使用 getIndex xs y
我想要 xs
中长度大于 y
的第一个子列表的索引。
输出是:
[[],[4],[4,3],[3,5,3],[3,5,5,6,1]]
aufgabe6: <<loop>>
为什么 getIndex
不起作用?
import Data.List
-- Die Sortierfunktion --
myCompare a b
| length a < length b = LT
| otherwise = GT
sortList :: [[a]] -> [[a]]
sortList x = sortBy myCompare x
-- Die Indexfunktion --
getIndex :: [[a]] -> Int -> Int
getIndex [] y = 0
getIndex (x:xs) y
| length x <= y = 1 + getIndex xs y
| otherwise = 0
where (x:xs) = sortList (x:xs)
main = do
print (sortList [[4],[3,5,3],[4,3],[3,5,5,6,1],[]])
print (getIndex [[4],[3,5,3],[4,3],[3,5,5,6,1],[]] 2)
最佳答案
让它终止
问题就在这种情况下
getIndex (x:xs) y
| length x <= y = 1 + getIndex xs y
| otherwise = 0
where (x:xs) = sortList (x:xs)
您混淆了哪个 (x:xs)
是哪个。你应该这样做
getIndex zs y
| length x <= y = 1 + getIndex xs y
| otherwise = 0
where (x:xs) = sortList zs
给予
Main> main
[[],[4],[4,3],[3,5,3],[3,5,5,6,1]]
3
*Main> getIndex [[],[2],[4,5]] 1
2
*Main> getIndex [[],[2],[4,5]] 5
3
这为您提供了排序列表中长度至少为 y
的第一个列表的编号,这实际上回答了“最多有多少个列表长度为 y
”的问题> 在原版中?”
我们如何找出其他事实?
如果你想从原始列表中获得位置,你可以使用它们的位置标记条目,使用 zip
:
*Main> zip [1..] [[4],[3,5,3],[4,3],[3,5,5,6,1],[]]
[(1,[4]),(2,[3,5,3]),(3,[4,3]),(4,[3,5,5,6,1]),(5,[])]
让我们创建一个实用函数来处理这些:
hasLength likeThis (_,xs) = likeThis (length xs)
我们可以这样使用它:
*Main> hasLength (==4) (1,[1,2,3,4])
True
*Main> filter (hasLength (>=2)) (zip [1..] ["","yo","hi there","?"])
[(2,"yo"),(3,"hi there")]
这意味着现在可以很容易地编写一个函数来为您提供长度大于 y
的第一个列表的索引:
whichIsLongerThan xss y =
case filter (hasLength (>y)) (zip [1..] xss) of
[] -> error "nothing long enough" -- change this to 0 or (length xss + 1) if you prefer
(x:xs) -> fst x
这给了我们
*Main> whichIsLongerThan [[4],[3,5,3],[4,3],[3,5,5,6,1],[]] 2
2
*Main> whichIsLongerThan [[4],[3,5,3],[4,3],[3,5,5,6,1],[]] 3
4
*Main> whichIsLongerThan [[4],[3,5,3],[4,3],[3,5,5,6,1],[]] 0
1
更短?
但我们可以做类似的技巧:
whichIsShorterThan xss y =
case filter (hasLength (<y)) (zip [1..] xss) of
[] -> error "nothing short enough" -- change this to 0 or (length xss + 1) if you prefer
(x:xs) -> fst x
所以你得到
*Main> whichIsShorterThan [[4],[3,5,3],[4,3],[3,5,5,6,1],[]] 2
1
*Main> whichIsShorterThan [[4],[3,5,3],[4,3],[3,5,5,6,1],[]] 1
5
*Main> whichIsShorterThan [[4],[3,5,3],[4,3],[3,5,5,6,1],[]] 0
*** Exception: nothing short enough
广义的
让我们把那里的共同主题拉出来:
whichLength :: (Int -> Bool) -> [[a]] -> Int
whichLength likeThis xss =
case filter (hasLength likeThis) (zip [1..] xss) of
[] -> error "nothing found" -- change this to 0 or (length xss + 1) if you prefer
(x:xs) -> fst x
所以我们可以做
*Main> whichLength (==5) [[4],[3,5,3],[4,3],[3,5,5,6,1],[]]
4
*Main> whichLength (>2) [[4],[3,5,3],[4,3],[3,5,5,6,1],[]]
2
关于list - Haskell <<循环>>,我们在Stack Overflow上找到一个类似的问题: https://stackoverflow.com/questions/16640928/