Case 1: earliest=-1d@d latest=-0d@d ... | timechart span=1h count as Samples, avg(duration) as avg
vs.
Case 2: earliest=-1d@d latest=-0d@d ... | timechart span=1d count as Samples, avg(duration) as avg
The value for Samples is the same in both cases (if the 24 separate Samples in Case 1 are added), but the value for avg is not (if the 24 separate averages in Case 1 are averaged).
Why is that?
If you are averaging an average (in case 2, you are taking an average of 24 averages), you will get a different value than if you average all of the original numbers together right away (case 1)
http://www.incontext.indiana.edu/2013/mar-apr/article3.asp
this link is just a quick example of some averaging of averages vs actual averages.
Count would remain same regardless of the bucketing is done. The average however depends upon the total/number of instances. Daily average will be same if the count is same for every hour for that day. Take this data sample
Hour duration
Hour1 100
Hour1 10
Hour1 10
Hour2 10
Hour2 10
Hour2 10
Hour2 10
Now the Average for the day (when span=1d) will be (100+10+10+10+10+10+10)=160/7=22.86
Average when span1d
Hour1=100+10+10=120/3=40
Hour 2=10+10+10+10=40/4=10
Average of hourly average= 40+10/2 (number of hours)=25.
If you are averaging an average (in case 2, you are taking an average of 24 averages), you will get a different value than if you average all of the original numbers together right away (case 1)
http://www.incontext.indiana.edu/2013/mar-apr/article3.asp
this link is just a quick example of some averaging of averages vs actual averages.