Question

这段代码对我来说似乎有点迟钝......有人想通过解释来帮助我吗？

uniq_c l = [ nl (tam l) i s | (s,i) <- uniq_c' l]

tam = maximum . map snd . uniq_c'

uniq_c' [] = []
uniq_c' (h:t) = let (list,rest) = span (==h) t
                    n = length list + 1
                in (h,n) : uniq_c' rest

nl tam n line = let l = length $ show n
                    l_tam = length $ show tam
                    n' = replicate (l_tam-l) " "
                in concat n' ++ show n ++ " " ++ line

Answer 1

那段代码不是很好。它不会重用任何现有的Haskell库。这是另一种选择。 group函数完成了工作;认识到重复发生。 length &&& head获取每个这样的子列表并进行计数和单词。最后uniq_c找到最大宽度，并使用printf格式化输出。

import Control.Arrow
import Data.List
import Text.Printf

uniq :: (Eq a) => [a] -> [(Int, a)]
uniq = map (length &&& head) . group

uniq_c :: [String] -> [String]
uniq_c l =
    let us = uniq l
        width = length . show . maximum . map fst $ us
    in  map (uncurry $ printf "%*d %s" width) us

Answer 2

uniq_c提供字符串出现在与特定字符串空格连接的字符串列表中的次数：

*A> uniq_c ["The","quick","brown","fox","fox"]
["1 The","1 quick","1 brown","2 fox"]

*A> uniq_c $ ["The","quick","brown","fox","fox","fox"] ++ (replicate 100 "fox")
["  1 The","  1 quick","  1 brown","103 fox"]

uniq_c'提供了(string,count)的元组列表。

tam找到最大计数（这将是任何特定字符串的最后一次出现）。

nl缩进计数，以使计数合理。

*A> mapM_ putStrLn $ uniq_c $ ["The","quick","brown","fox","fox","fox"] ++ (replicate 100 "fox")
  1 The
  1 quick
  1 brown
103 fox

Haskell uniq -c模拟器代码说明

2 个答案: