为什么在列表清单中应用“序列”会导致计算其笛卡尔乘积？ [英] Why does application of `sequence` on List of Lists lead to computation of its Cartesian Product?

问题描述

我的问题是关于 Prelude 中的序列函数，其签名如下所示：

  sequence：Monad m => [m a]  - > m [a] 
  

我明白这个函数对 List 可能 s。例如，在 [Just 3，Just 9] 上应用 sequence 给出 Just [3， 9] 。

我注意到在 List of <上应用 code> List s给出它的Cartesian Product。有人可以帮我理解这是怎么发生的？为什么会发生这种情况？

解决方案这是可行的，因为在Haskell中使用列表作为单子使他们成为模型的不确定性。考虑：

  sequence [[1,2]，[3,4]] 
  
 
 
 
根据定义，这与以下内容相同： 
 
 
  do x<  -  [1,2] 
y<  -  [3,4] 
 return [x，y] 
  
只读它为首先选择1和2之间，然后选择3和4之间的选择。列表monad现在会累积所有可能的结果 - 因此，答案 [[1,3]，[1,4]，[2,3]，[2,4] ] 。
 
 
 
（有关更加混淆的示例，请参阅）
 
My question is about the sequence function in Prelude, the signature of which is as follows:
sequence :: Monad m => [m a] -> m [a]
I understand how this function works for List of Maybes. For example, applying sequence on [Just 3, Just 9] gives Just [3, 9]. 

I noticed that applying sequence on List of Lists gives its Cartesian Product. Can someone please help me understand how/why this happens?
解决方案
This works because using lists as monads in Haskell makes them model indeterminism. Consider:
sequence [[1,2],[3,4]]
By definition this is the same as:
do x <- [1,2]
   y <- [3,4]
   return [x,y]
Just read it as "First a choice between 1 and 2, then a choice between 3 and 4". The list monad will now accumulate all possible outcomes - hence the answer [[1,3],[1,4],[2,3],[2,4]].

(for an even more obfuscated example, see here)

                        
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