SPARQL group by a substring and average - sparql

Question

I am querying a large data set (temperatures recorded hourly for nearly 20 years) and I'd rather get a summary, e.g. daily temperatures.
An example query is here:
http://www.boisvert.me.uk/opendata/sparql_aq+.html?pasteid=hu5rbc7W
PREFIX opensheff: <uri://opensheffield.org/properties#>
select ?time ?temp where {
?m opensheff:sensor <uri://opensheffield.org/datagrid/sensors/Weather_Mast/Weather_Mast.ic> ;
opensheff:rawValue ?temp ;
<http://purl.oclc.org/NET/ssnx/ssn#endTime> ?time .
FILTER (str(?time) > "2011-09-24")
}
ORDER BY ASC(?time)
And the results look like this:
time temp
"2011-09-24T00:00Z" 12.31
"2011-09-24T01:00Z" 11.68
"2011-09-24T02:00Z" 11.92
"2011-09-24T03:00Z" 11.59
Now I would like to group by a part of the date string, so as to get a daily average temperature:
time temp
"2011-09-24" 12.3 # or whatever
"2011-09-23" 11.7
"2011-09-22" 11.9
"2011-09-21" 11.6
So, how do I group by a substring of ?time ?

Answer 1

Eventually solved it. Running here:
http://www.boisvert.me.uk/opendata/sparql_aq+.html?pasteid=j8m0Qk6s
Code:
PREFIX opensheff:
select ?d AVG(?temp) as ?day_temp
where {
?m opensheff:sensor <uri://opensheffield.org/datagrid/sensors/Weather_Mast/Weather_Mast.ic> ;
opensheff:rawValue ?temp ;
<http://purl.oclc.org/NET/ssnx/ssn#endTime> ?time .
BIND( SUBSTR(?time, 1, 10) AS ?d ) .
}
GROUP BY ?d
ORDER BY ASC(?d)
We use BIND to set a new variable to the substring required, and then grouping and averaging by that variable is simple enough.