Question

transpose' :: [[a]] -> [[a]]
transpose' [[]]    = []
transpose' [[], _] = []
transpose' rows    = (map head' rows) : transpose' (map tail' rows)
  where
    head' (x:_) = x
    tail' (_:xs) = xs

matMult3 :: (Num a) => [[a]] -> [[a]] -> [[a]]
matMult3 a b = [ [ sum $ zipWith (*) ar bc | bc <- (transpose' b) ] | ar <- a ]

任何人都有任何想法，为什么这可能是一个非详尽的？我觉得有些东西不见了，但我太新了，不能让自己真正知道，我自我教导这可能就是为什么，任何建议都会受到赞赏。

Answer 1

如果您将空列表作为参数提供，则head'和tail'都不会定义会发生什么。您需要定义以下内容：

head' [] = ....
tail' [] = ....

Answer 2

我在Haskell中的表达不是很流畅，但是transpose'的以下版本适用于非粗糙矩阵：

transpose' :: [[a]] -> [[a]]
transpose' ([x]: xs) = [x: map singleTail' xs] where
    singleTail' [x] = x
transpose' rows = (map head rows) : transpose' (map tail rows)

，例如，

>>> transpose' [[1], [2], [3]]
[[1,2,3]]

>>> transpose' [[1,2,3]]
[[1], [2], [3]]

>>> transpose' [[1, 2], [2, 3], [3, 4]]
[[1,2,3],[2,3,4]]

但是参差不齐的矩阵失败了：

>>> transpose' [[1, 2], [2, 3], [3, 4, 5]]
:3:5-23: Non-exhaustive patterns in function singleTail'

我职能部门的非详尽模式＆＃39;

2 个答案: