SQL ACCESS 2010按组别排名前80％

Question

I'm trying to to calculate the top 80 percent of a time value per a unique identifier, so I need to group by that identifier. To which I am having trouble since the table I'm working with has 15 million records. A query that I found and used can manage to complete but crashes access once I try to scroll through the information.

SELECT TID, DATE, TRAVEL_TIME
FROM FH_2015
WHERE TRAVEL_TIME IN
(SELECT TOP 80 PERCENT TRAVEL_TIME
FROM FH_2015 AS A
WHERE A.TID = FH_2015.TID and A.DATE = FH_2015.DATE
ORDER BY TRAVEL_TIME ASC);

Answer 1

Assuming TID is a unique record identifier, this is all you should need:

SELECT TID, DATE, TRAVEL_TIME
FROM FH_2015
WHERE TID IN
(SELECT TOP 80 PERCENT
    TID
    FROM FH_2015
    ORDER BY TRAVEL_TIME DESC)

Note the DESC if you are actually looking for the top 80% of the TRAVEL_TIME.

And actually you could just do this:

SELECT TOP 80 PERCENT
TID, DATE, TRAVEL_TIME
FROM FH_2015
ORDER BY TRAVEL_TIME DESC