在Haskell中查找字符串中最短和最长的单词 [英] Finding the shortest and longest word of a string in Haskell

问题描述

我试图找到给定字符串中最短和最长的单词。如果输入的字符串是家庭移动电话，那么最长的单词应该是电话，最短的单词将被点击。我已经写了下列功能。我如何找到字符串的索引并从中找到最短和最长的单词？

  ---------将字符串转换为字符串列表并查找每个字的长度---- ---- 
 
 stringConvert :: String  - > [Int] 
 stringConvert x =地图（长度）（单词x）
 
 ----------查找最长的单词---------- -  
 
 findLongestWord :: String  - > Int 
 findLongestWord x = maximum（stringConvert x）
 
 ----------找到最短的单词----------- 
 
 findShortestWord :: String  - > Int 
 findShortestWord x = minimum（stringConvert x）
  

解决方案

<因为你正在寻找某个单词，所以你的函数的签名应该是

  findLongestWord :: String  - >字符串
 findShortestWord :: String  - >字符串
  

您已经实现的 stringConvert 正确的想法，但它有点问题，因为结果没有哪个单词与哪个长度相关的信息。将长度计算中的字符串拆分为单词列表可能更好，并且有一个方便的函数称为比较，它实际上不需要 stringConvert 函数。

  import Data.List（maximumBy）
 import Data .Ord（比较）
 
 findLongestWord :: String  - >字符串
 findLongestWord s = maximumBy（比较长度）（单词s）
  

maximumBy 类似于最大值，但它将第一个参数用作比较两个项目的函数。 比较是一个高阶函数，可用于转换类型为 a - >的函数。 b （其中 b 是可以比较的某种类型，即 Ord b => a - > b ，如果你已经熟悉类型约束）到两个 a 之间的比较函数（即类型为 a - > a - >订购）。

最短的单词可以以类似的方式找到。 b

I am trying to find the shortest and longest word in a given string. If the entered string is "house tap mobile telephone" then the longest word should be telephone and shortest will be tap. I have already written the following functions. How do i find the index of the strings and find the shortest and longest word out of it?

---------Converting string into a list of strings and finding length of each word--------

stringConvert :: String -> [Int]
stringConvert x = map (length) (words x)

----------Find the longest word-----------

findLongestWord :: String -> Int
findLongestWord x = maximum(stringConvert x)

----------Find the shortest word-----------

findShortestWord :: String -> Int
findShortestWord x = minimum(stringConvert x)

解决方案

Since you are looking for a certain word, the signatures of your functions should be

findLongestWord :: String -> String
findShortestWord :: String -> String

The stringConvert function you have implemented has the right idea, but it is a bit problematic because the result doesn't have the information which word is associated with which length. It might be better to separate splitting the string into a word list from the length calculation, and there is a handy function called comparing that actually removes the need for the stringConvert function altogether.

import Data.List (maximumBy)
import Data.Ord (comparing)

findLongestWord :: String -> String
findLongestWord s = maximumBy (comparing length) (words s)

maximumBy is similar to maximum but it takes as the first parameter a function that should be used to compare two items. comparing is a higher-order function that can be used to convert a function of type a -> b (where bis some type that can be compared, i.e. Ord b => a -> b if you are already familiar with type constraints) into a comparison function between two a's (i.e. a function of type a -> a -> Ordering).

The shortest word can be found in a similar manner.

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