haskell，如何使用元组创建小型数据库

发布于 2024-11-01 07:34:42 字数 760 浏览 5 评论 0原文

这是我的问题，我需要使用 2 元组（即对）列表构建一个数据库，表示有关拥有县和城市的一个国家的信息例如，c1 拥有 c2，c2 拥有 c3，c3 拥有 c4，但没有直接说明 c1（等）对 c4 的“间接”所有权。我需要编写一个函数“owns”，给定两个命名国家的字符串，返回一个布尔值，指示第一个字符串是否拥有第二个字符串（甚至是间接的，如上面的 c1 和 c4 的情况）。这是我的代码，我知道它不太正确，我对 haskell 非常陌生......因此，需要帮助......

lst = [("uk","scotland"),("scotland","aberdeen"),("china","hongkong"),("hongkong","kulong")]
owns :: String-> String -> Bool
owns a b 
    | n = lookup a (fromList lst)
        |if b==n 
            return true
        |otherwise m = lookup n (fromlist lst)
            if b==m
            return true
    | otherwise = False

我期望输出结果应该是这样的：

Main> owns "uk" "scotland"
True
Main> owns "uk" "aberdeen"
True
Main> owns "uk" "hongkong"
False

原文

here s my question, i need to construct a database representing information about one country owning county owning city, using a list of 2-tuples (i.e., pairs)
for example c1 owns c2, which owns c3, which owns c4, but where the "indirect" ownership of c4 by c1 (etc.) is not directly stated.
i need to write a function 'owns' which, given two strings naming country returns a Boolean indicating whether the first owns the second (even indirectly, as with the case of c1 and c4 above).
here s my code, and i know it s not quite right, i new very new to haskell ....therefore, need help please....

lst = [("uk","scotland"),("scotland","aberdeen"),("china","hongkong"),("hongkong","kulong")]
owns :: String-> String -> Bool
owns a b 
    | n = lookup a (fromList lst)
        |if b==n 
            return true
        |otherwise m = lookup n (fromlist lst)
            if b==m
            return true
    | otherwise = False

i m expecting the output result should be something like:

Main> owns "uk" "scotland"
True
Main> owns "uk" "aberdeen"
True
Main> owns "uk" "hongkong"
False

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软的没边 2024-11-08 07:34:44

owns parent child = parent == child || any (owns parent) [p | (p, c) <- lst, c == child]

您会想知道这是如何工作的。

首先，我们认识到这是一个递归问题。 c1 拥有 c2 拥有 c3 拥有 c4 等等。我们不知道这可能需要多少步骤。所以我们需要一个基本情况和一个递归情况。

基本情况是父==子。如果这是真的，那么总体答案就是真的。

现在，递归的情况。 any 接受一个函数和一个列表，如果列表中的任何成员使函数返回 True，则返回 True。

（我现在需要睡觉，如果需要的话稍后会再回来。）

owns parent child = parent == child || any (owns parent) [p | (p, c) <- lst, c == child]

You will want to know how this works.

First, we recognise that this is a recursion question. c1 owns c2 owns c3 owns c4 etc. We don't know how many steps this could take. So we need a base case and a recursive case.

The base case is parent == child. If this is True, then the overall answer is True.

Now, the recursive case. any takes a function and a list, and returns True if any member of the list makes the function return True.

(I need to sleep now, will come back to this later if needed.)

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