Extract variables from filenames and output to file


 
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# 1  
Old 04-13-2011
Extract variables from filenames and output to file

I need some help. I have a list of files (thousands) and would like to extract some variables from the file name and save that to a file

The list of files look like:
Quote:
e20012110129_xform_azisum_mlt_2.25_0.50.txt
e20012130535_xform_azisum_mlt_2.25_0.50.txt
e20012160231_xform_azisum_mlt_2.25_0.50.txt
e20012210636_xform_azisum_mlt_2.25_0.50.txt
e20012230332_xform_azisum_mlt_2.25_0.50.txt
e20012240738_xform_azisum_mlt_2.25_0.50.txt
e20012270434_xform_azisum_mlt_2.25_0.50.txt
e20012320839_xform_azisum_mlt_2.25_0.50.txt
I am trying to write the following script but I am stuck at how I can get thevariables 'doy' and 'yr' from each file and then combine into one file with two columns "yr doy' then write that to a file

Quote:
#! /bin/bash

# if there's an error, using a number greater than 0 will exit with that
# number
EXIT_ON_ERROR=1
IFS=$'\n'

dirctry=~/Research/selected
cd $dirctry/

# process the rest of files
ls *.txt | sort -rn > imputFiles.dat
while read fliz ; do
if [ -e "$filz" ] ; then
# Extract a part of the filename for naming the output file
filenme=$filz

doy=`echo ${filenme:5:3}`
yr=`echo ${filenme:1:4}`
hr=`echo ${filenme:8:2}`
min=`echo ${filenme:10:2}`
# I need to get "yr" and "doy" for each file combine into one file with two columns
and output that to a file like shown below

else
echo "ERROR: Couldn't open file '$filz'" >&2
if [ $EXIT_ON_ERROR -gt 0 ] ; then
exit $EXIT_ON_ERROR
fi
fi
done < imputFiles.dat
And the required output is
Quote:
2001 211
2001 213
2001 216
2001 221
2001 223
2001 224
2001 227
2001 232
Please any help and ideas will be highly appreciated
Thank you
# 2  
Old 04-13-2011
A single line of perl will do, let's say you save your file names to the file named 'data.txt'
Code:
 perl -pe  's/^.(\d{4})(\d{3}).*$/$1 $2/' data.txt

# 3  
Old 04-13-2011
Hi kevintse,

Thank you! That indeed does what I need. I don't fully understand what the fine detail of how this line does the job.

I have one extra question, what if I want to add another column to the output which is some value extracted from inside the file, would this be done within a command like this??

Thank you
# 4  
Old 04-13-2011
What I used in the one liner is called Regular Expression.
Code:
perl -pe  's/^.(\d{4})(\d{3}).*$/$1 $2/' data.txt

The slashes are just separators, the syntax is: s/Regex/replacement/
^ matches the very start of the string(a line in data.txt).
. matches a single character(any character).
(\d{4}) \d represents number from 0 to 9, 4 in the curly braces means the pattern will match 4 numbers, the outer braces capture the 4 numbers(yr you want), this is called a group in Regular Expression.
(\d{3}) is roughly the same as the previous pattern.
$1 $2 prints the 1st and 2nd groups captured by the pattern.

For your last question, yes, Perl can easily achieve what you want but may not be as easy as the previous command, shouldn't be complicated though
# 5  
Old 04-13-2011
Hi kevintse,

Thank you for that explanation and the time you have taken to help. Much appreciated. I am still figuring out how I will add another column there. One way I am thinking is if it is possible to put the 'filename' directly into this line command so that the command does not read the 'filename' from the file called data.

I thought something like this would work...
Quote:
perl -pe 's/^.(\d{4})(\d{3}).*$/$1 $2/' 'e20012110129_xform_azisum_mlt_2.25_0.50.txt'
so that the output is
Quote:
2001 211
then I can add another variable as a third column

Any advice will be appreciated
Thank you
# 6  
Old 04-14-2011
You may pipe the filename to the perl script:
Code:
echo 'e20012110129_xform_azisum_mlt_2.25_0.50.txt' | perl -pe 's/^.(\d{4})(\d{3}).*$/$1 $2/'

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