is create a function that takes in a list of values and a list of characters and coalesce the corresponding characters("atoms" I think they would technically be called) into a new list. Here is what I have so far; #lang racket (define (find num char) (if (= num 1) (car char) ;Problem here perhaps? (find (- num 1) (cdr char)))) (define (test num char) (if (null? num) '("Done") (list (find (car num) (test (cdr num) char))))) This however gives me an error, which for the

问题 I have a strange problem that couple of hours can't implement in Scheme. Let's say we have: (define x '( (Orlando (NY 3)) (Chicago (Montana 5) (Orlando 8)) ...and so on ... ) I want to transform it to '( (Orlando NY) (Chicago Montana Orlando) ...and so on ... ) Any help would be greatly appreciated. 回答1: You could also try (map (lambda (x) (cons (car x) (map car (cdr x)))) x) 来源： https://stackoverflow.com/questions/20749277/remove-integers-from-list

问题 I'm working on metacircular evaluator of 4.1.4 Running the Evaluator as a Program , building which with Racket: #lang racket (require (combine-in rnrs/base-6 rnrs/mutable-pairs-6)) (define (evaluate exp) (cond ; ... ((definition? exp) (display exp) (display " is a definition\n")) ; ... (else (display exp) (display " is something else\n")))) (define (definition? exp) (tagged-list? exp 'define)) (define (tagged-list? exp tag) (if (pair? exp) (eq? (car exp) tag) false)) (define (driver-loop)

问题 I'm trying to understand the relations between pair , cons , dotted tuples and proper list in PLT-Scheme 372. The detailed context of my question is as follows: After reading some textbook and doing trial-and-error, I have got the following understanding and intuitive ideas (I may be wrong...): all lists are pairs, e.g.: (pair? (list 'a 'b 'c)) => #t all conses are pairs, e.g.: (pair? (cons 'a (cons 'b 'c))) => #t some dot-separated tuples are pairs, e.g.: (pair? '(a . b)) => #t (pair? '(a .

问题 How to remove duplicate of a list? (running time is O(n log n) ) ex: '(4 6 1 1 2 3 3 5 6) => '(4 6 1 2 3 5) (define (re-dup lst) (cond ((empty? lst) empty) (else (define el (first lst)) (define el-free-lst (filter (lambda (x) (not (= el x))) (rest lst))) (cons el (re-dup el-free-lst))))) Is this right? 回答1: Your current solution is O(n^2) , because filter traverses the list once for each of the elements in the original list. It's possible to write an O(n) solution using a helper data

问题 I'm using DrScheme to work through SICP, and I've noticed that certain procedures (for example, square ) get used over and over. I'd like to put these in a separate file so that I can include them in other programs without having to rewrite them every time, but I can't seem to figure out how to do this. I've tried: (load filename) (load (filename)) (load ~/path-to-directory/filename) (require filename) (require ~/path-to-directory/filename) (require path-from-root/filename) None of these

问题 Although the following code works perfectly well in DrRacket environment, it generates the following error in WeScheme: Inside a cond branch, I expect to see a question and an answer, but I see more than two things here. at: line 15, column 4, in <definitions> How do I fix this? The actual code is available at http://www.wescheme.org/view?publicId=gutsy-buddy-woken-smoke-wrest (define (insert l n e) (if (= 0 n) (cons e l) (cons (car l) (insert (cdr l) (- n 1) e)))) (define (seq start end) (if

问题 I will use a simple example to illustrate my question. In Java, C, or any other OOP language, I could create a pie class in a way similar to this: class Apple{ public String flavor; public int pieces; private int tastiness; public goodness(){ return tastiness*pieces; } } What's the best way to do that with Scheme? I suppose I could do with something like this: (define make-pie (lambda (flavor pieces tastiness) (list flavor pieces tastiness))) (define pie-goodness (lambda (pie) (* (list-ref

问题 I've been working through the last few exercises ch 1 of SICP, where several of the exercises uses higher-order functions. Currently I'm trying to debug a problem in my solution to 1.45, which is raising an arity mismatch. The function which is raising the error is the result of twice applying an averaging operation to a fixed-point function solver. It would make my debugging efforts a lot easier if I could just dump some sort of representation of procedures, given that the procedure has been

问题 I have see the following code... The first call of (next-num) returns 1 , and the second returns 2 . (define next-num (let ((num 0)) (lambda () (set! num (+ num 1)) num))) (next-num) ; 1 (next-num) ; 2 What I can not understand is... num is created by let inside next-num , it is kind of a local variable... How does scheme know that each time next-num is called, the value of num is not erased by let ((num 0)) ; How does scheme know that it is always the same num that we modify whenever next