print a pattern when line doesn't start with "#" in awk

Question

I'd like to print "chr" at the begin of each line that doesn't start with "#" in my file.

input :

##toto
#titi
16
17

output :

##toto
#titi
chr16
chr17

I've tried with awk (awk '$1 ~ /^#/ ... ) or grep (grep "^[^#]" ...) but got no success. How can I accomplish this?

Answer 1

I think you need ^[^#] meaning, starting with a character that is not # and reconstruct those lines by prefixing with "chr"

awk '/^[^#]/{ $0 = "chr"$0 }1'

Answer 2

using sed

sed '/^#/! s/.*/chr&/'

Answer 3

command

awk '{if(!/^#/ && !/^$/){$0="chr"$0;print}else if (/^#/ && !/^$/){print $0}}' filename

output

##toto
#titi
chr16
chr17

python

#!/usr/bin/python
import re
k=re.compile(r'^#')
e=re.compile(r'^$')
p=open('filename','r')
for i in p:
    if not re.search(k,i)   and not re.search(e,i):
        print "chr{0}".format(i.strip())
    elif not re.search(e,i):
        print i.strip()

output

##toto
#titi
chr16
chr17