I have the following data strucure:
List<Item> Items = new List<Item>
{
new Item{ Id = 1, Name = "Machine" },
new Item{ Id = 3, Id_Parent = 1, Name = "Machine1"},
new Item{ Id = 5, Id_Parent = 3, Name = "Machine1-A", Number = 2, Price = 10 },
new Item{ Id = 9, Id_Parent = 3, Name = "Machine1-B", Number = 4, Price = 11 },
new Item{ Id = 100, Name = "Item" } ,
new Item{ Id = 112, Id_Parent = 100, Name = "Item1", Number = 5, Price = 55 }
};
I want to build a query that gets the sum of all children price in its parent (items are related by Id_Parent). For example, for Item Id = 100, I have 55, because thats the value of the its child .
For Item Id = 3 I have 21, becaue Item Id = 5 and Id = 9 all sum to that. So far soo good.
What I am strugling to get is for Item Id = 1 I should also have the sum = 21, because Id = 3 is a child of Id = 1 and it has a sum of 21.
Here is my code:
var result = from i in items
join item in item on i.Id_Parent equals item.Id
select new
{
Name = prod.Nome,
Sum =
(from it in items
where it.Id_Parent == item.Id
group it by new
{
it.Id_Parent
}
into g
select new
{
Sum = g.Sum(x => x.Price)
}
).First()
};
Help appreciated.
Create a recursive function to find all the children of a parent:
public static IEnumerable<Item> ItemDescendents(IEnumerable<Item> src, int parent_id) {
foreach (var item in src.Where(i => i.Id_Parent == parent_id)) {
yield return item;
foreach (var itemd in ItemDescendents(src, item.Id))
yield return itemd;
}
}
Now you can get the price for any parent:
var price1 = ItemDescendants(Items, 1).Sum(i => i.Price);
Note if you know that the children of an item are always greater in id value than their parent, you don't need recursion:
var descendents = Items.OrderBy(i => i.Id).Aggregate(new List<Item>(), (ans, i) => {
if (i.Id_Parent == 1 || ans.Select(a => a.Id).Contains(i.Id_Parent))
ans.Add(i);
return ans;
});
For those that prefer to avoid recursion, you can use an explicit stack instead:
public static IEnumerable<Item> ItemDescendentsFlat(IEnumerable<Item> src, int parent_id) {
void PushRange<T>(Stack<T> s, IEnumerable<T> Ts) {
foreach (var aT in Ts)
s.Push(aT);
}
var itemStack = new Stack<Item>(src.Where(i => i.Id_Parent == parent_id));
while (itemStack.Count > 0) {
var item = itemStack.Pop();
PushRange(itemStack, src.Where(i => i.Id_Parent == item.Id));
yield return item;
}
}
I included PushRange helper function since Stack doesn't have one.
Finally, here is a variation that doesn't use any stack, implicit or explicit.
public IEnumerable<Item> ItemDescendantsFlat2(IEnumerable<Item> src, int parent_id) {
var children = src.Where(s => s.Id_Parent == parent_id);
do {
foreach (var c in children)
yield return c;
children = children.SelectMany(c => src.Where(i => i.Id_Parent == c.Id)).ToList();
} while (children.Count() > 0);
}
You can replace the multiple traversals of the source with a Lookup as well:
public IEnumerable<Item> ItemDescendantsFlat3(IEnumerable<Item> src, int parent_id) {
var childItems = src.ToLookup(i => i.Id_Parent);
var children = childItems[parent_id];
do {
foreach (var c in children)
yield return c;
children = children.SelectMany(c => childItems[c.Id]).ToList();
} while (children.Count() > 0);
}
I optimized the above based on the comments about too much nested enumeration, which improved performance vastly, but I was also inspired to attempt to remove SelectMany which can be slow, and collect IEnumerables as I've seen suggested elsewhere to optimize Concat:
public IEnumerable<Item> ItemDescendantsFlat4(IEnumerable<Item> src, int parent_id) {
var childItems = src.ToLookup(i => i.Id_Parent);
var stackOfChildren = new Stack<IEnumerable<Item>>();
stackOfChildren.Push(childItems[parent_id]);
do
foreach (var c in stackOfChildren.Pop()) {
yield return c;
stackOfChildren.Push(childItems[c.Id]);
}
while (stackOfChildren.Count > 0);
}
@AntonínLejsek's GetDescendants is still fastest, though it is very close now, but sometimes simpler wins out for performance.
If you love us? You can donate to us via Paypal or buy me a coffee so we can maintain and grow! Thank you!
Donate Us With