Linux: How can i extract information from the files's name?

Question

I have some files(.txt), the name are:

run_freq100ampl0.6offs0.8_1
run_freq100ampl0.4offs1.1_1
run_freq100ampl0.2offs1.0_1
run_freq100ampl0.4offs1.2_1
run_freq100ampl0.2offs1.1_1
run_freq100ampl0.5offs0.8_1
run_freq100ampl0.2offs1.2_1
run_freq100ampl0.4offs0.8_1

I want get the number of ampl, like

0.6
0.4
0.2
0.4
0.2
0.5
0.2
0.4

How can i do it with 'find', 'grep' or 'cat' in Linux?

score 1 · Answer 1 · answered Oct 09 '13 at 13:32

1

Using grep:

grep -Po '(?<=ampl).*(?=offs)' input

Example:

$ grep -Po '(?<=ampl).*(?=offs)' <<< "run_freq100ampl0.2offs1.1_1"
0.2

Using sed:

$ sed 's/.*ampl\(.*\)offs.*/\1/' <<< "run_freq100ampl0.2offs1.1_1"
0.2

answered Oct 09 '13 at 13:32

devnull

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Thank you! But i want to show all values one time. How can i do it? – user2862989 Oct 09 '13 at 13:46
@user2862989 Either say `sed 's/.*ampl\(.*\)offs.*/\1/' filename` or `grep -Po '(?<=ampl).*(?=offs)' filename` – devnull Oct 09 '13 at 13:47

score 0 · Answer 2 · answered Oct 09 '13 at 15:32

0

Using awk

awk -F"ampl|off" '{print $2}'
0.6
0.4
0.2
0.4
0.2
0.5
0.2
0.4

Or using fixed position

awk '{print substr($0,16,3)}'

answered Oct 09 '13 at 15:32

Jotne

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score 0 · Answer 3 · answered Oct 09 '13 at 19:36

0

for f in *.txt
do
  f=${f%offs*}
  f=${f#*ampl}
  echo ${f}
done

answered Oct 09 '13 at 19:36

twalberg

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Linux: How can i extract information from the files's name?

3 Answers3