UNIX for Dummies Questions & Answers

Dear All,

I have a file which contains a column with age, which is represented in two following patterns

1. "007/A" or ''007/a" or ''7 /a" ..... In this case A or a means year and I would like to extract only the numeric values eg 7 in the above case if this pattern exits in a line of file.

2. "004/M" or "004/m" where M or m means month ...... for these lines I would like to first extract the numeric value of Month eg. 4 in this case and then convert it into a value of years which would be 0.33 eg 4 divided by 12.

Can someone please enlighten on this problem?

Appreciate your feedback.

Have a nice Sunday.

Cheers

Pawan

Please provide a sample input and desired output.

OK ... the input for example is :

Code:

"029/A" 
 "029/a" 
 "29 /A" 
 "029/M" 
 "029/m"

and the desired output should be

Code:

29
29
29
2.4
2.4

Thank you

Code:

perl -lne 'if(/(\d+)[ ]?\/a/i){$x=$1;$x=~s/^0*//;print"$x"}
if(/(\d+)[ ]?\/m/i){$y=$1;$y=~s/^0*//;printf "%0.1f\n",$y/12}' inputfile

AWK:

Code:

awk -F"/" '{gsub("\"","")}$2~"a|A"{printf ("%d\n",$1)}$2~"m|M"{printf ("%.1f\n",$1/12)}' file

Code:

awk -F'/|"' '$3~"m|M"{printf "%.1f\n", $2/12;next} {print $2+0}' infile