Question

我正在尝试执行以下操作：
我正在编写的函数列出了0-6中范围内的[(x,y),(i,j)...]元组列表我想返回[Int]，其中每个元素都是其中有多少元素列表中可以看到相应的数字例如[(0,1), (1,2), (2,3)]会返回[1, 2, 2, 1, 0, 0, 0]。

换句话说，一个0，两个1，两个2的1三，没有4s，5s或6s

countNumbers :: [(Int, Int)] -> [Int]  
countNumbers [] = [0, 0, 0, 0, 0, 0, 0]

但我不知道如何解决这个问题，因为我对Haskell有点新鲜编辑 - 我找到了解决方案 - 请告诉我是否有更简洁的编码方式！

type ProbabilityMatrix = (Int,Int,Int,Int,Int,Int,Int)
-- converts list of tuples into list of ints as suggested
tupleToList :: [(Int, Int)] -> [Int]  
tupleToList ((a,b):xs) = a : b : tupleToList xs  
tupleToList _          = []    

tupleToList2 :: [Int] -> ProbabilityMatrix -> ProbabilityMatrix  
tupleToList2 [] list = list  
tupleToList2 (x : xs) (zero, one, two, three, four, five, six)  
  | x == 0 =  tupleToList2 xs (zero + 1, one, two, three, four, five, six)  
  | x == 1 = tupleToList2 xs (zero, one + 1, two, three, four, five, six)  
  | x == 2 = tupleToList2 xs (zero, one, two + 1, three, four, five, six)  
  | x == 3 = tupleToList2 xs (zero, one, two, three + 1, four, five, six)  
  | x == 4 = tupleToList2 xs (zero, one, two, three, four + 1, five, six)  
  | x == 5 = tupleToList2 xs (zero, one, two, three, four, five + 1, six)  
  | x == 6 = tupleToList2 xs (zero, one, two, three, four, five, six + 1)  
  | otherwise = tupleToList2 xs (zero + 1, one, two, three, four, five, six)

Answer 1

如何为每个元组创建Int的结果列表，然后使用Sum函数将它们合并在一起。您至少需要两个带签名的功能：

tupleToList :: (Int, Int) -> [Int]

和

sumLists :: [Int] -> [Int] -> [Int]

第一个将检测元组中的两个项目，并为其生成相应的列表，例如tupleToList (4, 2) -> [0,0,1,0,1,0,0]。

第二个函数将通过在相同索引处对2个计数器求和来合并两个整数列表，例如： sumLists [0,1,1,0,1] [1,1,0,0,0] -> [1,2,1,0,1]。您可以递归执行此操作，直到最终得到1个问题的答案列表。

您为元组列表中的每个元素执行tupleToList（可能使用map），然后通过迭代执行sumLists 2个列表来合并结果列表（可能与foldl）

这种方法相当幼稚，可能会为更大的输入而缓慢运行。

Answer 2

您可以尝试此解决方案：

-- flattens a list of tuples into a list
flatten :: (Num a) => [(a, a)] -> [a]
flatten xs = concat [[a,b] | (a, b) <- xs]

-- updates list element at a position
update_list :: (Num a) => Int -> [a] -> [a]
update_list n xs = take n xs ++ [(xs !! n) + 1] ++ drop (n + 1) xs

-- performs count of numbers
count_numbers :: [(Int, Int)] -> [Int]
count_numbers xs = go (flatten xs) acc
    where go [] acc = acc
          go (x:xs) acc = go xs (update_list x acc)
          acc = replicate 7 0

首先展开列表：

*Main> flatten [(0,1), (1,2), (2,3)]
[0,1,1,2,2,3]

然后更新某个位置的列表：

*Main> update_list 1 (replicate 7 0)
[0,1,0,0,0,0,0]

与您的函数tupleToList2类似地执行列表元素的计数，其中存储和更新数字计数的累加器：

*Main> count_numbers [(0,1), (1,2), (2,3)]
[1,2,2,1,0,0,0]
*Main> count_numbers [(0,1), (1,2), (2,4)]
[1,2,2,0,1,0,0]
*Main> count_numbers [(0,0), (1,0), (3,6)]
[3,1,0,1,0,0,1]

计算列表中的整数出现次数

2 个答案: