Reputation: 2577
Codewise, what it the cleanest way to do this using linq? Below, I have a crude example where I want to find a matching class instance based on name.
class item
{
string name {get;set;}
int identifier {get;set;}
}
void DoSomething()
{
List<item> List1 = GetSampleItems();
List<item> List2 = GetOtherSampleItems();
for(int a=0;a<List1.count;a++)
{
for(int b=0;b<List2.count;b++)
{
if(List1[a].identifier == List2[b].identifier)
{
List1[a].name = List2[b].name;
}
}
}
}
Upvotes: 6
Views: 2251
Reputation: 476624
A better way is using a Dictionary<TK,TV>
:
Dictionary<int,item> l2dic = List2.ToDictionary(x => x.identifier);
item itm;
List1.ForEach(x => {
if(l2dic.TryGetValue(x.identifier,out itm)) {
x.name = itm.name;
}
});
Or as @Rawling says, use a foreach
loop instead:
Dictionary<int,item> l2dic = List2.ToDictionary(x => x.identifier);
item itm;
foreach(item x in List1) {
if(l2dic.TryGetValue(x.identifier,out itm)) {
x.name = itm.name;
}
}
Ideone demo (with slight modifications to your item
class).
This runs on average in linear time whereas your approach runs in quadratic time.
The assumption is however that the identifiers are unique: no two elements in the same list can have the same identifier.
A concluding note is that variables in general start with a lowercase character so list1
and list2
whereas classes and properties start with a capital one (this Item
, Identifier
and Name
).
Upvotes: 4
Reputation: 13248
This should work:
var query = from l1 in List1
join l2 in List2 on l1.identifier equals l2.identifier
select new
{
l1values = l1,
l2Name = l2.name
};
foreach(var item in query)
item.l1Values.name = item.l2Name;
Upvotes: 4
Reputation: 152556
Linq is for querying, not updating, so you'll still need to loop through the results to make the changes, but you can join
to match up the two lists like so:
var query = from l1 in List1
join l2 in List2
on l1.identifier equals l2.identifier
select new {l1, l2};
Now loop through the query to update the l1
items:
foreach(var item in query)
item.l1.name = item.l2.name;
As a side note, there's nothing wrong with the way you're doing it (other than you could break out of the inner loop if a match is found). If you understand how it works and the performance is acceptable, there's no compelling reason to change it.
Upvotes: 10