Question

我有一个类似：[[0,0,0,1,0],[1,0,0,1,0],[0,0,1,1,0],[0,0,0,0,0],[1,0,0,0,1]]的列表，我希望将此列表的头部[0,0,0,1,0]替换为[2,2,2,1,2]，以获取[[2,2,2,1,2],[1,0,0,1,0],[0,0,1,1,0],[0,0,0,0,0],[1,0,0,0,1]]。我怎么能这样做？

编辑：

我有此功能，返回[[0,0,0,1,0],[1,0,0,1,0],[0,0,1,1,0],[0,0,0,0,0],[1,0,0,0,1]]，我希望它返回[[2,2,2,1,2],[1,0,0,1,0],[0,0,1,1,0],[0,0,0,0,0],[1,0,0,0,1]]。

firstfunc :: (RandomGen g) => g -> Int -> Float -> [[Int]]
firstfunc rnd n p = makGrid $ map (\t -> if t <= p then 1 else 0) $ take (n*n) (randoms rnd)
  where makGrid rnd = unfoldr nextRow (rnd, n)
        nextRow (_, 0) = Nothing
        nextRow (es, i) = let (rnd, rest) = splitAt n es in Just (rnd, (rest, i-1))

Answer 1

change 0 = 2
change x = x

replace (x:rest) = map change x:rest

Answer 2

如果你想要一种改变列表头部的一般方法，你可以创建一个函数：

transformHead :: (a -> a) -> [a] -> [a]
transformHead _ [] = []
transformHead f (x:xs) = (f x):xs

然后你可以用它来转换firstFunc例如

的结果

transformHead (map (\x -> 2 - x)) $ firstfunc gen c p

只更换haskell列表的头部？

2 个答案: