我需要在一維封閉場上獲得細胞的鄰居
例如：
鄰居 [1,2,3,4,5,-1]（6 個元素）
必須回傳 [[-1,1,2],[1,2,3],[2,3,4],[3,4,5],[4,5,-1],[5,-1 ,1]]（6 個元素）
我的鄰居守則
neighbours :: [a] -> [[a]]
neighbours l = concat [[[last l, head l, last $ take 2 l]], neighbours' l, [[l !! (length l - 2), last l, head l]]] where
neighbours' :: [a] -> [[a]]
neighbours' (a:b:c:xs) = [a, b, c]:neighbours (b:c:xs)
neighbours' _ = []
main = print $ neighbours [1, 2, 3, 4]
回傳 [[4,1,2],[1,2,3],[4,2,3],[2,3,4],[4,3,4],[3,4,3], [3,4,2],[3,4,1]]（8 個元素），但預期為 [[4,1,2],[1,2,3],[2,3,4],[3 ,4,1]]（4 個元素）
如果我評論鄰居的 l 它會按預期回傳 [[4,1,2],[3,4,1]] （2 個元素）
如果你只留下鄰居的 l 它按預期回傳 [[1,2,3],[2,3,4]] （2 個元素）
2 2=4，但在這種情況下由于某種原因它是 8
為什么會發生？
附言
鄰居的創建串列中間
鄰居的 [1,2,3,4,5,-1] == [[1,2,3],[2,3,4],[3,4,5],[4,5,-1 ]]
[last l, head l, last $ take 2 l] 創建串列頭 [-1,1,2]
[我！！(length l - 2), last l, head l] 創建串列的最后一個元素 [5,-1,1]
uj5u.com熱心網友回復：
您的代碼有點難以掌握，因為您的兩個函式neighbour和neighbour'是相互遞回的，這有點不尋常。
您的代碼中的關鍵行是：
neighbours' (a:b:c:xs) = [a, b, c] : neighbours (b:c:xs)
如果我們假設這不是故意的，而您只是想寫：
neighbours' (a:b:c:xs) = [a, b, c] : neighbours' (b:c:xs)
----------------------------------------------- ---------
然后代碼按您的預期作業。
請注意，具有長（超過 80 個字符）的代碼行會使除錯變得非常困難。
建議代碼：
{-# LANGUAGE ScopedTypeVariables #-}
{-# LANGUAGE ExplicitForAll #-}
import qualified Data.List as L
neighbours :: [a] -> [[a]]
neighbours l = concat [
[[last l, head l, last $ take 2 l]],
neighbours' l,
[[l !! (length l - 2), last l, head l]]
]
where
neighbours' :: [a] -> [[a]]
neighbours' (a:b:c:xs) = [a, b, c] : neighbours' (b:c:xs)
neighbours' _ = []
-- neighbour is British English, neighbor is US English
neighbors :: [a] -> [[a]]
neighbors xs =
take count $ drop (count-1) allTriplets -- section of infinite list
where
count = length xs
allTriplets = map (take 3) (L.tails (cycle xs)) -- raw material
main :: IO ()
main = do
print $ "res1 = " (show $ neighbours [1, 2, 3, 4])
print $ "res2 = " (show $ neighbors [1, 2, 3, 4])
程式輸出：
"res1 = [[4,1,2],[1,2,3],[2,3,4],[3,4,1]]"
"res2 = [[4,1,2],[1,2,3],[2,3,4],[3,4,1]]"
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標籤：哈斯克尔