Haskell返回有多少输入大于其平均值 [英] Haskell returns how many inputs are larger than their average value

问题描述

我对haskell很陌生，编写了一个简单的代码来返回多少输入大于它们的平均值。我得到错误：
$ b

错误文件：.\AverageThree.hs：5 - 在应用程序中输入错误
*表达式：xyz
术语：x
类型：Int
* 不匹配：a - > b - > c

代码：

  averageThree :: Int  - > Int  - > Int  - > Float 
 averageThree x y z =（fromInteral x + fromInteral y + fromIntegral z）/ 3 
 
 howManyAverageThree :: Int  - > Int  - > Int  - > Int 
 howManyAverageThree x y z = length> averageThree 
  

任何人都可以帮助我？

首先，你不是应用函数 length 或 averageThree - 因此也不会将您的参数用于 howManyAverageThree 。

其次，长度的类型是 [a ] - > INT 。由于您在这里没有列表，您必须使用不同的函数，或者创建一个列表。

如果我正确理解了您所需的算法，那么您要需要做一些事情：

1. 应用 x y 和 z 至 averageThree 。


2. 使用 过滤器功能，将此计算的平均值与每个传入的参数进行比较;这将产生一个列表。


3. 查找结果列表的长度。

代码我冲了出来做到这一点：

  howManyAverageThree :: Int  - > Int  - > Int  - > Int 
 howManyAverageThree xyz = length $ filter（> avg）the_three 
其中avg = averageThree xyz 
 the_three = [fromIntegral x，fromIntegral y，fromInteral z] 
  
 $ b 
这充分利用了一些简洁的功能： 
 
 
 
 ，有时称为部分功能应用程序。这就是我用的（> avg）;通常，中缀函数> 接受两个相同类型的参数，并返回一个Bool  - 通过包装在圆括号中并提供一个表达式，我已经部分应用它，它允许它用作过滤器函数 
 
  其中关键字。我使用它来清理它，并使其更具可读性。
 
 过滤器函数，我在上面提到过。
 
 
 函数应用程序使用 $  。此运算符只是将函数应用程序从左关联变为右关联。
 
 
 
I'm very new to haskell, writing a simple code that returns how many inputs are larger than their average value. I got error:

  
ERROR file:.\AverageThree.hs:5 - Type error in application
  * Expression     : x y z
   Term           : x
   Type           : Int
  * Does not match : a -> b -> c
  
  
Code:
averageThree :: Int -> Int -> Int -> Float
averageThree x y z = (fromIntegral x+ fromIntegral y+ fromIntegral z)/3

howManyAverageThree ::Int -> Int -> Int -> Int
howManyAverageThree x y z  = length > averageThree

Anyone help me?
解决方案
The trouble you're having comes from a few places.

First, you aren't applying either function, length or averageThree - and hence also not using your arguments to howManyAverageThree.

Second, the type of length is [a] -> Int.  As you don't have a list here, you either have to use a different function, or make a list.

If I understand your desired algorithm correctly, you are going to need to do a few things:

Apply x y and z to averageThree.
Use the filter function, comparing this computed average with each passed in parameter; this will result in a list.
Find the length of the resulting list.
The code I dashed off to do this follows:
howManyAverageThree ::Int -> Int -> Int -> Int
howManyAverageThree x y z = length $ filter (> avg) the_three
    where avg = averageThree x y z
          the_three = [fromIntegral x,fromIntegral y,fromIntegral z]
This takes advantage of a couple of neat features:

Currying, sometimes called "partial function application". That's what I was using with (> avg); normally, the infix function > takes two parameters of the same type, and returns a Bool - by wrapping in parenthesis and providing an expression on one side, I have partially applied it, which allows it to be used as a filter function
The where keyword. I used this to clean it all up a little and make it more readable.
The filter function, which I mentioned above.
Function application using $.  This operator just changes the function application from left-associative to right-associative.


                        
这篇关于Haskell返回有多少输入大于其平均值的文章就介绍到这了，希望我们推荐的答案对大家有所帮助，也希望大家多多支持IT屋！