foldr和foldr1 Haskell [英] foldr and foldr1 Haskell

问题描述

考虑到这对我来说似乎是一个非常强大的资产，我正在试图深入折叠。然而，你可以帮助我：

  foldr（/）2 [1,2,3]  - （1 / （2 /（3/2））），结果0.75（其中2是基数）
 foldr1（/）[2,2,3]  - （1 /（2 /（3/2）） ），结果3.00 {其中2是基数）
  

我认为我严肃地监督褶皱。 Thx

解决方案

一般

  foldr1 op [y1，...，yn，x] 
  

基本上与

  foldr op x [y1，...，yn] 
  

只要 n> = 1 。也就是说，在你的特殊情况下

  foldr1（/）[2，2，3] 
  

会评估为 2 /（2/3）= 3 。

使用除法运算符，它有点无聊，因为

  foldr1（/ ）[y，y，x1，...，xn] 
 = y /（y /（foldr1（/）[x1，...，xn]））
 = foldr1（/）[ x1，...，xn] 
  

全部 n> = 1 。因此，在你的例子中，它应该返回结果

  foldr1（/）[3] 
 
 code> 

当然 3.0 ，它确实评估为 3.0 。

这里是一个不太简单的例子：

  foldr1（/）[2,3,5,7]  -  evals 0.47619047619047616 
  

在这里，它给出了与 2 /（3 /（5/7））相同的结果，这又是
与

  foldr（/）7 [2,3,5] 
  

我希望它能澄清评估顺序。

I am trying to dive deep in the folds, considering it seems a very powerful asset to me. However, can you help me with this:

foldr  (/) 2 [1,2,3] -- (1/(2/(3/2))), result 0,75 {where 2 is base)
foldr1 (/)   [2,2,3] -- (1/(2/(3/2))), result 3.00 {where 2 is base)

I think I am seriously overseeing an essential difference between the folds. Thx

解决方案

In general

foldr1 op [y1, ..., yn, x]

is essentially the same as

foldr op x [y1, ..., yn]

as long as n >= 1. That is, in your special case

foldr1 (/) [2, 2, 3]

will evaluate to 2/(2/3) = 3.

With the division operator, it's a bit boring, because

foldr1 (/) [y, y, x1, ... , xn]
  = y/(y/(foldr1 (/) [x1, ..., xn]))
  = foldr1 (/) [x1, ..., xn]

for all n >= 1. So, in your example, it should just return the result of

foldr1 (/) [3]

which is of course 3.0, and it does indeed evaluate to 3.0.

Here is less degenerate example:

foldr1 (/) [2, 3, 5, 7] --  evals to 0.47619047619047616

Here, it gives the same result as 2 / (3 / (5 / 7)), which is in turn the same as

foldr (/) 7 [2, 3, 5]

I hope it clarifies the order of evaluation a little bit.

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