Just use PROC TRANSPOSE and you don't need to worry about using ARRAYs.   First convert your not-wide and not-all dataset into a TALL dataset. data HAVE; input ord_veh _name_ $ Col1 $ Col2 $ Col3 $ Col4 $ Col5 $ ; datalines; 1 Capacity . . . . . 1 Weight . . . . . 2 Capacity . . . . . 2 Weight . . . . . 3 Capacity A A A A A 3 Weight B B B B B ; proc transpose data=have out=tall name=variable; by ord_veh _name_ ; var col1-col5; run; You could probably just stop here as this TALL structure is a much more useful format to work with.   But to get to your wanted dataset just add another TRANSPOSE. proc transpose data=tall out=want(drop=_name_); by ord_veh; id _name_ variable; var col1; run; If you don't like that the variables are named WEIGHTCOL`1 instead of WEIGHT1 then just add a data step to remove the text  COL from the value of variable named VARIABLE. data tall set tall; variable =substr(variable,4); run; ... View more

To convert a string to a number you need to use and informat, not a format.  Formats convert values to text. Informats convert text to values.   You can use the E8601DZ informat to convert that string into a datetime value.  You can use the TIMEPART() function to extract just the time of day component from the datetime value. Note that you cannot store that many decimal places in a datetime value.  The number of decimal digits required would be more than SAS can store exactly in the binary floating point format it uses to store numbers. 15 data test; 16 string = '2022-05-16T02:30:00.2159663Z'; 17 dt = input(string,E8601DZ30.); 18 time = timepart(dt); 19 format dt datetime29.7 time tod16.7 ; 20 put (_all_) (=/); 21 put dt = 32.8 ; 22 run; string=2022-05-16T02:30:00.2159663Z dt=16MAY2022:02:30:00.2159662 time=02:30:00.2159662 dt=1968287400.21596000 ... View more