haskell：递归函数，返回具有一定条件的元组列表中的字符（比较） [英] haskell: recursive function that return the char in a tuple list with certain condition(compare)

问题描述

我在这里得到了一个元组列表：

  [（0.5，'！'），（1，'*'），（1.5，'＃'）] 
  

我想要做的是输入一个数字n并与列表中每个元组的第一个数字进行比较

假设n = 0.1，当它比较0.5并且发现它小于0.5时，它将返回char'！'

假设n = 0.7，这就是> 0.5并继续比较，发现它< 1，那么它将返回char'*'

，并且比较整个列表并发现d仍然比最后一个大，它将返回char'n'

我一直在用这样的条件工作很多时间，但仍然无法得到这个，这里是我的代码：

  find :: Double  - > [（Double，Char）]  - > Char 
 find d [] =''
找到d xs 
 | d <= Double（xs [0]）= xs [0] 
 | d> Double（xs [0]）= find d tail（xs）
  

请使用递归！

解决方案

元组与Haskell中的数组不同

  find :: Double  - > [（Double，Char）]  - > char 
 find d [] =''
找到d（x：xs）
 | d <= fst x = snd x 
 |否则=找到d xs 
  

I'm learning recursive function in haskell that confused with such conditon:

I got a tuple list here:

[(0.5,'!'),(1,'*'),(1.5,'#')]

What I want to do is input a number n and compare with fist number in each tuple of the list

so suppose n=0.1, when it compared 0.5 and find it is smaller than 0.5, it will return char '!'

suppose n=0.7, which is > 0.5 and keep comparing, find that it < 1, then it will return char '*'

and after compare the whole list and find d is still bigger than the last one, it will just return char 'n'

I've been work with such condition lots of time,but still cannot get to this, here is my code:

find :: Double -> [(Double,Char)] -> Char
find d [] = ' '
find d xs
    | d <= Double(xs[0]) = xs[0]
    | d > Double(xs[0]) = find d tail(xs)

please use recursion!

解决方案

tuple is different from array in Haskell

find :: Double -> [(Double,Char)] -> Char
find d [] = ' '
find d (x:xs)
    | d <= fst x = snd x
    | otherwise = find d xs

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