2

I have ~2k files that I want to extract 12 of 26 columns and save the extracted information into a new file. I have been doing this for individual files using

cat "oldfile.csv" | grep -v "=" | cut -f 1-4,9,14,15,19,21,22,24,26 > newfile.csv

There must be a way to do all files with one Unix series of commands.

Any help is greatly appreciated

4

The shell can do loops:

for i in *.csv; do 
    grep -v "=" "$i" | cut -f 1-4,9,14,15,19,21,22,24,26 > "new.$i"
done

Although, if you have embedded field separators in your fields, this will bomb. But not because of the loop.

  • 1
    $i needs quoting, or it will break on filenames containing characters in IFS. – Chris Down Jan 30 '12 at 17:07
  • Spot on. I stand corrected. – Alien Life Form Jan 30 '12 at 17:13
1

If you want to provide particular files as argument to a script e.g. script named process_csv.sh

#!/bin/bash
for arg; do
    grep -v "=" "${arg}" | cut -f 1-4,9,14,15,19,21,22,24,26 > "${arg}.new"
done

Now you can run this script as following,

./process_csv.sh *.csv OR

./process_csv.sh first.csv fifth.csv seventh.csv OR

./process_csv.sh somefile1 somefile2

Before somebody points it out, in the above script, if you do not want to preserve the arguments you can use shift in the script.

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