Naming number to months

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# 1  
Old 01-06-2014
Naming number to months


I have many files of monthly means from model simulation such as 1,2,3,4.....12, corresponding to jan, feb, mar,....., dec.
e.g.,, .............,
i want to write a shell script such that it should conveet number into months and year such as..

how can i do it using shell script ?

# 2  
Old 01-06-2014
perl -npi.bak -e 'BEGIN{@mon=qw(null jan feb mar apr may jun jul aug sep oct nov dec);} s/avg_(\d).nc/avg_$mon[$1].nc/g;' monthly_means.dat

Remove the .bak to stop creating backups when you're satisfied it works as expected.
# 3  
Old 01-06-2014
Are you trying to rename these files then?

Well, as a starter I could suggest the case statement. Assuming that you are sh, ksh or bash it would be something like this:-
case month_num in
   1) month_name=jan ;;
   2) month_name=feb ;;
   3) month_name=mar ;;
   4) month_name=apr ;;
   5) month_name=may ;;
   6) month_name=jun ;;
   7) month_name=jul ;;
   8) month_name=aug ;;
   9) month_name=sep ;;
   10) month_name=oct ;;
   11) month_name=nov ;;
   12) month_name=dec ;;
   *) echo "Invalid month value" ;;

You will need to chop out the number from the incoming file name as month_num and the afterwards, you can use $month_name as you wish.

I hope that this helps. Let us know how you get on and if you need any more guidance.


Last edited by rbatte1; 01-06-2014 at 06:04 AM.. Reason: Spling mustike
# 4  
Old 01-06-2014

For example, I have 24 files corresponding to JAN to DEC 2001 and JAN to DEC 2002.

the files are avg_2, .......

I want to rename it as

avg_jan_2001, avg_feb_2001,...........,

I am trying it for first year as...

while [ $year -le $year_en ]
 for var in jan feb mar apr may jun jul aug sep oct nov dec
      echo $var_$year
     year = $(( year_st+1 ))

But it gives me only 2001 without end.

can anybody modify this script for two years ?


Last edited by Franklin52; 01-06-2014 at 07:33 AM.. Reason: Please use code tags
# 5  
Old 01-06-2014
A few issues here:-
  • Your value for year_en is 2001 so the range is 2001 to 2001
  • You increase the value year by adding one to the value year_st, which means you could have an infinite loop if year_st does not equal year_en
  • You should really use { and } to wrap your variables when you are using them, such that your echo becomes echo ${var}_${year}Does this give you anything further to work on?

    I hope that this helps you.

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