I have a table that has a value
field. The records have values somewhat evenly distributed between 0 and 100.
I want to query this table for n
records, given a target mean, x
, so that I'll receive a weighted random result set where avg(value)
will be approximately x
.
I could easily do something like
SELECT TOP n * FROM table ORDER BY abs(x - value)
... but that would give me the same result every time I run the query.
What I want to do is to add weighting of some sort, so that any record may be selected, but with diminishing pr开发者_StackOverflow中文版obability as the distance from x
increases, so that I'll end up with something like a normal distribution around my given mean.
I would appreciate any suggestions as to how I can achieve this.
why not use the RAND() function?
SELECT TOP n * FROM table ORDER BY abs(x - value) + RAND()
EDIT
Using Rand won't work because calls to RAND in a select have a tendency to produce the same number for most of the rows. Heximal was right to use NewID but it needs to be used directly in the order by
SELECT Top N value
FROM table
ORDER BY
abs(X - value) + (cast(cast(Newid() as varbinary) as integer))/10000000000
The large divisor 10000000000 is used to keep the avg(value)
closer to X while keeping the AVG(x-value)
low.
With that all said maybe asking the question (without the SQL bits) on https://stats.stackexchange.com/ will get you better results.
try
SELECT TOP n * FROM table ORDER BY abs(x - value), newid()
or
select * from (
SELECT TOP n * FROM table ORDER BY abs(x - value)
) a order by newid()
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