Sed help / extracting useful bits from a line with sed

Question

I need to parse some log files extracting only the numerical parts of interest.  Example line:
x264 [info]: SSIM Mean Y:0.9689320 (15.077db)
Two goals here:
1) Capture the number after the "Y:" --> 0.9689320
2) Capture the number in () without the db --> 15.077
I thought I'd attack it using sed to find anything up to Y: and delete it, then anything after a space and delete it, but find myself unable to do it.  Suggestions are welcomed.  I can do it with awk/sed combo but want to learn a better way.
awk '{print $5}' | sed 's/Y://'
Last edited by graysky (2011-04-10 17:35:21)

Answer 1

@graysky - As others have said use sed backreferences. Here is an another example.
sed 's/^.*:\([[:alnum:]]*\.[[:alnum:]]*\) [^\(]*(\(.*\)db)[^\)]*$/\1 \2/'
Note: disraptor's regex is much easier to read. I'd use it! I just whipped this up before I saw his
In:
echo "x264 [info]: SSIM Mean Y:0.9689320 (15.077db)" | sed 's/^.*:\([[:alnum:]]*\.[[:alnum:]]*\) [^\(]*(\(.*\)db)[^\)]*$/\1 \2/'
Out:
0.9689320 15.077
In awk: (One way of doing it!)
mawk '{gsub(/[Y\:\(\)db]/,""); print $5,$6}'
In:
echo "x264 [info]: SSIM Mean Y:0.9689320 (15.077db)" | mawk '{gsub(/[Y\:\(\)db]/,""); print $5,$6}'
Out:
0.9689320 15.077
This was quick but I hope it gets you on the right path
Edit:
How about just grep?
echo $(grep -o '[0-9]\.[0-9]*')
In:
echo "x264 [info]: SSIM Mean Y:0.9689320 (15.077db)" | echo $(grep -o '[0-9]\.[0-9]*')
Out:
0.9689320 5.077
Cheap, but I'd thought I would add it anyway!
P.S. Here is the full way to extract both numbers using awk's substr function as disraptor mentioned before.
mawk '{print substr($5,3),substr($6,2,6)}'
Last edited by harryNID (2011-04-11 20:08:31)