On 29/05/10 17:09, Tom Lane wrote:
> Jan Urbański <wulczer@wulczer.org> writes:
>> Now I tried to substitute some numbers there, and so assuming the
>> English language has ~1e6 words H(W) is around 6.5. Let's assume the
>> statistics target to be 100.
>
>> I chose s as 1/(st + 10)*H(W) because the top 10 English words will most
>> probably be stopwords, so we will never see them in the input.
>
>> Using the above estimate s ends up being 6.5/(100 + 10) = 0.06
>
> There is definitely something wrong with your math there. It's not
> possible for the 100'th most common word to have a frequency as high
> as 0.06 --- the ones above it presumably have larger frequencies,
> which makes the total quite a lot more than 1.0.
Upf... hahaha, I computed this as 1/(st + 10)*H(W), where it should be
1/((st + 10)*H(W))... So s would be 1/(110*6.5) = 0.0014
With regards to my other mail this means that top_stopwords = 10 and
error_factor = 10 would mean bucket_width = 7150 and final prune value
of 6787.
Jan
