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Anonymous
Not applicable

Average of a Count with Zeros

I'm trying to create a table that shows the average number of distinct employees that visit the gym per hour each day.  A day only consists of 3 hours.  My data consists of records that are created each time an employee swipes his badge, which looks like...

Sales_Person
Day
Hour
Frank12
Dana12
Dana12
Dana13
Dave1

3

Angela23

A straight table of distinct counts by hour and day would look like...

Hour
Day
Distinct Count
110
120
212
220
312
321

Note that the above table includes hour 1, even though nobody visited the gym during hour 1 on either day 1 or day 2.  Then taking the average of the distinct counts over the Day dimension...

Hour
Avg Daily Distinct Count
10
21
31.5

This is the table I'm looking for.  What dimensions, expressions, and other properties do I need to make this table?

1 Solution

Accepted Solutions
bgerchikov
Partner - Creator III
Partner - Creator III

Good exercise!

Try this one:

sum(

Aggr(

Count(DISTINCT Day),Sales_Person,Hour))

/

Count(DISTINCT TOTAL <Sales_Person> Day)

View solution in original post

7 Replies
bgerchikov
Partner - Creator III
Partner - Creator III

Hi,

Dimemsion will be Hour, Expression:

avg(

Aggr(DISTINCT

Count( Day),Sales_Person,Hour))

Please review attached.

Good luck!

bgerchikov
Partner - Creator III
Partner - Creator III

Hi Mark,

I'm wondering whether my suggestion was helpful. If yes, please don't hesitate to mark it accordingly

Thanks!

Anonymous
Not applicable
Author

Hi Boris,

Thank you for your response, but your suggestion and your example do not achieve the desired result.  In the example you posted (with different data), the averages are not correct.  Given the data you loaded, the final table should read:

Hour
avg(
10.3333
20.6667
31.6667
40

Take Hour 2, for instance.  Your expression returns 1 as the average number of visits for Hour 2 when the actual average is 2/3.  Think of it this way:  On Day 1, there were 2 distinct visitors during Hour 2.  On Day 2, there were 0 distinct visitors during Hour 2.  On Day 3, there were 0 distinct visitors during Hour 2.  The average number of visitors during Hour 2 over Days 1-3 is (2+0+0)/3 = 2/3.

bgerchikov
Partner - Creator III
Partner - Creator III

I see... Try this one:

sum(

Aggr(DISTINCT

Count(Day),Sales_Person,Hour))

/

Count(DISTINCT TOTAL <Sales_Person> Day)

this gives you desired result

Thanks!

Anonymous
Not applicable
Author

That's very close.  There's just one more detail to work out.  I tried adding a duplicate line to your inline table and the table calculations were off.  Here's what I mean...

The line I added was

Dana, 1, 1

which is exactly like another record in the inline table.  Since I only want to count visits by distinct employees in a given hour on a given day, this should not have changed the final average calculation.  But if you try it, you can see that the calculation for Hour 1 changes from 1/3 to 2/3 because it's counting both of Dana's visits in the same hour and day.  Do you know how we can fix this?

bgerchikov
Partner - Creator III
Partner - Creator III

Good exercise!

Try this one:

sum(

Aggr(

Count(DISTINCT Day),Sales_Person,Hour))

/

Count(DISTINCT TOTAL <Sales_Person> Day)

Anonymous
Not applicable
Author

Awesome job.  Thanks Boris!