scheme | 易学教程

Scheme: change value of an element in a list

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问题 I hate using SO as a way to find simple functions, but I really can't find a function like this anywhere: Given a list (1 2 3 4 5), I'd like the equivalent of (PHP's, Perl's, Python's) $a = array(1, 2, 3, 4, 5); $a[3] = 100; Which results in (1 2 3 100 5) Thanks! 回答1: You can write list-set! of Guile, like so: (define a (list 1 2 3 4)) ; a is '(1 2 3 4) (define (list-set! list k val) (if (zero? k) (set-car! list val) (list-set! (cdr list) (- k 1) val))) (list-set! a 2 100) ; a is '(1 2 100 4)

python爬虫3--urllib请求库之parse模块

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parse定义了处理URL的标准接口，实现URL的拆分，合并以及转换。 1.urlparse() url拆分 urlparse(urlstring，scheme=‘’，allow_ragments=True) scheme：默认协议，如果url不带协议的时候生效； allow_fragments：是否忽略fragment，如果忽略，会被解析成path，params或query的一部分。将url拆分为6部分： scheme：协议； netloc：域名； path：访问路径； params：参数； query：查询条件； fragment：锚点结果为元组，可用参数或索引取值。代码：运行结果： 2.urlunparse() url合并 urlunparse([scheme,netloc,path,params,query,frament]) 接受的参数为可迭代对象；个数必须为6个，否则报错代码：运行结果： 3.urlsplit() 和urlparse()相似，只是不再单独拆分params部分，将params合并到path中 4.urlunsplit() 和urlunparse()相似，唯一区别传入参数为5个 5.urljoin() base_url作为第一个参数，新连接作为第二个参数，该方法会分析base_url中的scheme，netloc，path三部分内容

Remove integers from list

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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

Implement SICP evaluator using Racket

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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)

Why does '(a . b . c) evaluate to (b a c) in PLT-Scheme 372?

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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 .

Error running scmutils by M-x mechanics in emacs

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问题 I have installed scmutils (for the book SICM) from the tar ball under /usr/local . I then put this in my .emacs : (defun mechanics () (interactive) (run-scheme "/usr/local/scmutils/mit-scheme/bin/scheme --library /usr/local/scmutils/mit-scheme/lib" )) which is mostly instruction from http://redsymbol.net/articles/using-gnu-emacs-with-scmutils/. But I get an error: /usr/local/scmutils/mit-scheme/bin/scheme: 1: /usr/local/scmutils/mit-scheme/bin/scheme: Syntax error: "(" unexpected Process

remove duplicate of a list in O(nlogn)

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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

Scheme - sum of list

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问题 I'm trying to implement a function which calc sum of list , its name is sum - (define (sum elemList) (if (null? elemList) (+ (car elemList) (sum (cdr elemList))) 0 ) ) The above implementation gives wrong result , for example - > (sum (list 1 2 3 4 )) 0 What I did wrong here ? 回答1: I think you swapped the then and the else part of the if : (define (sum elemList) (if (null? elemList) 0 (+ (car elemList) (sum (cdr elemList))) ) ) In the original function, for every non-empty list, 0 is returned

Scheme code cond error in Wescheme

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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

Methods and properties in scheme: is OOP possible in Scheme?

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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