ARRAYS- How to pick last 2 non-missing values and pick corresponding vars based on their positions

[ahhh]

Hi, Need help solving this-

My dataset looks like this:

data have;
input id value1 value2 value3 value4 value5 dt1 :date9. dt2 :date9. dt3 :date9. dt4 :date9. dt5 :date9.;
format dt1 dt2 dt3 dt4 dt5 date9.;
datalines;

1 300 400 60 . 40 01feb2015 03mar2016 21apr2016 . 21may2016
2 1200 300 40 40 . 05jan2012 04feb2012 15apr2013 23apr2014 .
3 30 60 100 . 200 09dec2011 10jan2011 15feb2011 . 16apr2013
4 . . 40 30 200 . . 12feb2011 16mar2011 17apr2012
5 200 200 200 20 20 10sep2010 18oct2010 13dec2010 23apr2013 26oct2014
;
run;

dt1 corresponds to value1, date2 corresponds to value2 etc..

I need to pull last two non-missing 'value' variables if they are under a certain value -say 100 and then their corresponding dates and calculate difference period in days between them 

This is what I want:

For ids 3 and 4: they don't qualify as atleast one of the last two non-missing values is greater than 100.

id lastnonmiss last2ndnonmiss dt_lastnonmiss dt_last2ndnonmiss timediffindays

1 40                 60                      21may2016       21apr2016            30

2 40                 40                      23apr2014        15apr2013             373

5 20                 20                      26oct2014         23apr2013             551

I tried using Coalesce function (which helped me pick the last non-missing and last 2nd nonmissing values) but unable to pick corresponding dates. Also, needed to include the initial condition that the last and 2nd last non-missing values should be less than 100. Please help ASAP! Much appreciated!!

[PeterClemmensen]

Can you show us the code you have tried so far?

Here is a brute force approach

data want;
   format lastnonmiss last2ndnonmiss dt_lastnonmiss dt_last2ndnonmiss timediffindays;

   set have;
   array v {5} value5-value1;
   array d {5} dt5-dt1;

   lastnonmiss=coalesce(of v[*]);
   dt_lastnonmiss=d[whichn(lastnonmiss, of v[*])];

   v[whichn(lastnonmiss, of v[*])]=.;

   last2ndnonmiss=coalesce(of v[*]);
   dt_last2ndnonmiss=d[whichn(last2ndnonmiss, of v[*])];

   if lastnonmiss <= 100 & last2ndnonmiss <= 100;

   timediffindays=dt_lastnonmiss-dt_last2ndnonmiss;

   format dt: date9.;
   keep lastnonmiss last2ndnonmiss dt_lastnonmiss dt_last2ndnonmiss timediffindays;
run;

Result:

lastnonmiss last2ndnonmiss dt_lastnonmiss dt_last2ndnonmiss timediffindays 
40          60             21MAY2016      21APR2016         30 
40          40             23APR2014      15APR2013         373 
20          20             26OCT2014      23APR2013         551 

[ahhh]

Thank you!

[KachiM · Accepted solution]

@ahhh 

Keep your VALUES in one array.

Keep your DATES in another array.

Have an array sized to 2 for holding the selected VALUES.

Similarly use an array sized to 2 to hold the corresponding DATES.

data have;
input id value1 value2 value3 value4 value5 dt1 :date9. dt2 :date9. dt3 :date9. 
                           dt4 :date9. dt5 :date9.;
format dt1 dt2 dt3 dt4 dt5 date9.;
datalines;

1 300 400 60 . 40 01feb2015 03mar2016 21apr2016 . 21may2016
2 1200 300 40 40 . 05jan2012 04feb2012 15apr2013 23apr2014 .
3 30 60 100 . 200 09dec2011 10jan2011 15feb2011 . 16apr2013
4 . . 40 30 200 . . 12feb2011 16mar2011 17apr2012
5 200 200 200 20 20 10sep2010 18oct2010 13dec2010 23apr2013 26oct2014
;
run;

data want;
   set have;
   array v value1 - value5;
   array d dt1 - dt5;
   array tv[2] _temporary_;
   array td[2] _temporary_;
   count = 0;
   do i = dim(v) to 1 by -1;
      if v[i] = . then continue;
      if v[i] Ge 100 then leave;
      count + 1;
      tv[count] = v[i];
      td[count] = d[i];
         
      if count = 2 then do;
         datediff = td[1] - td[2];
         output;
         leave;
      end;
   end;
drop count i;
run;

The output is:

[ahhh]

Thank you!