Putting the output into a table

Question

I know this is a FAQ but I've already read multiple pages about this and I can't get it to work. I'm using the grep command to locate a range of values from three different text files.

Command line:

grep -o "201[1-4].[0-9][ -9]" file1.txt file2.txt file3.txt

Output:

file1.txt:2012.69
file1.txt:2013.44
file2.txt:2013.17
file2.txt:2012.6    
file2.txt:2013.75
file2.txt:2013.08
file3.txt:2013.54
file3.txt:2013.9    

I want the output to have the names of the three files as the first row (header) and then all of the values found to be in each column. I think the problem is that I am unable to predict how many rows I would need.

Answer 1

You can use awk to join the output of grep to one row per file, then transpose with csvtool and optional column -nt to make the output nice (note, that -n is a Debian extension and might not be available for your platform).

grep -o "201[1-4].[0-9][ -9]" file1.txt file2.txt file3.txt \
| awk -F: '
  {
    if (!s[$1]){ s[$1]=$2 }
    else { s[$1]=s[$1]","$2 }
  }
  END {
    for (f in s){ print f","s[f] }
  }' \
| csvtool transpose -u " " - \
| column -nt

Output:

file1.txt  file3.txt  file2.txt
2012.69    2013.54    2013.17
2013.44    2013.9     2012.6
                      2013.75
                      2013.08

You might need to install csvtool.