Aggregates of counts over time

[senescence]

This is probably really easy but I can't work it out.

I have a table with a user id and a log in time.

I want to show the count of users who have logged in x number of times.

So I'd have 1,2,3 etc on the x axis and the count of users on the y axis.

Any help appreciated - thanks.

[Omega · Accepted solution]

@senescence

Try create a new Table = GROUPBY(Table1,Table1[User ID],"Count of login",COUNTX(CURRENTGROUP(),Table1[Login time]))

User ID by default will be summarized as count. 

Hope this helps.

[senescence]

@Omega

This works perfectly - thanks!

[Anonymous]

Hi @senescence,

Sorry I misunderstood  the question earlier, anyway the solution that @v-chuncz-msft provided should work I think

.

Regards,

L.Meijdam

[Anonymous]

Hi @senescence,

I recreated your problem, and the pictures below show how to get the visualisation you want.

(used a column chart)

(User ID are numbers in my example 1-7)

Hope this was what you wanted to know.

let me know if you have questions !

Regards,

L.Meijdam

[senescence]

Hi @Anonymous

Thanks - but this isn't quite what I want. I don't want to see individual user ids. I want to see the count of userids over the count of logintimes.

So. If I have the following:

User - Login Time

1 - 10.00 

2 - 11.00

3 - 12.00

4 - 13.00

5 - 14.00

6 - 15.00

1 - 16.00

2 - 17.00

3 - 18.00

1 - 19.00

2 - 20.00

1 - 19.00

The graph would show the count of logins in the x axis (1,2,3,4) and the count of users who have logged in that many times on the y axis (3,1,1,1).

[v-chuncz-msft]

@senescence,

You may follow the steps below.

1) add a calculated table

Table2 =
GENERATESERIES ( 1, 4, 1 )

2) add a measure

Measure =
COUNTROWS (
    FILTER (
        SUMMARIZE ( Table1, Table1[User], "Count", COUNTROWS ( Table1 ) ),
        [Count] = MAX ( Table2[Value] )
    )
)

[Omega]

hi @v-chuncz-msft,

Small question: If the count of logging changes, let's say a user have logged in 20 time, does that mean Generateseries() formula needs to be updated?