使用map将列表中的元素替换为(x-y)*(x-y)，其中y是初始元素

I have a bit of homework to do and I am a complete newbie to Haskell. The question I am having trouble with is to write a function which when given an integer x and a list of integers apply (x-y)*(x-y) to each element in the list and output the new list, with y being each element of the input list.

我有一些家庭作业要做，我对哈斯卡尔完全是个新手。我遇到的问题是编写一个函数，当给定一个整数x和一个整数列表时，它会将(x-y)*(x-y)应用到列表中的每个元素，并输出新的列表，y是输入列表的每个元素。

I have a very rough idea I will have to use the map function but I'm unsure how to go about it.

我有一个很粗略的想法，我将不得不使用map函数，但是我不确定该怎么做。

I have been looking at examples for squaring each element in a list and kind of understand how that works, but how I would implement the (x-y)*(x-y) with y being the current element completely baffles me.

我一直在寻找将列表中的每个元素平方的例子，并大致了解它是如何工作的，但是我如何实现(x-y)*(x-y)以y为当前元素完全令我困惑。

squares :: [Int] -> [Int]squares (x:xs) = x * x : squares xssquares [] = []

the exact question I have been set is,

我设定的确切问题是，

Write a function rela which takes as arguments an integer x and a list of integers. It returns a similar list, but where each element y has been replaced by (x-y)*(x-y), e.g.

编写一个函数rela，它接受整数x和整数列表作为参数。它返回一个类似的列表，但是每个元素y都被(x-y)*(x-y)替换，例如。
Main> rela 2 [3,5,7][1,9,25]

I have managed to get it working after reading through some books, but the code I have made misses out the first element in the list. Any explanation why?

在读完一些书之后，我已经设法让它工作了，但是我所写的代码漏掉了列表中的第一个元素。任何解释为什么?

equation1 :: Int -> Int -> Intequation1 x y = (x-y)*(x-y)rela :: Int -> [Int] -> [Int]rela x [] =[]rela x (y:ys) = [ equation1 x y | y <- ys ]

4 个解决方案

#1

First of all, you should probably create a separate function that does what you want.

首先，您可能应该创建一个独立的函数来执行您想要的操作。

e.g.

如。

f x y = (x-y)*(x-y)

Now, every time you create a function in Haskell with multiple parameters, it actually "curries" the function, which means that you get a new function when you apply the first argument to it.

现在，每当你在Haskell中创建一个具有多个参数的函数时，它实际上会“凝结”这个函数，这意味着当你对它应用第一个参数时，你会得到一个新的函数。

So, you would get a new function by doing this

这样就得到了一个新的函数

g = f 5

The expression f 5 is actually a function

表达式f5实际上是一个函数

And you can apply a number to 'g' and x will always be '5'

你可以给g加上一个数字x总是5

So if we want to create a function that takes two parameters, 'x' and 'y', and applies (x-y)*(x-y) to a list where y is the current element, then all we need to do is the following:

因此，如果我们想要创建一个包含两个参数的函数，即'x'和'y'，并将(x-y)*(x-y)应用到一个以y为当前元素的列表中，那么我们需要做的就是:

f x y = (x-y)*(x-y)squareDifference x = map (f x) [1,2,3,4]

Which you can use by calling squareDifference 5 or any other number as an argument

你可以通过调用squareDifference 5或任何其他数字作为参数来使用它吗

A more general version would allow you to pass in a list as well

一个更通用的版本也可以让你传递一个列表

squareDifference x xs = map (f x) xs

Which you would call by doing squareDifference 3 [1,2,3]

用squareDifference 3 [1,2,3]

#2

do you understand lambda functions?

你理解函数吗?

map (\val -> the function) xs

is what you need.

是你需要的。

currying is even better, but not as simple.

咖喱更好，但没那么简单。

edit:

编辑:

more conceptual...

更多的概念……

map iterates down a list applying a function.

map迭代一个应用函数的列表。

map (+ 3) xs

uses the currying technique mentioned above. you could also:

使用上面提到的当前技术。你也可以:

map (\x -> x + 3) xs

to accomplish the same thing.

为了完成同样的事情。

#3

Simple example:

简单的例子:

rela :: Int -> [Int] -> [Int]rela x = map (\y -> (x-y)*(x-y))

Or might you want any perversions? -) Here you are with Applicatives:

或者你想要任何变态?-)这是您的申请表:

import Control.Applicativerela :: Int -> [Int] -> [Int]rela x = map $ (*) <$> (x-) <*> (x-)

#4

Hello I guess you mean this:

你好，我猜你的意思是:

Prelude> let rela n = map (\ x -> (x - n)^2)Prelude> rela 2  [3,5,7] [1,9,25]

#1