How to grep two numbers from the same line at different places using bash?

Solution 1:

One possible way...

% grep 'solver.cpp:229' ExampleFile.txt | cut -d ' ' -f 3,6 | tr -d ','
2000 0.305721
2020 0.294722

Solution 2:

I lost grep but here it is with sed

$ sed -nr 's/.*Iteration ([0-9]+).*loss.*( [0-9]+.*)/\1\2/p' ExampleFile.txt
2000 0.305721
2020 0.294722

• -n don't print until we ask for something
• -r use ERE so I don't have to escape the () and + metacharacters
• s search and replace /old/new/
• .* matches any (or no) characters
• ([0-9]+) parentheses to keep this part of the pattern [0-9] a number + one or more occurrences of the preceding character.
• \1\2 backreferences to the patterns saved earlier with parentheses
• p print the bits we want to see

If the output is what you want, redirect it to your outfile:

sed -nr 's/.*Iteration ([0-9]+).*loss.*( [0-9]+.*)/\1\2/p' ExampleFile.txt > ResultFile.txt

Solution 3:

With awk specify Field separator as ',' comma and 'space' and match those lines which contain "Iteration" in, next print the columns #3 and #7 (or $NF as last column instead of $7)

awk -F'[, ]' '/Iteration/ {print $3,$7}' infile