使用foldr实现inits [英] Implementation of inits using foldr

问题描述

我必须通过使用map的foldr实现inits。我得到了大部分，但是我的结果列表中缺少空列表元素。

  inits :: [a] - > [[a]] 
 inits = foldr（\ xy  - > [x]：（map（x :) y））[] 
  

当调用此结果时：

*蓝图< inits [1,2,3]

[[1]，[1,2]，[1,2,3]]

现在我有点被卡住了，如果有人能指出我的错误的大方向，我会很高兴。

预先感谢

解决：

  inits :: [a]  - > [[a]] 
 inits = foldr（\ xy  - > []：（map（x :) y））[[]] 
  

解决方案使用 foldr fz 来编写一些东西，你需要考虑两件事情：

1. 基本情况 z ：应该 inits如果你有一个列表


2. 递归步骤 f > xs == x：xs'，你如何从 x 构造 inits xs $ b

   ---

   
   $ b $ b

   通过纸上的一些小例子可能会有所帮助。例如
   

   
   
   • 计算 inits [1] 递归：您有 x == 1 和 y == inits [] == [[]] 并且需要到达 [ []，[1]] 。
   
   
   • 递归地计算 inits [1,2] x == 1 和 y == inits [2] == [[]，[2]] 并且需要转到 [[]，[1]，[1,2]] 。
   
   
   
   

   I have to implement inits via foldr using map. I got most of it, however I'm missing the empty list element in my result list.

   inits :: [a] -> [[a]]
   inits = foldr ( \ x y -> [x] : (map (x:) y) ) []
   

   When called this results in:

   *Blueprint< inits [1,2,3]

   [[1],[1,2],[1,2,3]]

   I am a bit stuck now and would be glad if someone could point me in the general direction of my error.

   Thanks in advance

   Solved:

   inits :: [a] -> [[a]]
   inits = foldr ( \ x y -> [] : (map (x:) y) ) [[]]
   

   解决方案

   To write something with foldr f z you need to think about two things:

   1. the base case z: what should inits [] be?
   2. the recursive step f: if you have a list xs == x:xs', how can you construct inits xs from x and y == inits xs'?

   ---

   Working through some small examples on paper might help. e.g.

   • compute inits [1] recursively: you have x == 1 and y == inits [] == [[]] and need to get to [[], [1]].
   • compute inits [1, 2] recursively: you have x == 1 and y == inits [2] == [[], [2]] and need to get to [[], [1], [1, 2]].

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