Clean some properties value in List with Linq based on a value of the list

Question

I have an object "MyObject" with the properties (all string): "PropA", "PropB" and "PropC".

var List = new List();

I add some object in this list with the following value :

List.Add(new MyObject { PropA = "AA", PropB = "1", PropC = "TT"});
List.Add(new MyObject { PropA = "AA", PropB = "1", PropC = "TT"});
List.Add(new MyObject { PropA = "AA", PropB = "1", PropC = "TT"});

List.Add(new MyObject { PropA = "BB", PropB = "1", PropC = "TT"});
List.Add(new MyObject { PropA = "BB", PropB = "1", PropC = "TT"});

With linq, I'd like for each different "PropA" keep the first record but set to string.Empty the other. The result I'd like is a List with these values :

MyObject { PropA = "AA", PropB = "1", PropC = "TT"}
MyObject { PropA = "", PropB = "", PropC = "TT"}
MyObject { PropA = "", PropB = "", PropC = "TT"}
MyObject { PropA = "BB", PropB = "1", PropC = "TT"}
MyObject { PropA = "", PropB = "", PropC = "TT"}

I did with foreach but it's may be a bit cleaner in Linq, but the order of the result must be kept.

Answer 1

This would work for the specific case:

var list = 
 List.GroupBy(x => x.PropA)
        .SelectMany(grp => new MyObject[] { grp.First() }
        .Concat(grp.Skip(1)
          .Select(x => { x.PropA = String.Empty; x.PropB = String.Empty; return x; } )
           )
         );

LinqPad result:

As a side note, I don't think using Linq in this case is justified, it doesn't make the code faster or cleaner. One must use the available tools to write better, more performant, or cleaner code, but in this case, I don't think this is better than a foreach (at least a well thought foreach, and not a brute force one) in any possible way.

Answer 2

How about this one:

    var result = List.GroupBy(prop => prop.PropA)
        .SelectMany(group => new [] { group.First() }.Concat(group.Skip(1).Select(x => { x.PropA = x.PropB = ""; return x; }))).ToList();

Answer 3

var groups = List.GroupBy(obj => obj.PropA);
foreach (var group in groups)
{
    foreach (var item in group.Skip(1))
    {
        item.PropA = "";
        item.PropB = "";
    }
}

Answer 4

Yet another solution to this interesting question:

var result =
    list
    .GroupBy(x => x.PropA, (key, items) => new { Key = key, Items = items })
    .Select(x => x.Items.Select((item, index) =>
    {
        if (index == 0) return item;
        item.PropA = string.Empty;
        item.PropB = string.Empty;
        return item;
    }))
    .SelectMany(x => x)
    .ToList();

It modifies the original objects in each group but the first one. This time without Concat.

---

Sometimes it's all about performance. If someone concerned about this stumbles upon this question the answer is that in a case like this a simple loop is approx four times faster then the suggested linq queries:

For 1.000.000 items linq needs ~200ms and the following loop only ~45ms:

string prev = list.First().PropA;
foreach (var item in list.Skip(1))
{
    if (item.PropA == prev)
    {
        item.PropA = string.Empty;
        item.PropB = string.Empty;
    }
    else
    {
        prev = item.PropA;
    }
}

Answer 5

Without LINQ

HashSet<string> propA = new HashSet<string>();
HashSet<string> propB = new HashSet<string>();

for (int i = 0; i < list.Count; i++)
{
    if (!propA.Add(list[i].PropA))
    {
        list[i].PropA = string.Empty;
    }

    if (!propB.Add(list[i].PropB))
    {
        list[i].PropB = string.Empty;
    }
}

Answer 6

This?

string currentValue = "";
            List.OrderBy(x => x.PropA).ToList()ForEach(x =>
            {
                if (string.IsNullOrEmpty(currentValue))
                {
                    // Assuming PropA will never be null
                    currentValue = x.PropA;
                    // this is first element
                    return;
                }
                if (x.PropA == currentValue)
                {
                    x.PropA = "";
                }
                else
                {
                    currentValue = x.PropA;
                }

            });