搜索方案中的嵌套列表以找到一个数字 [英] search through nested list in scheme to find a number

问题描述

如何搜索嵌套列表以找到某个数字?

How would I search through the nested list to find a certain number?

例如列表为:

((1 2) (2 3) (3 4) (3 5) (4 5))

我正在寻找 1.

预期输出:

(1 2)

因为 1 在子列表中 (1 2).

since 1 is in the sub list (1 2).

推荐答案

首先创建函数来扁平化列表.像这样:

First of all create function to flatten list. Something like this:

> (flatten '((8) 4 ((7 4) 5) ((())) (((6))) 7 2 ()))
(8 4 7 4 5 6 7 2)

然后在普通列表中搜索您的号码.

And then search your number in the ordinary list.

这个问题看起来像作业，所以尝试自己开发这个功能，如果你不能做到 - 在这里发布你的代码，我会尽力帮助你.

This quesion looks like the homework so try to develop this function on your own and if you can not do it - post your code here and I'll try to help you.

更新

好的.据我了解，我们需要创建一个函数来获取对列表并返回另一个对列表，其中对的第一个元素等于某个数字.

Ok. As I understand we need to create function which get the list of pairs and return another list of pairs, where first element of pair equal some number.

例如:

(define data '((1 2)(2 3)(3 4)(3 5)(4 5)))
(solution data 3)
-> '((3 4) (3 5))

(define data '((1 2)(2 3)(3 4)(3 5)(4 5)))
(solution data 1)
-> '((1 2))

换句话说，我们需要根据某些条件过滤我们的对列表.在 Scheme 中有一个 过滤列表的功能.它需要一个列表来过滤和函数来决定 - 在结果列表中包含或不包含列表元素.

In the other words we need to filter our list of pairs by some condition. In the Scheme there is a function to filter list. It takes a list to filter and function to decide - to include or not the element of list in the result list.

所以我们需要创建这样的函数:

So we need to create such function:

(define (check-pair num p)
  (cond 
    [(= (first p) num) #t]
    [else #f]))

此函数获取一对(列表元素)，编号并决定 - 将此对包含或不包含到结果列表中.这个函数有2个参数，但是filter函数需要只有一个参数的函数，所以我们这样重写我们的函数:

This function get a pair (element of list), number and decide - incude or not this pair to result list. This function have 2 parameters, but the filter function require the function with only one parameter, so we rewrite our function such way:

(define (check-pair num)
  (lambda (p)
    (cond 
      [(= (first p) num) #t]
      [else #f])))

我创建了产生另一个函数的函数.这是柯里化.

I have created function wich produce another function. It is currying.

所以，我们必须创建我们的解决方案:

So, we have all to create our solution:

(define (check-pair num)
  (lambda (p)
    (cond 
      [(= (first p) num) #t]
      [else #f])))

(define (solution list num)
  (local 
    ((define check-pair-by-num (check-pair num)))
    (filter check-pair-by-num list)))

(define data '((1 2)(2 3)(3 4)(3 5)(4 5)))

(solution data 1)

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