Dne 25.6.2010 14:29, Michael Ludwig napsal(a):

> Svein Erling schrieb am 25.06.2010 um 11:16 (-0000):
>
>> [...] isn't it an aggregate function that requires 'going through'
>> records twice (first finding the average and then how much each value
>> deviates from this value)?
>
> Yes, exactly. Conceptually, it cannot be done in one pass.

Actually, it can.

STDDEV(x) = AVG(x^2) - (AVG(x))^2

AVG can be calculated in one pass.

Pepak