[英]Haskell Lists - Count fst of tuples and compare with length of list
我有以下清單:
[(Libri,50.0),(Proxis,20.0),(Proxis,45.0),(Amazon,45.0)]
而且我有一個稱為“ articles
列表。 我想用元組轉換上面的列表,這樣,如果我使(元篇文章)乘以元組的第一個元素出現在列表中,就只能得到它!
因此,例如,如果我有一個列表articles=["HP","Haskell"]
,則該列表應顯示:
[(Proxis,20.0),(Proxis,45.0)]
因為Proxis
出現了兩次!
編輯:
資料類型:
data Magasin = Proxis | Amazon | Libri deriving (Eq, Show)
type Article = String
type Prix = Float
data Entree = E Magasin Article Prix deriving (Eq, Show)
type Stock = [Entree]
這是我當前的代碼:
disponible::[Article]->Stock->[(Magasin,Float)]
disponible [] stk = []
disponible (art:reste) stk = (foldl(\acc (E m a p)->if a==art then (m,p):acc else acc) [] stk)++(disponible reste stk)
有任何想法嗎?
編輯:
disponible::[Article]->Stock->[(Magasin,Float)]
disponible articles stock = map(\(m,ls)->(m,sum $ map (\(E _ _ p)->p) ls)) $ filter ((==length articles).length.snd) magasinsArticles
where contientArticles = (filter (\(E _ a _)->a`elem`articles) stock)
magasins = foldl (\acc e-> if e`elem`acc then acc else (e:acc)) [] $ map (\(E m _ _)->m) contientArticles
magasinsArticles = map (\m->(m,filter(\(E m2 _ _)->m2==m) contientArticles)) magasins
找到了我想要的答案。
disponible::[Article]->Stock->[(Magasin,Float)]
disponible articles stock = map(\(m,ls)->(m,sum $ map (\(E _ _ p)->p) ls)) $ filter ((==length articles).length.snd) magasinsArticles
where contientArticles = (filter (\(E _ a _)->a`elem`articles) stock)
magasins = foldl (\acc e-> if e`elem`acc then acc else (e:acc)) [] $ map (\(E m _ _)->m) contientArticles
magasinsArticles = map (\m->(m,filter(\(E m2 _ _)->m2==m) contientArticles)) magasins
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