Rows per day

[Anonymous]

I would like to know how many rows I get per day per id

Input 

Document.deveui		Document.time
00-0b-57-ff-fe-b5-54-9a		2020-07-15T17:17:35.275759Z
00-0b-57-ff-fe-b5-54-9a		2020-07-17T19:30:17.743033Z
00-0b-57-ff-fe-b5-54-9a		2020-07-17T19:31:51.218990Z
00-0b-57-ff-fe-b5-54-9a		2020-07-17T19:31:52.638143Z
00-0b-57-ff-fe-b5-54-9a		2020-07-15T01:17:30.122525Z
00-0b-57-ff-fe-b5-54-ff		2020-07-15T00:17:36.310303Z
00-0b-57-ff-fe-b5-54-ff		2020-07-15T01:17:30.922525Z
00-0b-57-ff-fe-b5-54-ff		2020-07-17T19:31:51.218990Z
00-0b-57-ff-fe-b5-54-ff		2020-07-17T19:31:52.638143Z

Needed output 

		readings
00-0b-57-ff-fe-b5-54-9b		2
00-0b-57-ff-fe-b5-54-9b		3
00-0b-57-ff-fe-b5-54-ff		2
00-0b-57-ff-fe-b5-54-ff		2

[harshnathani]

Hi @Anonymous ,

Create a Calculated Column

Date = DATE(YEAR('Table'[Document.time]), MONTH('Table'[Document.time]), DAY('Table'[Document.time]))

Regards,
Harsh Nathani
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[amitchandak]

@Anonymous , you need to create a date column first

Date = [Document.time].date

then you have plot Document.deveui , Date, and count(Table[Document.deveui]) on a visual

[Greg_Deckler]

@Anonymous - Your sample source data does not seem to match your sample output. Do you want to average the number of rows an item has per day?

[lbendlin]

Convert Document.time to a date type or create a calculated column with the date value of Document.time.

Then you can add that new column to the visual and achieve the desired result.