如何在Haskell中返回名称列表? [英] How to return a list of names in Haskell?

问题描述

我正在编写一个简单的程序，处理英国各地的降雨.每个地方都有一个名称，以北和东为单位的位置以及降雨数字列表.

I am writing a simple program that deals with rainfall for places in the UK. Each place has a name, location in degrees north and east, and list of rainfall figures.

如何返回所有地点名称的列表，例如:[伦敦，加的夫，圣赫利尔].

How do I return a list of the names of all places, example: [London, Cardiff, St Helier].

错误:1.无法匹配类型2.列表理解

Error: 1. Couldn't match type 2. list comprehension

import Data.Char
import Data.List

type Place = (String, Int, Int, [Int])

testData :: [Place]
testData = [("London", 51.5, -0.1, [0, 0, 5, 8, 8, 0, 0]),
            ("Cardiff", 51.5, -3.2, [12, 8, 15, 0, 0, 0, 2]),
            ("St Helier", 49.2, -2.1, [0, 0, 0, 0, 6, 10, 0])]

listNames :: Place -> [String]
listNames details = [name | (name,north,east,[figure]) <- details]

推荐答案

您当前的解决方案存在几个问题:

There are several problems with your current solution:

type Place =(String，Int，Int，[Int])，但("London"，51.5，-0.1，[0，0，5，8，8，0，0])这里的问题是您已指定元组的两个中间字段为 Int s，但是您传入了 51.5 和 -0.1 ，它们是小数值.我建议将 Place 更改为: type Place =(String，Float，Float，[Int])(您也可以考虑使用记录).

type Place = (String, Int, Int, [Int]) but ("London", 51.5, -0.1, [0, 0, 5, 8, 8, 0, 0]) The problem here is that you have specified the two middle fields of the tuple to be Ints, but you pass in 51.5 and -0.1, which are fractional values. I would recommend changing Place to: type Place = (String, Float, Float, [Int]) (you could also look into using a record).

您的 listNames 函数的签名仅影响单个位置: listNames :: Place->[String] ，但实际上是要让它包含一个位置列表.您应该将其更改为 listNames :: [Place]->[String] .

Your listNames function's signature epxects only a single place: listNames :: Place -> [String], but you actually mean to have it take a list of places. You should change it to listNames :: [Place] -> [String].

您的列表理解使用一种严格的模式匹配，而您想要一个可以接受几乎所有内容的模式:模式匹配的 [figure] 部分仅匹配具有单个元素的列表，即绑定到 figure .确保您了解列表类型表示法 [a] 与列表构造函数 [1、2、3] 之间的区别.

Your list comprehension uses a restrictive pattern match, while you want one that accepts pretty much anything: the [figure] part of the pattern match only matches a list with a single element, which you are binding to figure. Make sure that you understand the difference between the list type notation [a] and the list constructor [1, 2, 3].

不仅如此，您仍然可以忽略地名以外的所有内容: [name |(名称，_，_，_)<-详细信息] .

Not only that, but you can disregard all but the place name anyway: [name | (name, _, _, _) <- details].

总的来说，您的代码将变为:

All together, your code would become:

type Place = (String, Float, Float, [Int])

testData :: [Place]
testData = [("London", 51.5, -0.1, [0, 0, 5, 8, 8, 0, 0]),
            ("Cardiff", 51.5, -3.2, [12, 8, 15, 0, 0, 0, 2]),
            ("St Helier", 49.2, -2.1, [0, 0, 0, 0, 6, 10, 0])]

listNames :: [Place] -> [String]
listNames details = [name | (name, _, _, _) <- details]

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