Find the Distinct Count of most common day for products used by day of the week (1-7) distinct count

zgoldflo

New Member
Joined
Dec 13, 2021
Messages
30
Office Version
  1. 365
Platform
  1. Windows
I have bike share data that indicates the day of the week a bike was used in number format 1-7 (indicates days of the week). I want to figure out the most common days of the week a specific route was most popular i.e. for each specific route which (number or) day was most common. I believe what I need to do is a distinct count. I have organized my data into a pivot table for the top ten monthly routes. I want to figure out throughout the month, which day of the week was most common for each route.

I have tried to get a distinct count, but for some reason, that option is grayed out and does not let me select it within the Pivot Table.

Here is the Pivot Table.

_12_2020.xlsx
H
10
Dec_2020_pt


Here is the source data.

_12_2020.xlsx
I
700:22:34
Dec_2020


Below is an image of the Distinct Count being grayed as an option in my Pivot Table
 

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  • Screenshot 2021-12-31 085912.jpg
    Screenshot 2021-12-31 085912.jpg
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Excel Facts

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Excel columns run from A to Z, AA to AZ, AAA to XFD. The last column is XFD.
Hi

Distinct count can only be used with a datamodel pivot table.
But you can't change a normal pivot table into a datamodel pivot table - you have to create it from scratch.
 
Upvote 0
Hi. What about the formula or ability to do that from the source data?

Also not sure what the difference is between the two pivot table mentioned above.
 
Upvote 0
So change the source data into a table? Then distinct count will do the trick for me?
 
Upvote 0
That was a question.

If I change the data source into a Table, create a Pivot Table from there, will I be able to then use Distinct Count to be able to solve the above question I asked?
 
Upvote 0
It seems to me you didn't follow my links and read the articles and viewed the picture.

An Excel table is always a good start but it does not able discount count in a pivot table.
 
Upvote 0

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