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Babu_s22's avatar
Babu_s22
New Member
1 year ago
Solved

Need help in DAX with logic calculation

I have a customer legal id detail table which have cust_id,leagal_id,Legal_id expiry_date,review_date. I have calculated a column expired_days and bucket based on group of 30 days. Now I  want to find the count of id moving from one bucket to another bucket over month and in the same how many are newly moved id count and how many are existing id count  by bucket over month.

 

Need Help
  • Anonymous's avatar
    Anonymous
    1 year ago

    Hi Babu_s22 

     

    Please try this:

    Here's the sample data:

     

    Then add a new table:

    Table 2 = 
	VAR _vtable = ADDCOLUMNS(
		'Table',
		"_ABC", VAR _Previous = CALCULATE(
			MAX('Table'[review_date]),
			FILTER(
				ALLSELECTED('Table'),
				'Table'[cust_id] = EARLIER('Table'[cust_id]) && 'Table'[review_date] < EARLIER('Table'[review_date])
			)
		)
		RETURN
			IF(
				_Previous <> BLANK() && 'Table'[Days_Buckets] <> CALCULATE(
					SELECTEDVALUE('Table'[Days_Buckets]),
					FILTER(
						ALLSELECTED('Table'),
						'Table'[cust_id] = EARLIER('Table'[cust_id]) && 'Table'[review_date] = _Previous
					)
				),
				1,
				0
			)
	)
	RETURN
		SUMMARIZE(
			SELECTCOLUMNS(
				_vtable,
				"_Month", FORMAT(
					'Table'[review_date],
					"YYYY-mm"
				),
				"_PreviousCount", COUNTX(
					FILTER(
						_vtable,
						MONTH([review_date]) = MONTH(EARLIER('Table'[review_date])) && [_ABC] = 1
					),
					[cust_id]
				),
				"_CurrentCount", COUNTX(
					FILTER(
						_vtable,
						MONTH([review_date]) = MONTH(EARLIER([review_date]))
					),
					[cust_id]
				)
			),
			[_Month],
			[_PreviousCount],
			[_CurrentCount]
		)

    The result is as follow:

     

     

    Best Regards

    Zhengdong Xu
    If this post helps, then please consider Accept it as the solution to help the other members find it more quickly.

2 Replies

  • lbendlin's avatar
    lbendlin
    Icon for Super User rankSuper User
    1 year ago

    https://www.daxpatterns.com/new-and-returning-customers/

     

    Please provide sample data that covers your issue or question completely, in a usable format (not as a screenshot).

    Do not include sensitive information. Do not include anything that is unrelated to the issue or question.

    Need help uploading data? https://community.fabric.microsoft.com/t5/Community-Blog/How-to-provide-sample-data-in-the-Power-BI-Forum/ba-p/963216

    Please show the expected outcome based on the sample data you provided.

    Want faster answers? https://community.fabric.microsoft.com/t5/Desktop/How-to-Get-Your-Question-Answered-Quickly/m-p/1447523

  • Anonymous's avatar
    Anonymous
    Not applicable
    1 year ago

    Hi Babu_s22 

     

    Please try this:

    Here's the sample data:

     

    Then add a new table:

    Table 2 = 
	VAR _vtable = ADDCOLUMNS(
		'Table',
		"_ABC", VAR _Previous = CALCULATE(
			MAX('Table'[review_date]),
			FILTER(
				ALLSELECTED('Table'),
				'Table'[cust_id] = EARLIER('Table'[cust_id]) && 'Table'[review_date] < EARLIER('Table'[review_date])
			)
		)
		RETURN
			IF(
				_Previous <> BLANK() && 'Table'[Days_Buckets] <> CALCULATE(
					SELECTEDVALUE('Table'[Days_Buckets]),
					FILTER(
						ALLSELECTED('Table'),
						'Table'[cust_id] = EARLIER('Table'[cust_id]) && 'Table'[review_date] = _Previous
					)
				),
				1,
				0
			)
	)
	RETURN
		SUMMARIZE(
			SELECTCOLUMNS(
				_vtable,
				"_Month", FORMAT(
					'Table'[review_date],
					"YYYY-mm"
				),
				"_PreviousCount", COUNTX(
					FILTER(
						_vtable,
						MONTH([review_date]) = MONTH(EARLIER('Table'[review_date])) && [_ABC] = 1
					),
					[cust_id]
				),
				"_CurrentCount", COUNTX(
					FILTER(
						_vtable,
						MONTH([review_date]) = MONTH(EARLIER([review_date]))
					),
					[cust_id]
				)
			),
			[_Month],
			[_PreviousCount],
			[_CurrentCount]
		)

    The result is as follow:

     

     

    Best Regards

    Zhengdong Xu
    If this post helps, then please consider Accept it as the solution to help the other members find it more quickly.