计算字符串Haskell中字符的出现次数 [英] Count occurrences of character in a string Haskell
问题描述
尝试确定如何计算 char
的出现次数是 string
。我将它存储在 [char,count]
列表中。
countChars :: String - > [(Char,Int)]
我是新手,正在学习Haskell,因此非常感谢所有帮助。
给你一个美味
> ('a',4),('d',3),('e',1),((地图(头部&长度).group。sort) 'f',3),('s',2)]
/ p>
您可以轻松定义您的头&&&如果该语法是不熟悉的话,那么它就是一个长度。
> head_and_length x =(head x,length x)
组可以递归写入
group [] = []
group(x:xs)=(x:ys):group zs
where(ys ,zs)=(takeWhile(== x)xs,dropWhile(== x)xs)
采取尽可能多的匹配元素,并递归应用于其余的元素,直到没有剩下。请注意,这个定义需要相同的元素是连续的,这就是为什么需要排序的原因。
编写一个复杂函数是不推荐的,最好实现小特性分开(也许使用现有的功能),可以独立测试并创建这些作为最终解决方案的组合。
Trying to determine how to count the occurrences of a char
is a string
. I was it to be stored in a list [char,count]
.
countChars :: String -> [(Char, Int)]
I'm new and learning Haskell so any help is much appreciated.
to give you a taste
> (map (head &&& length) . group . sort) "asdfasdfaeadf"
[('a',4),('d',3),('e',1),('f',3),('s',2)]
after bunch or imports.
You can easily define your head &&& length, if that syntax is unfamiliar.
> head_and_length x = (head x, length x)
group can be written recursively
group [] = []
group (x:xs) = (x:ys) : group zs
where (ys,zs) = (takeWhile (==x) xs, dropWhile (==x) xs)
take as many matching elements as possible and recursively apply on the remaining elements until nothing left. Note that this definition requires same elements to be contiguous, that's why a sort is required.
Writing a complex function is one piece is not recommended, it's better to implement small features separately (and perhaps use the existing functions) which can be independently tested and create a composition of these as a final solution.
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