我如何总结Haskell中列表中的值？ [英] How can I sum up the values in lists within lists in Haskell?

问题描述

当我试图学习Haskell只是为了好玩，我没有其他的可能，只能问你一个初学者的愚蠢问题^^。
我偶然发现如何找到列表中的列表总和，例如[[1,3]，[4,7]，[2,5]] = [4,11,7]，到目前为止我还没有找到解决方案。有人知道Haskell是如何工作的吗？

解决方案

考虑函数类型，它可能会有所帮助。

您有 [[Int]] ，并且您想要一个函数 [[Int] ] - > [Int] 获得 [Int]

你需要的函数是 map sum :: [[Int]] - >以下代码运行在 ghci

$中b
$ b

首先您列出了整数列表。注意在这个例子中，你需要用 [[Int]] 来引导 ghci 来获得你想要的类型，而不是令人困惑的通用 [[1,3]，[4,7]，[2,5]] :: Num t => [[t]]

  Prelude>让xs = [[1,3]，[4,7]，[2,5]] :: [[Int]] 
  

接下来，您可能已经知道 sum ，让我们为 [Int] 而不是通用的 sum ::（Foldable t，Num a）=> t a - >一个，这将使类型更容易阅读

  Prelude> let sumInts = sum :: [Int]  - >诠释
  

接下来，让我们测试 sumInts 某些元素 xs = [[1,3]，[4,7]，[2,5]]

 前奏> ：t sumInts [1,3] 
 sumInts [1,3] :: Int 
 Prelude> sumInts [1,3] 
 4 
  

现在你可以做 sum 列表中的元素，可以将其用于整个列表中，您可以使用 map

 前奏> ：t map 
 map ::（a  - > b） - > [a]  - > [b] 
  

让我们看看您是否传递 sumInts 到地图你会得到什么类型？

  Prelude> ：t map sumInts 
 map sumInts :: [[Int]]  - > [Int] 
  

这应该适用于 xs :: [[Int]] 但首先检查类型以确定

  Prelude> ：t map sumInts xs 
 map sumInts xs :: [Int] 
  

现在做计算

  Prelude> map sumInts xs 
 [4,11,7] 
  

由于 sumInts = sum ，这也意味着 sum 也可以工作

 前奏> ：t map sum xs 
 map sum xs :: [Int] 
 Prelude> map sum xs 
 [4,11,7] 
  

注1： map sum real type是 map sum ::（Foldable t，Num b）=> [t b] - > [b] ）在上一个例子中，它是通过xs的类型 [[Int]] 来推断的 [[Int ]] - > [Int]

As I try to learn Haskell just for fun and the experience I have no other possibility than to ask you the "dumb" questions of a total beginner ^^. I stumbled across the problem how it might be possible to find the sums of lists whithin lists e.g. [[1,3],[4,7],[2,5]] = [4,11,7] and I have found no solution so far. Does anybody have a clue how this might work in Haskell?

解决方案

Think about function types, it may help.

You have [[Int]] and you want a function [[Int]] -> [Int] to get [Int]

The function you want is map sum :: [[Int]] -> [Int]

The following codes was run in ghci

First you have list of list of ints. Note that in this example you need to guide ghci with [[Int]] to get the type you want instead of the confusing generic [[1,3],[4,7],[2,5]] :: Num t => [[t]]

Prelude> let xs = [[1,3],[4,7],[2,5]] :: [[Int]]

Next, you might already know sum, let make one that specific for [Int] instead of generic sum :: (Foldable t, Num a) => t a -> a, this will make the type easier to read

Prelude> let sumInts = sum :: [Int] -> Int

Next, let test the sumInts on some element of xs = [[1,3],[4,7],[2,5]]

Prelude> :t sumInts [1,3]
sumInts [1,3] :: Int
Prelude> sumInts [1,3]
4

Now you can do sum on an element of a list, to do that to entire list you can use map

Prelude> :t map
map :: (a -> b) -> [a] -> [b]

Let see if you pass sumInts to a map what type you will get?

Prelude> :t map sumInts 
map sumInts :: [[Int]] -> [Int]

This should work with xs :: [[Int]] but let check type first to make sure

Prelude> :t map sumInts xs
map sumInts xs :: [Int]

Now do the computation

Prelude> map sumInts xs
[4,11,7]

Since sumInts = sum , this also mean sum would work too

Prelude> :t map sum xs
map sum xs :: [Int]
Prelude> map sum xs
[4,11,7]

Note 1: map sum real type is map sum :: (Foldable t, Num b) => [t b] -> [b]) in the last example it was inferred by type [[Int]] of xs to [[Int]] -> [Int]

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