[SOLVED] replace a pattern using sed

Issue

I have a file containing a lot of text and some digits describing numbers < 1 with three digits of precision. I'd like to replace those numbers with roughly equivalent integer percentages (numbers 0-99).

0.734 -> 73
0.063 -> 6
0.979 -> 97

It would be great to round properly, but not required.

I've tried the following and several variants and I can't seem to get even one match:

sed -e 's/0\.(\d\d)/&/' myfile.txt

Which I understand to mean, match the digit 0, then the decimal, capture the next two digits and have sed replace the whole match with the captured part.

Even if I get this to work, I don't know how to handle the 0.063 -> 6 case. Sure would appreciate a helping hand on this.

Solution

sed does support character class but use the longer POSIX name. Digits are [[:digit:]]. It's shorter to just write [0-9].

Try this:

sed -E 's/0\.([0-9][0-9]).*/\1/;s/^0//' myfile.txt

The -E flag tells it to use modern regex. There are actually 2 commands here, separated by ;:

s/0\.([0-9][0-9]).*/\1/: put the two digits following 0 and a dot into a capture group and replace the whole string with this capture group.

s/^0//: remove the leading zero from the string after the above.

Answered By - Code Different

Answer Checked By - Mary Flores (WPSolving Volunteer)