Question

我有一个元组列表

居：

    public ActionResult Create()
    {
        //will contain the list of CategorytNames and it will bind this on the Create.
        //cshtml view as given below
        ViewBag.CategoryID = new SelectList(db.Categories, "Id", "CategoryName");
        return View();
    }

public ActionResult Edit(int id = 0)
    {
        FoodMenuItem foodmenuitem = db.FoodMenuItems.Find(id);
        if (foodmenuitem == null)
        {
            return HttpNotFound();
        }


        //checking what the selected value was using LINQ query and then using that value inside SelectList.
        //Now, DropDownList will look at the provided default selected value and displays it
        int selected = (from cat in db.FoodMenuItems
                           where cat.ID == id
                           select cat.CategoryID).First();
        ViewBag.CategoryID = new SelectList(db.Categories, "Id", "CategoryName", selected);
        // ViewBag.CategoryID = new SelectList(db.Categories, "Id", "CategoryName", foodmenuitem.CategoryID);
        return View(foodmenuitem);
    }

元组的长度并不总是相同的。在示例中是5但可以更改

我找到了一个元素（部分元组）。

元素的长度并不总是相同。

我正在寻找元组中元素的位置。

要查找的元素的长度始终小于或等于元组的长度。

要查找的项目的顺序是“正确的”，这意味着类型（'m'，'o'）的案例不存在

居：

[('a','b','c','d','e'),
 ('f','g','h','i','l'),
 ('m','n','o','p','q'),
 ('r','s','t','u','v'),
 ('z', 'aa', 'ab', 'ac', 'ad'),
 .....]

Answer 1

正如@tobias_k建议的那样，最简单的方法是将元组和子元素转换为字符串，并使用in运算符进行子字符串检查：

l = [('a','b','c','d','e'),
          ('f','g','h','i','l'),
          ('m','n','o','p','q'),
          ('r','s','t','u','v'),
          ('z', 'aa', 'ab', 'ac', 'ad'),
          ...]

def subtuple_index(tuples, t):
    def tuple2str(t): return ',{},'.format(','.join(t))
    t = tuple2str(t)
    for i,x in enumerate(map(tuple2str, tuples)):
        if t in x: return i
    return -1


>>> subtuple_index(l, ('m','n'))
2
>>> subtuple_index(l, ('z','aa','ab'))
4
>>> subtuple_index(l, ('m','o'))
-1
>>> subtuple_index(l, ('z','a'))
-1

Answer 2

如前所述，将元组转换为字符串要容易得多：

tuple_list = [('a','b','c','d','e'),
              ('f','g','h','i','l'),
              ('m','n','o','p','q'),
              ('r','s','t','u','v'),
              ('z', 'aa', 'ab', 'ac', 'ad'),
              ...]

def find_tuple(tuplist, *partial_elements):
     for tup in tuplist:
         tup_string = ',{0},'.format(','.join(tup))
         if ',{0},'.format(','.join(partial_elements)) in tup_string:
             return tuplist.index(tup)

find_tuple(tuple_list, 'm', 'n') -> 2
find_tuple(tuple_list, 'z', 'aa', 'ab') -> 4

请注意，这只会返回它在找到部分元素的列表中找到的第一个元组的索引。

在元组列表中查找部分元组

2 个答案: