List 在给定位置编辑列表？

事情是这样的。 有人开了连锁餐厅。每个餐厅都是一个三元组，带有“餐厅名称”、多张桌子和一个三元组列表，其中包含菜单（价格、“食谱名称”、“特殊配料”）：

所以，我得到了如下结果：

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")])]

我的部分任务（我陷入困境的部分）是：选择餐厅列表，添加、替换、更改订单、编辑菜单、餐厅和连锁餐厅列表

因此，问题是：在给定餐厅名称位置、菜单位置和新菜谱的情况下，创建一个函数，允许从菜单中编辑一个3元组（菜谱）

示例：编辑配方12（2222，“SSSS”、“LLLL”）

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，但从来没有用三元组作为参数。。。我需要它/

这是我的密码：

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") 

---

但在我的代码中，我必须插入一个带有元组的列表作为参数。问题是：是否可以只插入一个元组作为参数而不插入一个列表？

这个问题很模糊，但这可能会有所帮助

实际上，您不能编辑元组、列表等。相反，您应该从旧的元组、列表等构建新的元组、列表等。例如，您想在

[Int]

位置

n

插入一些

Int

：

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

然后：

---

不管那个参数是3元组还是其他类型的值。。。它的工作方式应该与您以前解决此问题的方式相同。因此，如果您提供失败的代码，即使它没有编译，社区的响应也会更好。允许我们给出建议并指出更简单的方法。这个问题很模糊。你到底希望你的函数做什么。已添加我的代码。：）如果您有tuple

t

，您可以通过

[t]

或

返回t

从中创建列表。

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

> insert 5 100 [1..10]
[1,2,3,4,5,100,6,7,8,9,10]