Haskell - 过滤最后一个元素 [英] Haskell - Filter Last Element

问题描述

我想过滤不满足属性的列表的最后一个元素。一个例子是：

I want to filter the last element of a list that does not satisfy a property. An example would be

smallerOne :: a->Bool 
smallerOne x = x < 1 

函数filterLast应给出

The function filterLast should give

filterLast smallerOne [1, -2, -3, -4, 5] 
[1, -2, -3, -4] //Filters the last element that is not smaller than 1 

这里是我的代码（我是一个初学者，非常抱歉， p>

Here is my code (I am a beginner so sorry for the bad attempt)

filterLast :: (a -> Bool) -> [a] -> [a] 
filterLast p [] = [] 
filterLast p xs 
    | p (last xs) = filterLast p (init xs) : last xs 
    | otherwise = init xs 

感谢您的帮助

Thank you for your help

推荐答案

smallerOne :: (Ord a, Num a) => a -> Bool 
smallerOne x = x < 1 

filterLast :: (a -> Bool) -> [a] -> [a]
filterLast p (x:[]) = if (p x) then x:[] else []
filterLast p (x:xs) = x : filterLast p xs

这是解决您的问题的方法。现在，还有一些解释：

This is the solution to your problem. Now, also some explanation:

smallerOne ::（Ord a，Num a）=> a - > Bool

smallerOne :: (Ord a, Num a) => a -> Bool

您必须包含类别约束 Ord 和 Num < 比较运算符来处理数字。你可以在这里阅读更多： http://learnyouahaskell.com/types-and-typeclasses#typeclasses -101

you must include the class constraint Ord and Num, because your are trying to ordonate numbers through this < comparison operator. You can read more here: http://learnyouahaskell.com/types-and-typeclasses#typeclasses-101

filterLast 通过模式匹配实现，Haskell非常聪明，并且会进入与给定列表匹配的分支，因此他将进入该分支：

filterLast is implemented through pattern matching, Haskell is pretty smart and will enter the branch that will match the given list, so he will enter this branch:

filterLast仅当有一个元素时，p（x：[]）= if（px）then x：[] else []

filterLast p (x:[]) = if (p x) then x:[] else []

，即你的最后一个元素。

你说你是初学者，我真的推荐你这个教程， http://learnyouahaskell.com/chapters ，它非常酷。

You said you are a beginner, I really recommend you this tutorial, http://learnyouahaskell.com/chapters, it is pretty cool.

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