Warning: file_get_contents(/data/phpspider/zhask/data//catemap/6/haskell/8.json): failed to open stream: No such file or directory in /data/phpspider/zhask/libs/function.php on line 167

Warning: Invalid argument supplied for foreach() in /data/phpspider/zhask/libs/tag.function.php on line 1116

Notice: Undefined index: in /data/phpspider/zhask/libs/function.php on line 180

Warning: array_chunk() expects parameter 1 to be array, null given in /data/phpspider/zhask/libs/function.php on line 181
Haskell 我想把网站的评级列入名单,我该怎么做?_Haskell - Fatal编程技术网

Haskell 我想把网站的评级列入名单,我该怎么做?

Haskell 我想把网站的评级列入名单,我该怎么做?,haskell,Haskell,---对用户评级进行四舍五入的函数 userRate :: Film -> Float userRate (_,_,_,[],[]) = 0 userRate (_,_,_,likes,dislikes) = fromIntegral(length likes)/fromIntegral(length likes + length dislikes)* 100 --该函数使用列表理解来过滤网站评级为75及以上的影片 websiteRating :: Film -> Int webs

---对用户评级进行四舍五入的函数

userRate :: Film -> Float
userRate (_,_,_,[],[]) = 0
userRate (_,_,_,likes,dislikes) = fromIntegral(length likes)/fromIntegral(length likes + length dislikes)* 100
--该函数使用列表理解来过滤网站评级为75及以上的影片

websiteRating :: Film -> Int
websiteRating film = round(userRate (film)) 
我想在列表中显示网站评级,我该怎么做?
谢谢

也许你正在寻找这样的东西,整个结果是一组由电影和评级组成的配对

Title: Silence
Director: Martin Scorsese
Year:2016
Website Rating : 76



Title: The Terminal
Director: Steven Spielberg
Year:2004
Website Rating : 76%
评级::[Film]->[(Film,Int)]
电影分级=[(电影,分级)
|胶卷=75]

您如何显示与每部影片相关的其他内容?有一个函数可以获得电影的评级,还有一个函数可以显示你所关心的电影的属性。将第一个与第二个结合使用?在我的电影类型中,我有(String,String,Int,[喜欢],[不喜欢])。如何在元组末尾添加函数
Title: Silence
Director: Martin Scorsese
Year:2016
Website Rating : 76



Title: The Terminal
Director: Steven Spielberg
Year:2004
Website Rating : 76%
rating:: [Film] -> [(Film, Int)]
rating films = [ (film, rating) 
   | film <- films , let rating = websiteRating film, rating >=75 ]