如何在Haskell中汇总扩展列表的中间元素? [英] How can I sum the middle elements of an expanding list in Haskell?

问题描述

到目前为止，我知道如何从末端扩展列表，但是由于第一个条件(使单例增加一倍)，它们最终变得倍增.代码像这样有意义吗?

So far I know how to expand a list from its ends, but they end up getting doubled because of the first condition, which is to double a singleton. Would it make sense for the code to be like this:

sumExpand :: [Integer] -> [Integer]

sumExpand l = expand l []
  where
    expand [] a     = a
    expand (x:[]) a = x: expand [] (x:a)
    expand (x:xs) a = expand (x:a) (expand xs a)

对我来说，要处理其输出:

And for me to work on its output:

[1,1,2,2,3,3] from [1,2,3]
instead of [1,3,5,3]

后一个是我的愿望吗?这是我找到两个元素列表的临时解决方案的方法:

The latter being my desire? Here's how I got to a temporary solution for a list of two elements:

expand (x:xs) a = x: tail (expand (map (x+) xs) (last xs:a))

---

输出:

*Main> sumExpand [1,2]
[1,3,2]
*Main> sumExpand [1,2,3]
[1,7,4,3]

---

基本上，我希望算法像这样工作: [a，b，c] =>[a，a + b，b + c，c]

推荐答案

基本上，所有您想要计算输入列表和其转换版本之间的按组件和:

Basically, all you want to compute component-wise sums between your input list and a shifted version of it:

a   b   c   d   e
    a   b   c   d   e
---------------------------
a  a+b b+c c+d d+e  e

用0( 0:x 和 x ++ [0] )填充每个空插槽，而您只需要 zipWith

Fill each empty slot with a 0 (0:x and x++[0]), and you just need zipWith

> (\x -> zipWith (+) (0:x) (x++[0])) [1,2,3]
[1,3,5,3]

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