oiy*_*yio 1 haskell function data-structures
我有一些函数(charfreq,wordfreq,charcount,wordcount,parerror),我想在dataStructure中使用给定的字符串.但我怎么能这样做?我正在尝试这些和许多方式,但我得到了所有的错误.谢谢你提前.
data StrStat = StrStat { charfreq :: [( Char , Int )]
, wordfreq :: [([ Char ] , Int )]
, charcount :: Int
, wordcount :: Int
, parerror::Maybe Int
}
analyze :: [Char] -> StrStat
analyze x = StrStat { charfreq = (charfreq x) {-this line gives error-}
, wordfreq = (wordfreq x)
, charcount = (charcount x)
, wordcount = (wordcount x)
, parerror = (parerror x)
}
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错误信息是: Syntax error in input (unexpected `=')
analyze :: [Char] -> StrStat
analyze x = StrStat { "charfreq" = (charfreq x)
, "wordfreq" = (wordfreq x)
, "charcount" = (charcount x)
, "wordcount" = (wordcount x)
, "parerror" = (parerror x)
}
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当我尝试前一个时,我在同一行得到了同样的错误
我得到的第一个版本的错误是
Couldn't match expected type `StrStat' with actual type `[Char]'
In the first argument of `charfreq', namely `x'
In the `charfreq' field of a record
In the expression:
StrStat
{charfreq = (charfreq x), wordfreq = (wordfreq x),
charcount = (charcount x), wordcount = (wordcount x),
parerror = (parerror x)}
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这对我来说是有意义的,因为你正在应用你的getter(data StrStat例如,在你的声明中定义的所有charfreq :: StrStat -> [( Char , Int )])被调用类型的数据[Char],而不是StrStat值.
的charfreq =和如用于设置的各个领域的基于关键字的参数StrStat,并且需要被给予适当的值(例如[(Char, Int)]在他们的RHS).
我猜你要做的是构造一个StrStat值,你可以通过构造适当的值来做:
import Control.Arrow
import Data.List
data StrStat = StrStat { charfreq :: [( Char , Int )]
, wordfreq :: [([ Char ] , Int )]
, charcount :: Int
, wordcount :: Int
, parerror::Maybe Int
}
freq :: Ord a => [a] -> [(a, Int)]
freq = map (head &&& length) . group . sort
analyze :: [Char] -> StrStat
analyze x = StrStat { charfreq = freq x
, wordfreq = freq $ words x
, charcount = length x
, wordcount = length $ words x
, parerror = Nothing
}
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