Clash Royale CLAN TAG

C# Linq that includes sorting and grouping

This is my code:

public class Photos
{
public long PhotoLabel { get; set; }
public int UserID { get; set; }
}

List<Photos> photolist = new List<Photos>();
var result1 = photolist.OrderByDescending(p => p.PhotoLabel).ThenBy(r => r.UserID).ToList();


If I display the contents now, this is what I get (First sorted in descending order of PhotoLabel and then sorted by UserID:

|------|---------------------|---------------------|
| Row | UserID | PhotoLabel |
|----------------------------|---------------------|
| 1 | 92 | 20180729181046 |
|----------------------------|---------------------|
| 2 | 92 | 20180729181041 |
|----------------------------|---------------------|
| 3 | 92 | 20180729181037 |
|----------------------------|---------------------|
| 4 | 88 | 20180729174415 |
|----------------------------|---------------------|
| 5 | 88 | 20180729174405 |
|----------------------------|---------------------|
| 6 | 04 | 20180729174358 |
|----------------------------|---------------------|
| 7 | 1 | 20170924183847 |
|----------------------------|---------------------|
| 8 | 1 | 20170921231422 |
|----------------------------|---------------------|
| 9 | 1 | 20170920194624 |
|----------------------------|---------------------|
| 10 | 32 | 20170820114728 |
|----------------------------|---------------------|
| 11 | 32 | 20170820114725 |
|----------------------------|---------------------|
| 12 | 32 | 20170820114421 |
|----------------------------|---------------------|
| 13 | 32 | 20170820114416 |
|----------------------------|---------------------|
| 14 | 1 | 20170225151023 |
|----------------------------|---------------------|
| 15 | 1 | 20170225151000 |
|----------------------------|---------------------|
| 16 | 1 | 20170225150957 |
|----------------------------|---------------------|


From the sorted table above, this is what I want to achieve:

Display groups of UserIDs and PhotoLabels where UserIDs appear 3 or more times in one group (eg: rows 4 and 5 where UserID=88 and row 6 where UserID=04 should be eliminated since the UserID=88 appears just twice in the group and UserID=04 appears only once in the group).

Display only the top most group of UserIDs and exclude any repeating UserIDs (eg: rows 7,8 and 9 displays the UserID=1 group. Don't display any other UserID=1 group such as rows 14,15 and 16.

The expected result from query should be:

|------|---------------------|---------------------|
| Row | UserID | PhotoLabel |
|----------------------------|---------------------|
| 1 | 92 | 20180729181046 |
|----------------------------|---------------------|
| 2 | 92 | 20180729181041 |
|----------------------------|---------------------|
| 3 | 92 | 20180729181037 |
|----------------------------|---------------------|
| 7 | 1 | 20170924183847 |
|----------------------------|---------------------|
| 8 | 1 | 20170921231422 |
|----------------------------|---------------------|
| 9 | 1 | 20170920194624 |
|----------------------------|---------------------|
| 10 | 32 | 20170820114728 |
|----------------------------|---------------------|
| 11 | 32 | 20170820114725 |
|----------------------------|---------------------|
| 12 | 32 | 20170820114421 |
|----------------------------|---------------------|
| 13 | 32 | 20170820114416 |
|----------------------------|---------------------|


Thank you so much in in advance! :-)

2 Answers
2

Please try this!

List<Photos> photolist = photolist.GroupBy(p => p.UserID)
.Where(g => g.Count() > 2)
.Select(p => new
{
UserId = p.UserID,
PhotoLabel = p.PhotoLabel

})
.ToList();


With the two syntaxes:

var minCount = 3;
var topN = 10;

var res = result1.GroupBy(p => p.UserID)
.Where(g => g.Count() >= minCount)
.Select(p => p)
.Take(topN);

var res2 = (from p in result1
group p by p.UserID into g
where g.Count() >= minCount
select p).Take(topN);


Notice that I used yout result1 set as it it already sorted. If I was to start from photolist, I would also need to apply the sorting, because of the TOP N requirement.

result1

photolist

By clicking "Post Your Answer", you acknowledge that you have read our updated terms of service, privacy policy and cookie policy, and that your continued use of the website is subject to these policies.