Converting CSV to Column separated values but placing the values most right - shell

Question

Concept is a debian running computer with an SH script file in retrieving data.
I have searched on how to convert data from csv back to columns and rows. This was fairly easy.
for instance, i have this:
And using this command: column -t -s, > this will gives me this:
Date SkyT_Min SkyT_Min_Time SkyT_Max SkyT_Max_Time SkyT_Mean AmbT_Min AmbT_Min_Time AmbT_Max AmbT_Max_Time
2019-09-19 -22.8 07:29:48.00 -1.6 12:27:57.00 -16.77 5.9 05:15:11.00 23.4 14:28:49.00
2019-09-25 -15.8 11:49:40.00 9.1 20:17:11.00 1.07 14.7 02:47:10.00 21.7 11:46:38.00
2019-09-26 -9.6 02:59:29.00 10.8 11:09:18.00 5.66 16.4 20:58:37.00 23.4 14:08:58.00
So overall that is already great!
But how do i get it to get the values in each column to start on the far right, instead on the left?
So i would get something like this:
Thank you very much for your answer.
UPDATE:
I have found the answer to my question:
--R
But it gives me errors: "invalied option -- 'R'
So if anyone has any idea. :)

Answer 1

You can work with awk:
-F tells awk the column separator
"%20s " tells awk to reserve 20 characters for each column and to concat a whitespace to each entry
awk -F "," '{for(i=1;i<=NF;i++){printf "%20s ", $i}; printf "\n"}' inputfile.csv
Output:
Date SkyT_Min SkyT_Min_Time SkyT_Max SkyT_Max_Time SkyT_Mean AmbT_Min AmbT_Min_Time AmbT_Max AmbT_Max_Time
2019-09-19 -22.8 07:29:48.00 -1.6 12:27:57.00 -16.77 5.9 05:15:11.00 23.4 14:28:49.00
2019-09-25 -15.8 11:49:40.00 9.1 20:17:11.00 1.07 14.7 02:47:10.00 21.7 11:46:38.00
2019-09-26 -9.6 02:59:29.00 10.8 11:09:18.00 5.66 16.4 20:58:37.00 23.4 14:08:58.00
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