Selecting ID's with 3 consecutive Observations

[SNG1]

Hi All,

I have a Data Set which kind of looks like below - 

Loan_ID	Sys_Date	Payment	Scheduled Payment	Flag
4	8-Feb-18	$135.96	$135.96	Y
4	22-Feb-18	$135.96	$135.96	Y
4	8-Mar-18	$135.96	$135.96	Y
4	22-Mar-18	$135.96	$135.96	Y
4	5-Apr-18	$135.96	$135.96	Y
4	19-Apr-18	$135.96	$135.96	Y
4	3-May-18	$135.96	$135.96	Y
4	17-May-18	$135.96	$135.96	Y
4	31-May-18	$135.96	$135.96	Y
4	14-Jun-18	$135.96	$135.96	Y
4	28-Jun-18	$135.96	$135.96	Y
4	12-Jul-18	$135.96	$135.96	Y
4	26-Jul-18	$4,702.18	$135.96	Y
5	15-Feb-18	$0.00	$137.52	N
5	1-Mar-18	$137.52	$137.52	Y
5	15-Mar-18	$137.52	$137.52	Y
5	29-Mar-18	$137.52	$137.52	Y
5	12-Apr-18	$0.00	$137.52	N
8189	17-Aug-18	$51.44	$51.44	Y
8189	31-Aug-18	$51.44	$51.44	Y
8189	14-Sep-18	$0.00	$51.44	N
8189	28-Sep-18	$51.44	$51.44	Y
8189	12-Oct-18	$0.00	$51.44	N
8189	26-Oct-18	$0.00	$51.44	N
8189	9-Nov-18	$51.44	$51.44	Y
8189	15-Nov-18	$202.88	$51.44	Y
8189	23-Nov-18	$51.44	$51.44	Y
8189	7-Dec-18	$51.44	$51.44	N
8189	21-Dec-18	$51.44	$51.44	N
9999	17-Aug-18	$99.30	$99.30	Y
9999	31-Aug-18	$0.00	$99.30	N
9999	5-Sep-18	$99.30	$99.30	Y
9999	14-Sep-18	$99.30	$99.30	Y
9999	28-Sep-18	$0.00	$99.30	N
9999	29-Sep-18	$99.30	$99.30	Y
9999	12-Oct-18	$99.30	$99.30	Y
9999	16-Oct-18	$49.30	$99.30	Y
9999	25-Oct-18	($149.30)	$99.30	N
9999	26-Oct-18	$99.30	$99.30	Y
1111	28-Sep-18	$50.00	$50.00	Y
1111	29-Sep-18	$50.00	$50.00	Y
1111	12-Oct-18	$30.00	$50.00	N
1111	16-Oct-18	$0.00	$50.00	N
1111	25-Oct-18	$50.00	$50.00	Y
1111	26-Oct-18	$0.00	$50.00	N

Loan_ID,

Sys_Date,

Payment (which is the Paid Amount),

Scheduled Payment (which is the scheduled Payment amount for the Loan),

Flag (Y = Successful Payment; N = Unsuccessful Payment).

From the data set below I only need the ID's which have 3 consecutive payments (given the condition that PAYMENT >= SCHEDULED_PAYMENT)

For Example from the above data I only need to keep the following ID's (Loan_id 4,5 & 8189)

I need to keep loan_ID 4 as it meets 3 consecutive Payments (I.E. Each Payment is >= Scheduled Payment and as per Y in the flag field they were all Successful)

I need to keep loan_ID 5 AS THIS ALSO HAS 3 CONSECUTIVE PAYMENTS (i.e. Three times FLAG = Y and payment >= scheduled payment)

I need to keep 8189 this also has an instance of three consecutive payments (i.e. i.e. Three times FLAG = Y and payment >= scheduled payment)

I need to EXCLUDE 9999 as this does not meet my criteria, you can see that even though there are times FLAG = Y  but one of the amounts in Payments ($49.30) is < Scheduled Payment ($99.30) therefore this has to be excluded.

i NEED TO EXCLUDE 1111 AS IT DOES NOT MEET THE 3 Consecutive payments criteria

I have tried different variations but I am unable to get to the final result which should only Show 

Loan_ID

4

5

8189

Any help in solving this dilemma is much appreciated!

[ballardw]

You should provide an example of what you expect the actual output data set to look like given that input data. I am not sure if you are expecting 3 records or many more.

Also, can you try to clean up your definition of "consecutive". Listing examples doesn't necessarily provide the rule(s) needed.

You say:

From the data set below I only need the ID's which have 3 consecutive payments 
(given the condition that PAYMENT >= SCHEDULED_PAYMENT)

but then wander off with Flag values in the examples. So you should data the flag condition a bit more explicitly.

[SNG1]

Thanks ballardw for the response - 

The actual output data just needs to be the Loan_ID that meets the condition - 

3 Consecutive Payments where each payment is >= scheduled payment 

I just created the FLAG (i.e. Y or N) so that it would be easier to understand.

Basically I only need to keep the Loan_ID that has 3 consecutive Y and for each of the 3 the Payment should be >= scheduled payment

So my output should be only the Loan_id'S that satisfy the above and for this data set it is 4,5,8189

Hope this helps

[Patrick]

Below (untested) code should do the job. Source table "have" must be pre-sorted by loan_id sysdate.

data want(keep=loan_id);
  set have;
  by loan_id sys_date;
  if first.loan_id then count=0;
  if count<3 then
    do;
      if flag='Y' and payment>=scheduled_payment then count+1;
      else count=0;
    end;
  if last.loan_id and count=3 then output;
run;

[ChrisNZ]

No data is given, so this untested, but should do roughly what you want:

data WANT ;   
  COUNT=0;
  do until(last.ID); 
    set HAVE;
    by ID notsorted;
    if FLAG='Y' & PAYMENT >= SCHEDULED then COUNT+1;
    else if COUNT <3 then COUNT=0;              
  end;       
  do until(last.ID); 
    set HAVE;
    by ID notsorted;
    if COUNT >= 3 then output; 
  end;
run;

[Edit: code altered slightly]