The following query works, but I want to get the same result without using grp.Sum(). Can we do it?

from item in (await VehicleReplaceCostDataAsync())
                group item by (item.type, item.size, item.ADA, item.eseq) into grp
                orderby (grp.Key.eseq, grp.Key.size, grp.Key.ADA)
                select new VehicleReplacementCost
                {
                    type = grp.Key.type,
                    size = grp.Key.size,
                    ADA = grp.Key.ADA,
                    count = grp.Sum(x => x.count),
                    cost = grp.Sum(x => x.cost),
                    Fcount = grp.Sum(x => x.Fcount),
                    Fcost = grp.Sum(x => x.Fcost),
                    eseq = grp.Key.eseq,
                }).ToList();

CodePudding user response：

Perhaps by using .Aggregate()? [docs]

count = grp.Aggregate(0, (a, b) => a   b.count)

CodePudding user response：

Thanks for the answer from Astrid. It looks like a good one, but I didn't test it. My colleague gave this solution instead by using yield:

var groups = costs
                .GroupBy(type => (type.SystemId, type.Type, type.Size, type.ADA, type.Eseq))
                .OrderBy(group => (group.Key.SystemId, group.Key.Eseq, group.Key.Size, group.Key.ADA));

 

            foreach (var group in groups)
            {
                var result = new ProgramGuideVehicleCostRow
                {
                    SystemId = group.Key.SystemId,
                    Type = group.Key.Type,
                    Size = group.Key.Size,
                    ADA = group.Key.ADA,
                };
                foreach (var row in group)
                {
                    result.Cost  = row.Cost;
                    result.Fcost  = row.Fcost;
                    result.Count  = row.Count;
                    result.Fcount  = row.Fcount;
                }
                yield return result;
            }