[SOLVED] 'sed' or 'awk' help

upendra_35 · 10-22-2012, 02:36 PM

Hi,

I would like to do the following. Please help me doing this either with awk or sed.

Code:

CUFF.8.1 => CUFF.8_seq1
CUFF.9.1 => CUFF.9_seq1
CUFF.8.2 => CUFF.8_seq2

So basically what all i want to do is convert .1 into _seq1 and all .2 into _seq2 and so on. I know how to use sed but when i do the below it subsitutes the first "." but not the second "."

PHP Code:



sed  's/.1/_seq1/'

Didier Spaier · 10-22-2012, 02:55 PM

Code:

sed 's/.1/_seq1/g'

upendra_35 · 10-22-2012, 02:58 PM

Quote:

Originally Posted by Didier Spaier

Code:

sed 's/.1/_seq1/g'

When I add 'g' it does substitute at both positions. I guess 'g' is for global right?

CUFF_seq1_seq1

Didier Spaier · 10-22-2012, 04:03 PM

Right.

sycamorex · 10-22-2012, 04:07 PM

Edit: sorry - I forgot about the dot.

If I understand you correctly and all the numbers are 1-digit numbers, this should cover all your cases:

Code:

sed 's/[0-9]/_seq&/g' file

millgates · 10-22-2012, 04:08 PM

Hi, what about

Code:

sed 's/\(.*\)\./\1_seq/'

upendra_35 · 10-22-2012, 10:34 PM

Quote:

Originally Posted by millgates

Hi, what about

Code:

sed 's/\(.*\)\./\1_seq/'

This is exactly what i wanted. Thanks for your help.

Upendra

David the H. · 10-23-2012, 03:02 AM

First, understand that "." has a special meaning in regular expressions. It stands for "any character". So '.1' will match any single character followed by a one, i.e. 'a1', ',1', ' 1', etc.

To make it literal you have to either backslash escape it or enclose it in [] brackets.

Second, in sed's 's///' substitution expression, the entire left side match is replaced by the right side string. So if you want to carry over any values you have to use regex backreference capturing.

millgates' reply shows one working version, although I would give it a more precise matching expression. His version just matches everything from the beginning of the line up to the first literal period. If the line contained a period before the one you wanted, it would insert the _seq in the wrong place.

I would also add one more thing. It's a gnu extension that allows you to enable extended regex features through backslashing. A cleaner and more readable way to get the same effect is to use the -r option, and just enable ext-re globally.

Code:

sed -r 's/(CUFF[.][0-9])[.]([0-9])/\1_seq\2/'

Add a 'g' to the end if more than one string can exist on a single line, as mentioned.

See the grep man page for a good explanation of basic vs. extended regular expressions. And when you have a chance, take some time to study up on regex in general. You'll be glad you did.

Here are a few regular expressions tutorials:
http://mywiki.wooledge.org/RegularExpression
http://www.grymoire.com/Unix/Regular.html
http://www.regular-expressions.info/

millgates · 10-23-2012, 03:09 AM

Quote:

Originally Posted by David the H.

millgates' reply shows one working version, although I would give it a more precise matching expression. His version just matches everything from the beginning of the line up to the first literal period. If the line contained a period before the one you wanted, it would insert the _seq in the wrong place.

Actually, it matches everything up to the last period, since the .* is greedy. But I agree it deserves a better matching expression.

David the H. · 10-23-2012, 03:10 AM

Oh, by the way, if we can guarantee that there are no other numbers on each line before the pattern that you want, the expression can be made much easier.

Code:

sed 's/[0-9]/_seq&/2'

'&' can be used in the RHS to mean the entire LHS match, and a little-documented feature of sed allows you to specify exactly which instance on the line to substitute. So we just tell it to match and replace the second number found on the line.

Edit @millgates: Thanks for the correction. I guess I'm not thinking too clearly right now.

Edit again:

Actually, scratch my substitution above. I forgot we have to remove the period as well. Can't do that with a simple match.