处理复杂输入(具有嵌套值)

如果您只是想缩短代码,请查看 Stack Exchange community for code golfing

如果我们认为代码太多,可能不是因为我们需要缩短代码,而是因为我们需要缩短代码 . 实现这一目标需要经验和良好实践。我们要做的是分离出明显正确的简单概念,然后以明显正确的方式将它们结合起来。方法包括自上而下的设计(将解决方案分解为更小的部分)和自下而上的设计(从更小的部分构建到解决方案)及其混合。

一个自下而上的作品直接打动了我,那就是对一系列数字求和的任务。这在哈斯克尔的前奏曲中有一个定义,叫做 sum :: (Num a, Foldable t) => t a -> a . 在最终解决方案的某个地方,我们将使用这个。

另一种方法是简化问题。我们可能会被问题的措辞引入歧途。仔细检查后,我们可能会发现一个等效且更简单的短语。

5
1 2 3 4 5
2
-100 100

1 2 3 4 5
-100 100

现在我们有了由行返回分隔的数字列表,其中每个数字由一个空格分隔。

在这一点上,我们有一个明确的方法来分解解决方案在自上而下的方式。首先,我们简化输入。其次,我们解析数字列表。第三,我们对列表进行汇总。第四,我们打印总数。因此,这是我们解决方案的框架:

simplifyInput :: String -> [String]

parseNumberList :: String -> [Integer]

-- Note we can use `sum` from Prelude to sum the number lists.

printSums :: [Integer] -> IO ()

main :: IO ()
main = getContents >>= printSums . fmap (sum . parseNumberList) . simplifyInput

现在,这只是实施解决方案中每个明显部分的问题。

simplifyInput :: String -> [String]
simplifyInput = dropEveryOther . drop 1 . lines
  where
  dropEveryOther :: [a] -> [a]

以书面形式 simplifyInput

dropEveryOther :: [a] -> [a]
dropEveryOther [] = []
dropEveryOther (x:y:xs) = y : dropEveryOther xs

然后继续。。。

parseNumberList :: String -> [Integer]
parseNumberList = fmap read . words

printSums :: [Integer] -> IO ()
printSums = putStr . unlines . fmap show

simplifyInput :: String -> [String]
simplifyInput = dropEveryOther . drop 1 . lines
  where
  dropEveryOther :: [a] -> [a]
  dropEveryOther [] = []
  dropEveryOther (_:y:xs) = y : dropEveryOther xs

parseNumberList :: String -> [Integer]
parseNumberList = fmap read . words

printSums :: [Integer] -> IO ()
printSums = putStr . unlines . fmap show

main :: IO ()
main = getContents >>= printSums . fmap (sum . parseNumberList) . simplifyInput

Alec在其中一条评论中发布了我的原始代码的超级压缩版本。我决定发布一个小故障,以防有人迷路,不知道里面发生了什么:)

以下代码段前面需要有有效的导入:

import Control.Monad
import Control.Applicative

所以我们从亚历克的版本开始:

main = readLn >>= flip replicateM_ (getLine >> sum . map read . words <$> getLine >>= print)

flip 函数以删除一组括号:

main = readLn >>= (`replicateM_` (getLine >> (print =<< sum . map read . words <$> getLine)))

replicateM_ 复制项_ ,我们可以用λ替换is:

main = readLn >>= \n -> replicateM_ n (getLine >> (print =<< sum . map read . words <$> getLine))

printBatchResult = print =<< sum . map read . words <$> getLine
main = readLn >>= \n -> replicateM_ n (getLine >> printBatchResult)

我们可以翻转 print =<<

printBatchResult = sum . map read . words <$> getLine >>= print
main = readLn >>= \n -> replicateM_ n (getLine >> printBatchResult)

printBatchResult = sum . map read . words <$> getLine >>= print
handleBatch = getLine >> printBatchResult
main = readLn >>= \n -> replicateM_ n handleBatch

再说一次:

sumLine = sum . map read . words
printBatchResult = sumLine <$> getLine >>= print
handleBatch = getLine >> printBatchResult
main = readLn >>= \n -> replicateM_ n handleBatch

sumLine = sum . map read . words
handleNumbersLine = sumLine <$> getLine
printBatchResult = handleNumbersLine >>= print
handleBatch = getLine >> printBatchResult
main = readLn >>= (\n -> replicateM_ n handleBatch)

最后一次:

sumLine = sum . map read . words
handleNumbersLine = sumLine <$> getLine
printBatchResult = handleNumbersLine >>= print
handleBatch = getLine >> printBatchResult
handleAllBatches n = replicateM_ n handleBatch
main = readLn >>= handleAllBatches

我们可以替换 <$> 具有 fmap :

sumLine = sum . map read . words
handleNumbersLine = fmap sumLine getLine
printBatchResult = handleNumbersLine >>= print
handleBatch = getLine >> printBatchResult
handleAllBatches n = replicateM_ n handleBatch
main = readLn >>= handleAllBatches

sumLine line = (sum . map read . words) line
handleNumbersLine = fmap sumLine getLine
printBatchResult = handleNumbersLine >>= \sum -> print sum
handleBatch = getLine >> printBatchResult
handleAllBatches n = replicateM_ n handleBatch
main = readLn >>= \numberOfBatches -> handleAllBatches numberOfBatches

sumLine :: String -> Int
sumLine line = (sum . map read . words) line

handleNumbersLine :: IO Int
handleNumbersLine = fmap sumLine getLine

printBatchResult :: IO ()
printBatchResult = handleNumbersLine >>= \sum -> print sum

handleBatch :: IO ()
handleBatch = getLine >> printBatchResult

handleAllBatches :: Int -> IO ()
handleAllBatches n = replicateM_ n handleBatch

main = readLn >>= \numberOfBatches -> handleAllBatches numberOfBatches

一些最终意见:

• >>= bind 函数from monad将一个monad转换为另一个(或相同的)并转换其值。在里面 main 它需要的功能 IO Int ,转换lambda和返回 IO ()
• >> -(用于 handleBatch )忽略左参数(一行中有多少数字(可以说)是不必要的),只返回右参数-这是一个处理带有数字的行的函数。