Hi SAS users

Many Thanks for helping me out in my last post.

I have a tricky programming problem that I would like to ask.

Not sure if using array will be a better option here here it's.

Attach is a data set where we have account number and debt outstanding due in number of days (current due -- 150 days).

accno Curr_due 30Days 60Days 90Days 120Days 150day last_val

153553 0.18 0 0 0 0 0 0.18 167743 0 0 0 0 0 0.26 0.26 198358 0 0.4 0 0.13 0 0 0.13 451646 0 0 0 0 0.66 0 0.66 477051 0 0 3.6 3.83 0 0 3.83 1106116 0 0 289.9 0 0 0 289.9

The aim is to create a last_val column where it will search thru all the current_due--onefifty_days horizontally and look for the last value that is not zero for each accno and place that value in the last_value column. Sound simple but I having problem programming it.

Below is part of my working to achieve this but it proving more difficult than I first thought. I need to have a stopping criteria that will stop the array from looping if it found the last debt aging variable.

Say

accno i debt_aging 153553 1 0.18 153553 2 0 153553 3 0 153553 4 0 153553 5 0

then the loop will stop at i=1 and put 0.18 in the last_val column.

libname tmp 'C:\';

data dataIn(keep = accno thirty_days sixty_days ninety_days onetwenty_days onefifty_days); set tmp.paul; run;

data Array_1(keep = accno myTestVar i); set dataIn; array testArray (5) thirty_days sixty_days ninety_days onetwenty_days onefifty_days; do i = 1 to 5; myTestVar = testArray(i); if myTestVar > 0 then output; end; run;

data array_2(keep = accno myTestVar i); set dataIn; array testArray (5) thirty_days sixty_days ninety_days onetwenty_days onefifty_days; do i = 1 to 5; myTestVar = testArray(i); output; end; run;

Not sure if array is the best way to go about solving this problem, but any suggestion is most welcome.

Many Thanks in advance

Best Regards

Paul