How can I extract specific part of a file?

Question

I have multiple files containing several different lines. Among the lines, I am interested only in the ones starting with a specific pattern, such as:

USER1    Info> :FERRARI:RED,:LAMBORGHINI:ORANGE,:MASERATI:BLUE,...

In this example (with spaces as indicated).

From these lines I only need the car's make info (without the colours) so the output should look like this:

FERRARI, LAMBORGHINI and MASERATI

The car's make are always different in each file. In the example I put 3 car's make, but in each file they could be less or more. Is there an easy way to do this in bash or python? Thanks

Answer 1

To extract every word (not contains comma) which is between the two colons

grep -oHnE ":[^,]*:" files* | awk 'BEGIN{FS=":"} {x=$1$2; a[x]=a[x]","$4} END{for(x in a) print a[x]}' | sed 's/^,//'

Assuming we have file car_info.txt

cat car_info.txt
USER1    Info> :FERRARI:RED,:LAMBORGHINI:ORANGE,:MASERATI:BLUE
USER1    Info> :FERRARI:RED,:LAMBORGHINI:ORANGE
USER1    Info> :FERRARI:RED,:LAMBORGHINI:ORANGE,:MASERATI:BLUE


grep -oHnE ":[^,]*:" car_info.txt 
car_info.txt:1::FERRARI:
car_info.txt:1::LAMBORGHINI:
car_info.txt:1::MASERATI:
car_info.txt:2::FERRARI:
car_info.txt:2::LAMBORGHINI:
car_info.txt:3::FERRARI:
car_info.txt:3::LAMBORGHINI:
car_info.txt:3::MASERATI:

grep

• -o prints only the matched parts of a matching line
• -H prints filenames
• -n prints line numbers
• -E to support extended-regex

Now the strategy is to make the lines with same filename:line to be shown in one line

awk 'BEGIN{FS=":"} {x=$1$2;a[x]=a[x]","$4} END{for(x in a) print a[x]}'

• BEGIN{FS=":"} setting the field separator to :
• x=$1$2 put filename and line in x
• create a key-value a with key=x and adding the 4th field to it's value in each line with same x
• for(x in a) print a[x] print the values of the key-value a

sed 's/^,//' is for removing , from the begining of lines

Answer 2

awk -F':' /^USER1.\*Info/' {print $2" "$4" "$6}' < infile