The easy way is to use cts:count() and "item-frequency" with 
cts:triples(). But it's not very fast.

The more complicated way is to turn on the "SPARQL Value Frequencies" 
trace event, which will output something like this:

2014-03-21 17:29:10.386 Info: [Event:id=SPARQL Value Frequencies] 
sessionKey=3215158362941807382 values=
2014-03-21 17:29:10.386 Info:   () = count:4000038, sUniq:187066, 
pUniq:67, oUniq:494586
2014-03-21 17:29:10.386 Info: 
<http://www.w3.org/1999/02/22-rdf-syntax-ns#type> = freq:187057, s: 
(count:0, pUniq: 0, oUniq: 0), p: (count:187057, sUniq: 187057, oUniq: 
12), o: (count:0, sUniq: 0, pUniq: 0)
2014-03-21 17:29:10.386 Info:   <http://xmlns.com/foaf/0.1/Document> = 
freq:10, s: (count:0, pUniq: 0, oUniq: 0), p: (count:0, sUniq: 0, oUniq: 
0), o: (count:10, sUniq: 10, pUniq: 2)

The value you want is the one after "() = count:", in this case 4000036. 
Divide that by 3 to get the number of triples in all stands of the 
database (so 1333346 in this case). That number will include triples in 
deleted documents that haven't been merged out of the stands yet.

John

On 24/03/14 18:10, Clark Richey wrote:
> Is there an easy way to determine the number of total triples in the DB?
> I have triples insides of other documents.
>
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-- 
John Snelson, Lead Engineer                    http://twitter.com/jpcs
MarkLogic Corporation                         http://www.marklogic.com
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