Question

就是这样。有人有一连串的餐馆。每个餐厅都是一个带有“餐馆名称”的3元组，多个餐桌和一个带有菜单的3元组列表（价格，“食谱名称”，“特殊成分”）：

("restaurant name", number of tables, [(price, "recipe name" , "special ingredient")])

所以，我有类似的东西：

chainRestaurants = [("Food and friends",20,[(2.5,"Steak","lemon"),(3.5,"Vegetarian Meals","tomato"),(4.0,"Italian Beef","banana")]),("All in",10,[(2.5,"Stracotto","Garlic"),(3.0,"Roast Beef","Butter"),(3.3,"Veal Chops","Pepper")]),("Orange",25,[(4.5,"Turkey","Mustard"),(5.1,"Chicken","egg"),(6.0,"Chicken Salad","fruit")])]

我的部分工作（我被困住的部分）是：选择餐馆列表并添加，更换，更改订单，编辑菜单，餐馆和餐馆连锁列表。

因此，问题：根据餐馆名称位置，菜单位置和新配方，制作一个允许从菜单中编辑3元组（食谱）的功能。

示例：editRecipe 1 2（2222，“SSSSSS”，“llllll”）

输出：

[("Food and friends",20,[(2.5,"Steak","lemon"),(2222,"SSSSSS","llllll"),(4.0,"Italian Beef","banana")]),("All in",10,[(2.5,"Stracotto","Garlic"),(3.0,"Roast Beef","Butter"),(3.3,"Veal Chops","Pepper")]),("Orange",25,[(4.5,"Turkey","Mustard"),(5.1,"Chicken","egg"),(6.0,"Chicken Salad","fruit")])]

我已经解决了这个问题，但从来没有以3元组作为参数......我需要它！：/

我的代码是：

chainRestaurants = [("Food and friends",20,[(2.5,"Steak","lemon"),(3.5,"Vegetarian Meals","tomato"),(4.0,"Italian Beef","banana")]),("All in",10,[(2.5,"Stracotto","Garlic"),(3.0,"Roast Beef","Butter"),(3.3,"Veal Chops","Pepper")]),("Orange",25,[(4.5,"Turkey","Mustard"),(5.1,"Chicken","egg"),(6.0,"Chicken Salad","fruit")])]

menu1 = [(2.5,"Steak","lemon"),(3.5,"Vegetarian Meals","tomato"),(4.0,"Italian Beef","banana")]
restaurant1 = ("Food and friends",20,menu1)

menu2 = [(2.5,"Stracotto","Garlic"),(3.0,"Roast Beef","Butter"),(3.3,"Veal Chops","Pepper")]
restaurant2 = ("All in",10,menu2)

menu3 = [(4.5,"Turkey","Mustard"),(5.1,"Chicken","egg"),(6.0,"Chicken Salad","fruit")]
restaurant3 = ("Orange",25,menu3)

chainRestaurants1 = [restaurant1] ++ [restaurant2] ++ [restaurant3]


editRecipe restaurantposition menuposition newrecipe | menuposition == 0 = error "No such thing" 
                                                 | restaurantposition == 1 &&   menuposition > length menu1 = error "No such thing" 
                                                 | restaurantposition == 2 && menuposition > length menu2 = error "No such thing" 
                                                 | restaurantposition == 3 && menuposition > length menu3 = error "No such thing" 

editRecipe restaurantposition menuposition newrecipe = case restaurantposition of 1 -> [("Food and friends",20, take (menuposition-1) menu1 ++ newrecipe ++ drop (menuposition) menu1)] ++ [restaurant2] ++ [restaurant3]
                                                                                2 -> [restaurant1] ++ [("Food and friends",20, take (menuposition-1) menu2 ++ newrecipe ++ drop (menuposition) menu2)] ++ [restaurant3]
                                                                                3 -> [restaurant1] ++ [restaurant2] ++ [("Orange",25, take (menuposition-1) menu3 ++ newrecipe ++ drop (menuposition) menu3)]
                                                                                otherwise -> error "No such thing"

但是这段代码需要一个列表（newrecipe）作为参数，但我需要一个3元组。：/

ex: editRecipe 1 2 [(5.1,"xxxxxxxxx","ccccccccccc")]

我需要：

editRecipe 1 2 (5.1,"xxxxxxxxx","ccccccccccc")

但是使用我的代码，我必须插入一个以元组为目标的列表作为参数。问题是：是否可以只插入一个元组作为参数而不是列表？

Answer 1

问题很模糊，但也许这会有所帮助。

实际上，你不能编辑元组，列表等。相反，你应该从旧构建构建新元组，列表等。例如，您想在位置Int的{{1}}中插入一些[Int]：

然后：

insert :: Int -> Int -> [Int] -> [Int]
insert n x xs = before ++ [x] ++ after
  where
    (before, after) = splitAt n xs

编辑给定位置的列表？

1 个答案: