递归函数内的枚举数 [英] Numeration count inside a recursive function

问题描述

我想做类似的事情：

 >枚举[banana，potato，ice] 
 [（1，banana），（2，potato），（3，ice）] 
  

我写道：

  enumerate :: [String]  - > [（Int，String）] 
 enumerate [] = [] 
  

/管理int计数器？有没有办法做到这一点没有支持功能？

更新：我知道 Zip 函数。但为了学习的目的，我想实现我自己的zip功能。

更新2：目前代码

这是我迄今为止所做的，使用支持功能。考虑到：

1）我想实现我自己的 zip 函数;

2 ）我不想改变函数struct：

  enumerate :: [String]  - > [（Int，String）] 
 
 enumerate :: [String]  - > [（Int，String）] 
枚举[] = [] 
枚举列表= aux 1列表
 
 
 aux :: Int-> [String]  - > [（Int，String）] 
 aux _ [] = [] 
 aux i （x：xs）= [（i，x）] ++ aux（i + 1）xs 
  

有没有可能改进这个功能？由于我不想再添加最后一个函数，所以我认为支持函数是唯一的出路，对吗？

解决方案

>不要害怕写一个支持函数，其实看到它是一个机会：为什么任意的起始值1？为什么没有函数？

 > enumerateFrom 42 [banana，potato，ice] 
 [（42，banana），（43，potato），（44，ice）] 
  

一旦你有了， enumerate 很容易。

编辑：
或者给你的 aux 函数一个真实的名字，恕我直言 enumerateFrom 很好，或者将它移动到一个 where 子句，如果你已经知道了。然后听chi，使用 x：... 而不是 [x] ++ ...

I want do do something like:

>enumerate ["banana", "potato", "ice"]
[(1, "banana"), (2, "potato"), (3, "ice")]

I wrote:

enumerate :: [String] -> [(Int, String)]
enumerate [] = []

How may I control/manage the int counter ? Is there a way to do that without a support function ?

Update: I know about Zip function. But for matter of study, I want to implement my own zip function.

Update 2: Currently Code

This is what I did so far, using a support function. Considering that:

1) I want to implement my own zip function;
2) I do not want to change the function struct:

   enumerate :: [String] -> [(Int, String)]  

enumerate :: [String]->[(Int,String)]
enumerate [] = []
enumerate list = aux 1 list


aux :: Int->[String]->[(Int, String)]
aux _ [] = []
aux i (x:xs) = [(i, x)] ++ aux (i+1) xs

Is it possible to improve this function ? As I don't want to add one more last to the function, so I think support function is the only way to go, right ?

解决方案

Don't be afraid to write a support function, in fact, see it as opportunity: Why the arbitrary starting value 1? Why not have a function

>enumerateFrom 42 ["banana", "potato", "ice"]
[(42, "banana"), (43, "potato"), (44, "ice")]

Once you have that, enumerate is easy.

Edit: Either give your aux function a real name, IMHO enumerateFrom is good or move it into a where clause if you know that already. And listen to chi, use x : ... instead of [x] ++ ...

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