How to remove "-" and a space from the beginning in a bash script?

Question

I have an output that looks as below

- 0.1-1
- 0.1-2
- 0.1-3
- 0.1-6
- 0.1-7
- 0.1-9

How to use grep or something else so as to remove "-" and a space from the beginning.

0.1-1
0.1-2
0.1-3
0.1-6
0.1-7
0.1-9

grep just finds stuff. it doesn't "change" it. you need other tools. — Marc B, Sep 27 '16 at 14:58

score 2 · Answer 1 · answered Sep 27 '16 at 14:57

2

With sed:

sed -e 's/^- //' input.txt

Or with GNU grep:

grep -oP '^- \K.*' input.txt

answered Sep 27 '16 at 14:57

redneb

`-P` is available on fewer builds of GNU grep than `-o` is (introduced in a later release, and requires an extra compile-time option that modifies the set of library dependencies and thus isn't likely to be available on embedded systems). – Charles Duffy Sep 27 '16 at 15:02

Avinash Raj · Answer 2 · 2016-09-27T15:00:19.477

1

You may use grep also,

grep -oE '[0-9].*' file

edited Sep 27 '16 at 15:00

answered Sep 27 '16 at 14:59

Avinash Raj

1

That still prints the whole line, not only the subpart matched; you'd need GNU extensions for that. – Charles Duffy Sep 27 '16 at 14:59
added -o flag.... – Avinash Raj Sep 27 '16 at 15:00
1

*nod*. Worth noting, though, that that isn't in [the POSIX standard for `grep`](http://pubs.opengroup.org/onlinepubs/009604499/utilities/grep.html), so its presence on any particular platform isn't guaranteed. – Charles Duffy Sep 27 '16 at 15:01

score 1 · Answer 3 · answered Sep 27 '16 at 15:05

1

With awk:

awk '{print $2}' file

answered Sep 27 '16 at 15:05

dawg

galfisher · Answer 4 · 2016-09-27T15:02:49.757

0

You can use cut to remove the first two columns of every line:

cut -c3- input.txt

edited Sep 27 '16 at 15:02

answered Sep 27 '16 at 15:01

galfisher

4 Answers4