[SOLVED] sed - stream editor and regex [0-9]*

fanoflq · 02-05-2016, 09:08 PM

I do not understand why this sed operation did not repeat "123":

$ echo "abc 123" | sed 's/[0-9]*/& &/'
abc 123

I was expecting output of:
abc 123 123

since 123 was matched,
and I have replacement for 123 as /& &/ = 123 123.

Or this
abc 123 abc 123

since "a" in abc is matches [0-9]*.

Obviously I misunderstood has regex matched numbers using pattern [0-9]*.
My understanding is that [0-9]* will match any set of digits starting from zero or more single digit like: 3, 00, 10, 234, a, ab, ...

What did I missed?
Thanks.

syg00 · 02-05-2016, 09:40 PM

Quote:

Originally Posted by fanoflq

My understanding is that [0-9]* will match any set of digits starting from zero or more single digit

When testing regex, use the -n switch on sed, and "p" to print your records. Then you know when you aren't doing what you want. (you're just getting the echo'd data here).

Also when using regex, use the -r switch.
That should get you started.

AlucardZero · 02-05-2016, 09:40 PM

Not quite. [0-9]* matches zero or more 0-9's. Which matches first at the front of the string.

Code:

~$ echo "abc 123" | sed 's/[0-9]*/fjfg/'
fjfgabc 123

Try:

Code:

$ echo "abc 123" | sed 's/[0-9][0-9][0-9]/& &/'
abc 123 123

makyo · 02-05-2016, 10:04 PM

Hi.

Code:

echo "abc 123" | sed 's/[0-9]\{1,\}/& &/'
abc 123 123

For systems like:

Code:

OS, ker|rel, machine: Linux, 3.16.0-4-amd64, x86_64
Distribution        : Debian 8.3 (jessie) 
sed (GNU sed) 4.2.2

Best wishes ... cheers, makyo

fanoflq · 02-05-2016, 10:18 PM

Why do you want to use -n?
From man page:
-n, --quiet, --silent
suppress automatic printing of pattern space

$ echo "abc 123" | sed -n 's/[0-9]*/& &/p'
abc 123 <=== Is this the first pattern match?
$ echo "abc 123" | sed 's/[0-9]*/& &/p'
abc 123
abc 123

Why two different results?

fanoflq · 02-05-2016, 10:36 PM

Pattern [0-9]* will be a valid match to anything.
So regex should return these:
a
ab
abc
abc ( there is a space after abc)
abc 1
abc 12
abc 123

b
bc
....

c
c + space
c 1
....
So I really should expect an infinite match.
This is where I not sure about correctness of sed's regex.

astrogeek · 02-05-2016, 10:41 PM

The problem as noted already is that the '*' matches zero or more digits at the beginning, so the output is 'nothing space nothing' (the zero digits that were matched, twice, separated by a space) followed by the rest of the string.

Use '+' instead...

Code:

echo "abc 123" |sed 's/[0-9]\+/& &/'
abc 123 123

Note the escaped '\+' which is not necessary with the -r option ('+' with -r).

fanoflq · 03-03-2017, 04:59 PM

I went back and lookup this post while reading sed again.

Quote:

Originally Posted by astrogeek

The problem as noted already is that the '*' matches zero or
more digits at the beginning, so the output is
'nothing space nothing' (the zero digits that were matched,
twice, separated by a space) followed by the rest of the string.

Use '+' instead...

Code:

echo "abc 123" |sed 's/[0-9]\+/& &/'
abc 123 123

Note the escaped '\+' which is not necessary with the -r option ('+' with -r).

I finally got it.

The digits were matched and from man sed:

Code:

s/regexp/replacement/
Attempt to match regexp against the pattern space.  
If successful, replace that  portion  matched  with 
replacement.   

The replacement may contain the special character & 
to refer to that portion of the pattern space which matched,
 and the special escapes \1 through \9 to refer to the 
corresponding  matching sub-expressions in the regexp.

This is how I finally understood it.
Since & is 123 when a match is found,
then the effective replacement of found
match (123) in this case is /& &/ = /123 123/
This "123 abc" becomes "123 123 abc".
Thanks.