我正在尝试创建一个 Haskell 函数,它接受两个字符串作为参数,第一个是我们想要在第二个参数中定位的字符串,并返回一个元组列表,其中包含每次出现的开始和结束索引。例如,
indexTuples :: String -> String -> [(Int, Int)]
indexTuples "aa" "foobaarfoobaar"
Output: [(4,5), (11,12)]
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到目前为止,我已经制作了一个查找索引的帮助函数(我试图不使用除 Prelude 方法之外的任何额外方法,而是自己实现)。
我的辅助函数接受一个字符串和一个字符并返回索引,如下所示:
findPos :: (Num a1, Enum a1, Eq a2) => [a2] -> a2 -> [a1]
findPos str c = [index | (x, index) <- zip str [0..], x == c]
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(我在这里找到了这个解决方案。)该函数将带有无限数字列表的字符串压缩为元组,然后选择字符等于参数的元组c
并返回每个元组的索引。这给了我这个输出:
Ok, one module loaded.
ghci> findPos "blablabla" 'b'
[0,3,6]
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但我该如何做到这一点,以便它接受两个字符串呢?像这样:
ghci> findPos "blablabla" "bl"
[(0,1), (3,4), (6,7)]
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我尝试将c
char 对象更改为字符串对象,但随后在ghci
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我正在尝试编写一个接受字符串的函数,然后将字符串作为字符串单词的列表返回(例如单词内置函数),到目前为止,我已经写了
ord :: String -> [String]
ord [] = []
ord xs = let
ys = groupBy (\x y -> y /= ' ') xs
in filter (not . null) ys
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我认为这将摆脱列表中的空字符串,但我只会得到此输出
输入:
ord “aa b c - dd”
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输出:
["aa"," b"," "," "," "," "," "," "," c"," "," "," -"," "," "," "," dd"]
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当这是我想要的输出时:
[“aa”, ”b”, ”c”, ”-“, ”dd”]
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如果我尝试写,我会得到相同的结果
ord :: String -> [String]
ord [] = []
ord xs = filter (not . null) ys …
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