count distinct user with above 100% efficiency

[Heny]

I have this expression  Sum(norm*total_pages_count)/sum(total_time) that give me efficiency percentage

now I would like to count users on a monthly scale that are above 100%

I've tried 

count({<user={"=Sum(norm*total_pages_count)/sum(total_time)>0.1"}>} distinct user)

but I guess I'm totally off with my expression

any help?

[Kushal_Chawda]

@Heny  Your expression looks fine. Above 100% means >1

count({<user={"=Sum(norm*total_pages_count)/sum(total_time)>1"}>} distinct user)

[Heny]

I tried with 1, not getting any results

[Kushal_Chawda]

@Heny  share sample data with expected output

[Heny]

thanks for your help, but my data is a little complicated

I'll try to work around it

[Kushal_Chawda]

@Heny  Meanwhile you can try below as well as output of set analysis may differ depending on the context of different dimensions and charts

count(distinct aggr(if( Sum(norm*total_pages_count)/sum(total_time)>1,user),user))

[Heny]

ok, so I tried the above.

It is better than what I wrote but I still am not getting complete results

this is the data for November: (I added my efficiency expression to a variable)

user	$(vEfficiancy)	if(SUM(NORM*TOTAL_COUNT)/SUM(TOTAL_TIME*24*3600)>1,1,0)
A	108.1%	1
B	106.2%	1
C	105.4%	1
D	105.4%	1
E	102.7%	1
F	102.3%	1
G	99.7%	0
H	97.2%	0
I	97.0%	0

now, I should get a result of 9 but I'm getting only a count of 2 for November,

this is how my line chart looks:

[Heny]

sorry, I meant result of 6 not 9

[Kushal_Chawda · Accepted solution]

@Heny  try below

count(distinct aggr(if( Sum(norm*total_pages_count)/sum(total_time)>1,user),user,Month))

[Heny]

great! This works