Using Sed to delete characters between the 10th character and the first period after the 10th

Question

I want to remove all characters between the 10th character of each line and the first period that comes after.

Say I have input that looks like

0123456789 foo. foo....
1234566789 bar. bar...
0912309299 foobar. foobar..

Output will be something like

0123456789. foo....
1234566789. bar...
0912309299. foobar..

Is there a sed command that does this?

Thanks

Answer 1

This is one way to make it:

$ sed -r 's/^([0-9]{10})[^.]*/\1/' file
0123456789. foo....
1234566789. bar...
0912309299. foobar..

Or also (to get 10 first chars in general, no matter if they are a number or not):

sed -r 's/^(.{10})[^.]*/\1/'

Explanation

• -r allow extended regex.
• sed 's/find/replace/' is the basic substitution: replace find with replace once. In this case, we use the following:
• ^([0-9]{10})[^.]* catch first 10 digits. Also, catch all characters up to a dot ..
• \1 print back the first block.

This is a schema of the substitution:

0123456789 foo. foo....
1234566789 bar. bar...
0912309299 foobar. foobar..
^^^^^^^^^^       ^^^^^^^^^^
     |    ^^^^^^^     |
     |         |      |
^([0-9]{10})  [^.]*   |
      |               |
      |          |----
^^^^^^^^^^-------------
0123456789. foo....
1234566789. bar...
0912309299. foobar..

Answer 2

I think this should work.

sed -P 's/(?<=^.{10}).*?\.//'

EDIT: This is obviously wrong, as pointed out by Kent below, I mistakenly thought that sed supported -P for perl compatible regexes, this sent me down a rabbit hole of figuring out how to get perl compatible regexes in sed, one solution offered by kent was ssed but I didn't want to do that. long story short, here's the perl script:

perl -pe 's/(?<=^.{10}).*?(?=\.)//' example.txt
0123456789. foo....
1234566789. bar...
0912309299. foobar..

and the post that sullied my hithertoo clean mind with perl knowledge. I'd never checked out perl but that's a handy post for when you want to extend your bash-fu just that little bit beyond sed or awk. As is obviously pointed out by the answer above me, none of this is necessary in this case, just for interest.

Answer 3

For your data this awk gives correct output.

awk '{$2="";sub(/ /,".")}1' file
0123456789. foo....
1234566789. bar...
0912309299. foobar..

Just remove second field and change first space to a .