如何在C＃中将列表列表转换为字典？

Question

I have a list of boxes(unique, with id) and in each box there are certain items(unique, with id). Example:

list<box> boxes = new list<boxes>;

where,

class box
{   ...
    list<item> items = new list<item>;
    ...
}

• Box1: item1, item2, ...
• Box2: item4, item5, ...
• Box3: ...

I need to find the item details, given the item id. For this I current do something like this:

foreach (box b in boxes)
{
    foreach (item i in b.items)
    {
        if (i.id == searchId)
            return i;
    }
}

Question is: How can I convert this list data structure to a dictionary data structure?

As I have keys (Id), so I think using a dictionary would be a better choice?

Answer 1

If it's possible that same items can exist in several boxes, you can select all items, and group them by id, then select first item from each group as a value for dictionary:

Dictionary<int, item> items = boxes.SelectMany(b => b.items)
                                   .GroupBy(i => i.id)
                                   .ToDictionary(g => g.Key, g.First());

If all items have unique ids:

var items = boxes.SelectMany(b => b.items)                     .
                 .ToDictionary(i => i.id);

---

Getting item will look like:

if (items.ContainsKey(searchId))
    return items[searchId];

As @Douglas stated, to avoid double lookup it's better to use TryGetValue method:

item i;
if (items.TryGetValue(searchId, out i))
    return i;

NOTE: Linq alternative without dictionary will be (it does exactly same as your code - enumerates boxes and their items for each search):

var item = boxes.SelectMany(b => b.items).FirstOrDefault(i => i.id == searchId);

So, if you don't want to hold dictionary with items between searches, or if you need to execute single search, then you can use this solution.

Answer 2

Assuming all your items are unique:

var dictionary = boxes.SelectMany(box => box.items).ToDictionary(item => item.id);

You can then look up values using:

item i;
if (dictionary.TryGetValue(searchID, out i))
    return i;