简体   繁体   中英

C# get distinct and max value

This is my datatable

X       Y   P
Str1    C1  10
Str1    C1  5
Str1    C1  1
Str1    C1  2
Str2    C1  25
Str2    C1  4
Str1    C2  411
Str1    C2  2356
Str2    C2  12
Str2    C2  33

I am trying to get distinct rows based on X and Y columns and get max value of P column using linq to dataset

 X      Y   P
Str1    C1  10
Str2    C1  25
Str1    C2  2356
Str2    C2  33

-

            var query0 = (from row in tablo.AsEnumerable() 

        select new
        {
            X = row.Field<string>("X"),
            Y = row.Field<string>("Y"),

        }).Distinct();

How can i do? Thanks.

You need GroupBy here:-

var result = tablo.AsEnumerable().
                  .GroupBy(row => new 
                            { 
                               X= row.Field<string>("X"), 
                               Y = row.Field<string>("Y") 
                            })
                  .Select(x => new 
                              {
                                 X = x.Key.X,
                                 Y = x.Key.Y,
                                 P = x.Max(z => z.Field<int>("P"))
                              });

You could try something like this:

 var result =  tablo.AsEnumerable() 
                    .Select(row=> new
                    {
                       X = row.Field<string>("X"),
                       Y = row.Field<string>("Y"),
                       P = row.Filed<int>("P")
                    }).GroupBy(i=>new {X = i.X, Y = i.Y})
                      .Select(gr=>new{
                          X = gr.Key.X,
                          Y = gr.Key.Y
                          P = gr.Max(z => z.P)
                    });

Essentially, we first declare a sequence of an anoymous type objects, that have three properties, X , Y and P . Then we group by this sequence based on X and Y . Last, we pick up the maximum price for each group.

The technical post webpages of this site follow the CC BY-SA 4.0 protocol. If you need to reprint, please indicate the site URL or the original address.Any question please contact:yoyou2525@163.com.

 
粤ICP备18138465号  © 2020-2024 STACKOOM.COM