sed command that replaces number and word by two

Question

1) sed command which puts two of the first 'w' on each line. E.g "hewor" = "hewwor".

and

2) sed command which puts two of the first digit on each line E.g "hew0r" = "hew00r"

For first one I got

$ sed s/w/ww/

for the second one I don't understand how to replicate the same digit for example I got

$ sed s/[0-9]/00/

would work but it would have to be zero each time. How do I get the same digit?

How about reading the manual... You'll learn how to reference the matched portion, regardless of what's in the LHS. ;) — don_crissti
– don_crissti, Commented Jan 22, 2018 at 0:13
My bad, I meant $sed s/[0-9]/00/ as you can see, it has to be zero. I'm looking for a way for the digit to be the one found in the [0-9]. — Tinler
– Tinler, Commented Jan 22, 2018 at 0:20
"sed s/[0-9]/\0\0/" does the exact same as "sed s/[0-9]/00/" or am I missing something? — Tinler
– Tinler, Commented Jan 22, 2018 at 0:42
Yes. It becomes hew4r -> hew00r. I want hew44r. Also, why should I use that instead of "sed s/[0-9]/00/"? For this and the first example? — Tinler
– Tinler, Commented Jan 22, 2018 at 0:51
OK. in gnu sed using \0 refers to the previously matched regex. In bsd sed this is done using &. So in your case since \0 does not work, you can use sed 's/[0-9]/&&/'. Actuall using & will work even in gnu sed. — George Vasiliou
– George Vasiliou, Commented Jan 22, 2018 at 1:02

John Smith · Accepted Answer · 2018-01-22 00:25:00Z

2

You need to use the sed's feature, called groups (may be not the best reference, try to search for other tutorials). In you case the solution is

sed 's/\([0-9]\)/\1\1/' input_file.txt

the regexp for the first group $[0-9]$ will match any digit, and the part \1\1 says to replace the first group with itself repeated twice.

answered Jan 22, 2018 at 0:25

John Smith

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