Question

我正在编写一些代码以在haskell中使用任意基数。它们将存储为代表数字的整数列表。

我几乎设法使其工作，但是我遇到了将元组列表[[a_1，b_1），...，（a_n，b_n）]转换为单个列表的问题，该列表定义如下：

对于所有i，L（a_i）= b_i。如果没有i使得a_i = k，则a（k）= 0

换句话说，这是数组中值的（位置，值）对的列表。如果某个位置没有相应的值，则应将其设置为零。

我已经读过（https://wiki.haskell.org/How_to_work_on_lists），但我认为这些方法都不适合该任务。

baseN :: Integer -> Integer -> [Integer]
baseN n b = convert_digits (baseN_digits n b)

chunk :: (Integer, Integer) -> [Integer]
chunk (e,m) = m : (take (fromIntegral e) (repeat 0))

-- This is broken because the exponents don't count for each other's zeroes
convert_digits :: [(Integer,Integer)] -> [Integer]
convert_digits ((e,m):rest) = m : (take (fromIntegral (e)) (repeat 0)) 
convert_digits []       = []

-- Converts n to base b array form, where a tuple represents (exponent,digit).
-- This works, except it ignores digits which are zero. thus, I converted it to return (exponent, digit) pairs.
baseN_digits :: Integer -> Integer -> [(Integer,Integer)]

baseN_digits n b | n <= 0 = [] -- we're done.
          | b <= 0 = [] -- garbage input. 
          | True   = (e,m) : (baseN_digits (n-((b^e)*m)) b)
                     where e = (greedy n b 0)      -- Exponent of highest digit
                           m = (get_coef n b e 1)  -- the highest digit

-- Returns the exponent of the highest digit.
greedy :: Integer -> Integer -> Integer -> Integer
greedy n b e | n-(b^e) <  0  = (e-1) -- We have overshot so decrement.
             | n-(b^e) == 0  = e     -- We nailed it. No need to decrement. 
             | n-(b^e) >  0  = (greedy n b (e+1)) -- Not there yet.

-- Finds the multiplicity of the highest digit
get_coef :: Integer -> Integer -> Integer -> Integer -> Integer
get_coef n b e m | n - ((b^e)*m) < 0  = (m-1) -- We overshot so decrement.
                 | n - ((b^e)*m) == 0 = m    -- Nailed it, no need to decrement.
                 | n - ((b^e)*m) > 0  = get_coef n b e (m+1) -- Not there yet.

您可以调用“ baseN_digits n base”，它将为您提供相应的元组数组，需要将其转换为正确的输出

Answer 1

这是我一起扔的东西。

 f = snd . foldr (\(e,n) (i,l') -> ( e , (n : replicate (e-i-1) 0) ++ l')) (-1,[])
 f . map (fromIntegral *** fromIntegral) $ baseN_digits 50301020 10 = [5,0,3,0,1,0,2,0]

我想我理解您的要求（？）

编辑：也许更自然

f xs = foldr (\(e,n) fl' i -> (replicate (i-e) 0) ++ (n : fl' (e-1))) (\i -> replicate (i+1) 0) xs 0

将位置，值元组列表转换为单个列表

1 个答案: