haskell - 使用 sortBy 对列表进行排序

Question

我正在尝试根据每个元组中的第 4 个元素对我的元组列表进行排序。第四个元素包含一个人名字符串。我想将包含相同名称的元组放在一起。排序前的元组列表示例如下:

[("A",100,"Q",3,"Todd",2.0),
 ("B",203,"R",3,"Rachel",1.66),
 ("B",273,"F",1,"Mike",2.66),
 ("A",200,"P",1,"Rachel",0.0),
 ("A",549,"D",3,"Todd",2.0),
 ("B",220,"S",3,"Todd",4.0),
 ("B",101,"M",3,"Jon",3.33),
 ("A",999,"N",3,"Rachel",1.33)]

我也希望它看起来像:

[("A",100,"Q",3,"Todd",2.0),
 ("A",549,"D",3,"Todd",2.0),
 ("B",220,"S",3,"Todd",4.0),
 ("B",203,"R",3,"Rachel",1.66),
 ("A",200,"P",1,"Rachel",0.0),
 ("A",999,"N",3,"Rachel",1.33),
 ("B",273,"F",1,"Mike",2.66),
 ("B",101,"M",3,"Jon",3.33)]

我需要让所有包含 Todd 的元组彼此相邻，以此类推。名称显示的顺序无关紧要，只要它们彼此相邻即可。

sortedList= show . sortBy byName . (map stringToTuple) . (map words) . lines

这是我调用排序依据的代码行。我知道我需要创建一个函数 byName 以某种方式确定元组是否共享一个通用名称。

任何帮助引导我进入编写 byName 方法的正确方向将不胜感激。谢谢

Answer 1

从 sortBy 类型开始:

> :t sortBy
sortBy :: (a -> a -> Ordering) -> [a] -> [a]

这意味着 byName 需要具有类型 a -> a -> Ordering 。在这种情况下， a 是一个元组，其第五个元素的类型为 String ； byName 将忽略其他字段。所以你需要定义一个函数，比如

type MyType = (String, Int, String, Int, String, Double)
byName :: MyType -> MyType -> Ordering
byName (_, _, _, _, a, _) (_, _, _, _, b, _) = ...

我将把 ... 替换为正确的表达式作为练习。

(Recall that Ordering is a type with three values, LT , EQ , and GT , where byName a b == LT if a < b , byName a b == EQ if a == b , and byName a b == GT if a > b . In your case, two tuples will compare as equal as long as they have同名。听起来您实际上并不关心 byName 是否返回 LT 或 GT 否则。)