calculation time differente between : [expression1]/[expression2] and column(1) / column(2)

[amien]

is this the same or are the results of expression1 allready in the cache?

[#2]

Hi Amien.

Please can you explain very well your preoccupation?

Thanks

[amien]

let say i have 1 dimension and two expressions:

DIM : Date

EXPR 1 : SUM(value1) --> label is EXPR1

EXPR 2 : SUM(value2) --> label is EXPR2

now i want to calculate the ratio between those expressions.

i have two options:

1. use the name of the label of both expressions also expression. in this case [EXPR1] / [EXPR2]

2. use column(1) / column(2).

my question is, with [EXPR1] is the calculation done all over again or is it the same asl column(1)? i assume with column(1) it just picks the pre-calculated result of SUM(value) instead of calculating it all over again.

my expressions are quite heavy and i dont want the expressions to be calculated twice. its a performance/architect question

[ashfaq_haseeb]

Hi Amien

[Expression1]/[Expression2] and column(1)/column(2)

gives you exactly same result.

Hope that helps

Regards

ASHFAQ

[#5]

Hello Amien,

for my experience, it's quite the same. Both are aliases for the same column(s) and the expression(s) behind it. So independent from your type of "calling" them, QV will have cached intermediate results or not.

So I think the more exciting question is how to reduce online calculation time. The time users are waiting for a response. I am sure you did work hard to reduce it, but think (again) about preaggregation in your load-scripts and/or reducing data, split one qvw into two parts with the according data reduction. Ask your users, what kind of data, what granularity, time-scope (are really the last twenty years necessary), ... they really need. Often they are telling much more, not exactly knowing their true requests.

HtH

Roland

[amien]

Thanks Roland

[rwunderlich]

Amien wrote:my question is, with [EXPR1] is the calculation done all over again or is it the same asl column(1)? i assume with column(1) it just picks the pre-calculated result of SUM(value) instead of calculating it all over again

That's a very interesting question! It might be worth running it by QT support and seeing if they can get an answer from R&D.