how to extract substring and numbers only using gr

Question

I have a text file containing both text and numbers, I want to use grep to extract only the numbers I need for example, given a file as follow:

miss rate 0.21  
ipc 222  
stalls n shdmem 112

So say I only want to extract the data for miss rate which is 0.21. How do I do it with grep or sed? Plus, I need more than one number, not only the one after miss rate. That is, I may want to get both 0.21 and 112. A sample output might look like this:

0.21 222 112

Cause I need the data for later plot.

Answer 1

Use awk instead:

awk '/^miss rate/ { print $3 }' yourfile

To do it with just grep, you need non-standard extensions like here with GNU grep using PCRE (-P) with positive lookbehind (?<=..) and match only (-o):

grep -Po '(?<=miss rate ).*' yourfile

Answer 2

I believe

sed 's|[^0-9]*\([0-9\.]*\)|\1 |g' fiilename

will do the trick. However every entry will be on it's own line if that is ok. I am sure there is a way for sed to produce a comma or space delimited list but I am not a super master of all things sed.

Answer 3

The grep-and-cut solution would look like:

to get the 3rd field for every successful grep use:

grep "^miss rate " yourfile | cut -d ' ' -f 3

or to get the 3rd field and the rest use:

grep "^miss rate " yourfile | cut -d ' ' -f 3-

Or if you use bash and "miss rate" only occurs once in your file you can also just do:

a=( $(grep -m 1 "miss rate" yourfile) )
echo ${a[2]}

where ${a[2]} is your result.

If "miss rate" occurs more then once you can loop over the grep output reading only what you need. (in bash)

Answer 4

You can use:

grep -P "miss rate \d+(\.\d+)?" file.txt

or:

grep -E "miss rate [0-9]+(\.[0-9]+)?"

Both of those commands will print out miss rate 0.21. If you want to extract the number only, why not use Perl, Sed or Awk?

If you really want to avoid those, maybe this will work?

grep -E "miss rate [0-9]+(\.[0-9]+)?" g | xargs basename | tail -n 1

Answer 5

If you really want to use only grep for this, then you can try:

grep "miss rate" file | grep -oe '\([0-9.]*\)'

It will first find the line that matches, and then only output the digits.

Sed might be a bit more readable, though:

sed -n 's#miss rate ##p' file

Answer 6

Using the special look around regex trick \K with pcre engine with grep :

grep -oP 'miss rate \K.*' file.txt

or with perl :

perl -lne 'print $& if /miss rate \K.*/' file.txt