问题
With the following data
string[] data = { "a", "a", "b" };
I'd very much like to find duplicates and get this result:
a
I tried the following code
var a = data.Distinct().ToList();
var b = a.Except(a).ToList();
obviously this didn't work, I can see what is happening above but I'm not sure how to fix it.
回答1:
When runtime is no problem, you could use
var duplicates = data.Where(s => data.Count(t => t == s) > 1).Distinct().ToList();
Good old O(n^n) =)
Edit: Now for a better solution. =) If you define a new extension method like
static class Extensions
{
public static IEnumerable<T> Duplicates<T>(this IEnumerable<T> input)
{
HashSet<T> hash = new HashSet<T>();
foreach (T item in input)
{
if (!hash.Contains(item))
{
hash.Add(item);
}
else
{
yield return item;
}
}
}
}
you can use
var duplicates = data.Duplicates().Distinct().ToArray();
回答2:
Use the group by stuff, the performance of these methods are reasonably good. Only concern is big memory overhead if you are working with large data sets.
from g in (from x in data group x by x)
where g.Count() > 1
select g.Key;
--OR if you prefer extension methods
data.GroupBy(x => x)
.Where(x => x.Count() > 1)
.Select(x => x.Key)
Where Count() == 1
that's your distinct items and where Count() > 1
that's one or more duplicate items.
Since LINQ is kind of lazy, if you don't want to reevaluate your computation you can do this:
var g = (from x in data group x by x).ToList(); // grouping result
// duplicates
from x in g
where x.Count() > 1
select x.Key;
// distinct
from x in g
where x.Count() == 1
select x.Key;
When creating the grouping a set of sets will be created. Assuming that it's a set with O(1)
insertion the running time of the group by approach is O(n)
. The incurred cost for each operation is somewhat high, but it should equate to near linear performance.
回答3:
Sort the data, iterate through it and remember the last item. When the current item is the same as the last, its a duplicate. This can be easily implemented either iteratively or using a lambda expression in O(n*log(n)) time.
来源:https://stackoverflow.com/questions/3138651/lambda-expression-to-find-difference