PenWin wrote:

> 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.

Isn't it rather:

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

?

--
Milan Babuskov

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