You might want to look at the uniq and sort applications. ./yourscript.ksh | sort | uniq (FYI, yes, the sort is necessary in this command line, uniq only strips duplicate lines that are immediately after each other) EDIT: Contrary to what has been posted by Aaron Digulla in relation to uniq‘s commandline options: Given the … Read more
They are essentially equivalent to each other (in fact this is how some databases implement DISTINCT under the hood). If one of them is faster, it’s going to be DISTINCT. This is because, although the two are the same, a query optimizer would have to catch the fact that your GROUP BY is not taking … Read more
I believe this is what you want: table.groupby(‘YEARMONTH’).CLIENTCODE.nunique() Example: In [2]: table Out[2]: CLIENTCODE YEARMONTH 0 1 201301 1 1 201301 2 2 201301 3 1 201302 4 2 201302 5 2 201302 6 3 201302 In [3]: table.groupby(‘YEARMONTH’).CLIENTCODE.nunique() Out[3]: YEARMONTH 201301 2 201302 3
You can use the DISTINCT keyword within the COUNT aggregate function: SELECT COUNT(DISTINCT column_name) AS some_alias FROM table_name This will count only the distinct values for that column.
MusiGenesis’ response is functionally the correct one with regard to your question as stated; the SQL Server is smart enough to realize that if you are using “Group By” and not using any aggregate functions, then what you actually mean is “Distinct” – and therefore it generates an execution plan as if you’d simply used … Read more
You’re looking for a group by: select * from table group by field1 Which can occasionally be written with a distinct on statement: select distinct on field1 * from table On most platforms, however, neither of the above will work because the behavior on the other columns is unspecified. (The first works in MySQL, if … Read more
SELECT DISTINCT a,b,c FROM t is roughly equivalent to: SELECT a,b,c FROM t GROUP BY a,b,c It’s a good idea to get used to the GROUP BY syntax, as it’s more powerful. For your query, I’d do it like this: UPDATE sales SET status=”ACTIVE” WHERE id IN ( SELECT id FROM sales S INNER JOIN … Read more
You are so close! All you need to do is select BOTH the home and its max date time, then join back to the topten table on BOTH fields: SELECT tt.* FROM topten tt INNER JOIN (SELECT home, MAX(datetime) AS MaxDateTime FROM topten GROUP BY home) groupedtt ON tt.home = groupedtt.home AND tt.datetime = groupedtt.MaxDateTime
What if I want to obtain a distinct list based on one or more properties? Simple! You want to group them and pick a winner out of the group. List<Person> distinctPeople = allPeople .GroupBy(p => p.PersonId) .Select(g => g.First()) .ToList(); If you want to define groups on multiple properties, here’s how: List<Person> distinctPeople = allPeople … Read more