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Not applicable

Calculate distinct values ignoring one dimension

Dear all,

I want to make a chart with the count of the distinct values in my dataset over a certain time period. So I have an expression count( distinct VALUE) and a dimension MONTH.

However, when I define such a chart I get the count of distinct VALUE per MONTH and not over the total period. I only want to see 1 entry of VALUE in my entire chart. How can I realise such a thing?

Thanks!

1 Solution

Accepted Solutions
swuehl
MVP
MVP

You could use something like

=count (distinct aggr(ID, ID))

as expression (see attached sample). aggr() function will list your IDs in load order and determine based on this if a values is already read in. So to make it work, your load order of IDs should be also in chronological order.

Hope this helps,

Stefan

edit:

even a

count( aggr(ID,ID))

should / could be enough, the distinct we need is implicitely part of the aggr() function.

View solution in original post

8 Replies
swuehl
MVP
MVP

Remove the dimension?

CELAMBARASAN
Partner - Champion
Partner - Champion

Hi,

     Try with dollar expansion like below

     use your expression as $(=count( distinct VALUE))

Celambarasan

Not applicable
Author

Thanks for your fast replies. However:

@swuehl: then my chart wouldn't have a dimension...

@Celambarasan Adhimulam: that ignores my dimension completely. I still want the count to be calculated for every MONTH but it shouldn't include the VALUES already counted in the previous MONTHs

jagan
Luminary Alumni
Luminary Alumni

Hi,

Try this expression

=Count( distinct TOTAL VALUE)

Regards,

Jagan.

Not applicable
Author

That gives me the same result as using the dollar sign expansion

swuehl
MVP
MVP

You could use something like

=count (distinct aggr(ID, ID))

as expression (see attached sample). aggr() function will list your IDs in load order and determine based on this if a values is already read in. So to make it work, your load order of IDs should be also in chronological order.

Hope this helps,

Stefan

edit:

even a

count( aggr(ID,ID))

should / could be enough, the distinct we need is implicitely part of the aggr() function.

jagan
Luminary Alumni
Luminary Alumni

Hi,

Can you attach sample file.

Regards,

Jagan.

Not applicable
Author

swuehl: thanks! that does the work perfectely!!