racket

问题 I'm having a problem transforming a vector like this: #(#(1 2 3)#(1 2 3)#(1 2 3)#(1 2 3)#(1 2 3))) Into one like this: #(#(1 1 1 1 1) #(2 2 2 2 2) #(3 3 3 3 3)) I wrote a piece of test code but the output is wrong. I went into the debugger and I think I know which line of code cause the problem. I can't seems to find a way to make it work. Any help is greatly appreciated. (define (test) (let* ((table #(#(1 2 3)#(1 2 3)#(1 2 3)#(1 2 3)#(1 2 3))) (counter 5) (size 3) (new-table (make-vector

问题 I want to write a function in Racket which takes an amount of money and a list of specific bill-values, and then returns a list with the amount of bills used of every type to make the given amount in total. For example (calc 415 (list 100 10 5 2 1)) should return '(4 1 1 0 0) . I tried it this way but this doesn't work :/ I think I haven't fully understood what you can / can't do with set! in Racket, to be honest. (define (calc n xs) (cond ((null? xs) (list)) ((not (pair? xs)) (define y n)

问题 After spending some time looking at the documentation, searching the web, and experimenting at the prompt, I haven't succeeded at plotting points using Racket. Could someone post an example of how I'd go about plotting (0 1 2 3 4 5) for my x coordinates and (0 1 4 9 16 25) for the y coordinates. I think 1 good example will clear the problem up. 回答1: Based on the first example of the doc, and given that the function you want to plot already exists in Racket, it's as simple as: (require plot)

问题 I have an experiment for my project, basically, I need to embedded some s-expression into the code and make it run, like this, (define (test lst) (define num 1) (define l (list)) `@lst) ; oh, this is not the right way to go. (define lst `( (define num2 (add1 num)) (displayln num2))) I want the test function be like after test(lst) in racket code: (define (test lst) (define num 1) (define l (list)) (define num2 (add1 num) (displayln num2)) How can I do this in racket? Update The reason I would

问题 I am writing a small hangman game in scheme and am getting a very weird issue that almost seems like a language specific one. In my game I have a variable that holds the number of mistakes allowed and on each recursive call of my game loop, I "let" the value to a new one if it needs to be changed. Here is some code to help visualize how I run the game loop. guessed_list - a list of string characters containing old guesses and one new guess (ex. '("a" "x" "b") where "a" is the new guess) game

问题 How can I create a macro so that S:N or [S:N] returns a range of numbers starting with S and ending with N (step 1). Basically, it should be able to use it in place of 'in-range'. I tried to create something similar to Curly brackets {} to replace 'begin' in Racket but could not. Edit: I tried following as suggested by @soegaard : my-top.rkt: #lang racket (define-syntax-rule (my-top S:N) (range S N) ) (provide (rename-out [my-top #%top])) test.rkt: #lang racket (require "my-top.rkt") (1:42)

问题 I'm curious about the return value of define in Scheme. So I wrote the following lines in Racket #lang r5rs (display (define a 3)) And get the error define: not allowed in an expression context in: (define a 3) I have 2 questions about this: Does it mean that define has no return value? According to R5RS, define is not an expression. It's a program structure. Is it true that only expressions have return values, and other forms don't? 回答1: "If a tree falls in a forest and no one is around to

问题 Evaluating this code (C-c C-c): #+begin_src scheme (andmap + '(1 2 3) '(4 5 6)) #+end_src leads to the following babel error: ERROR: Unbound variable: andmap The cause: babel evaluated the code with Guile instead of Racket. How can I tell Babel to execute code using Racket, not Guile? 回答1: http://terohasu.net/blog/2011-09-08-on-racket-support-in-emacs-org-mode.html describes a way: When configuring Emacs to set things up I wasn’t familiar with Babel or any of the solutions for evaluating

问题 Lets say I want to do the following: (define (foo lst x) (filter function lst) but function takes in 2 arguments (and function was given to me), one being the list lst it will use, and the other being x . Syntactically, how would I change that line to pass in the second argument? Sorry I am new to Scheme/DrRacket. 回答1: Try this, using curry: (define (foo lst x) (filter (curry function x) lst)) That is, assuming that function takes as first parameter x and as second parameter each one of the

问题 I'm trying to understand what the difference is between cons and append in terms of how the lists are being stored underneath. consider the case: (define x '(hi, hello)) (define y '(randomvalue1, randomvalue2)) if I'm assuming that calling (cons x y) on these two lists will generate a dotted pair where the first pointer points to x and the second points to y . all in all neither x or y has been changed. Only thing added memory wise is the dotted pair. However when I call append with the two