Remove specific characters in filename

Question

Is there any easy solution how to trim suffix in my filename? Problem is, that my suffix length is vary. Only the same string in filename is _L001.

See the example:

NAME-code_code2_L001_sufix
NAME-code_L001_sufix_sufix2_sufix3
NAME-code_code2_code3_L001_sufix_sufix2_sufix3

I need to output everything before _L001:

NAME-code_code2
NAME-code
NAME-code_code2_code3

I was thinking do something like this (when suffix is fixed length):

echo NAME-code_code2_L001_sufix | rev | cut -c 12- | rev

But of course my suffix length is vary. Is there any bash or awk solution?

Thank you.

sorry, it was typo. I just edited it. – Paul Oct 06 '16 at 12:10 — Paul, Oct 06 '16 at 12:10

Inian · Accepted Answer · 2016-10-06T12:08:35.480

Using pure string manipulation technique:-

$ string="NAME-code_code2_L001_sufix"; printf "%s\n" "${string%_L001*}"
NAME-code_code2

For all the lines int the file, you can do the same by bash, by reading the file in-memory and performing the extraction

# Setting a variable to the contents of a file using 'command-substitution'
$ mystringfile="$(<stringfile)"                 

# Read the new-line de-limited string into a bash-array for per-element operation
$ IFS=$'\n' read -d '' -ra inputArray <<< "$mystringfile"

# Run the sub-string extraction for each entry in the array
$ for eachString in "${inputArray[@]}"; do printf "%s\n" "${eachString%_L001*}"; done

NAME-code_code2
NAME-code
NAME-code_code2_code3

You can write the contents to a new-file by modifying the printf in the for loop as

printf "%s\n" "${eachString%_L001*}" >> output-file

score 2 · Answer 2 · answered Oct 06 '16 at 11:52

2

You can use _L001 as field separator in awk and print first field:

awk -F '_L001' '{print $1}' file

NAME-code_code2
NAME-code
NAME-code_code2_code3

answered Oct 06 '16 at 11:52

anubhava

761,203
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pan-mroku · Answer 3 · 2016-10-06T12:08:32.427

1

I would propose sed.

sed 's|\(.*\)_L001.*|\1|'

example:

$ for LINE in NAME-code_code2_L001_sufix NAME-code_L001_sufix_sufix2_sufix3 NAME-code_code2_code3_L001_sufix_sufix2_sufix3; do echo "$LINE"|sed 's|\(.*\)_L001.*|\1|';done
NAME-code_code2
NAME-code
NAME-code_code2_code3

edited Oct 06 '16 at 12:08

answered Oct 06 '16 at 11:50

pan-mroku

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Sorry, it was a typo :-) – Paul Oct 06 '16 at 12:02

score 1 · Answer 4 · answered Oct 06 '16 at 12:33

1

Here is grep solution: This will print lines from the start till _L001 is seen.

grep -oP '^.*?(?=_L001)' inputfile
NAME-code_code2
NAME-code
NAME-code_code2_code3

answered Oct 06 '16 at 12:33

P....

17,421
2
32
52

score 1 · Answer 5 · answered Oct 06 '16 at 14:03

Many ways to do this:

# Here is your Input text.
bash$> cat a.txt
NAME-code_code2_L001_sufix
NAME-code_L001_sufix_sufix2_sufix3
NAME-code_code2_code3_L001_sufix_sufix2_sufix3
bash$>

# Desired output using perl.
bash$> cat a.txt |perl -nle 'if (/^(.+)_L.*$/){print $1}'
NAME-code_code2
NAME-code
NAME-code_code2_code3
bash$>

# Desired output using sed.
bash$> cat a.txt |sed 's#\(.*\)_L001_.*#\1#g'
NAME-code_code2
NAME-code
NAME-code_code2_code3
bash$>

# Desired output using cut
bash$> cat a.txt |cut -f1 -d "L"|sed 's/_$//g'
NAME-code_code2
NAME-code
NAME-code_code2_code3
bash$>

score 1 · Answer 6 · answered Oct 06 '16 at 14:07

1

You can also use string substitution, something like:

for i in NAME-code_code2_L001_sufix NAME-code_L001_sufix_sufix2_sufix3 NAME-code_code2_code3_L001_sufix_sufix2_sufix3
do
    echo ${i%_L001*}
done

answered Oct 06 '16 at 14:07

M. Modugno

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Remove specific characters in filename

6 Answers6