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?
I think you need ^[^#] meaning, starting with a character that is not # and reconstruct those lines by prefixing with "chr"
awk '/^[^#]/{ $0 = "chr"$0 }1'
!/^#/ { ... } (but this matches empty lines too)
awk '/^[^#]/ ?
/^[^#]/ ignores lines starting with #, then do the opposite, remove the negation. as /^#/
using sed
sed '/^#/! s/.*/chr&/'
s/^/chr/ (doesn't need .*+&)
sed 's/^[^#]/chr&/' may be preferable to avoid adding the prefix to empty lines if any. Or sed 's/^[[:digit:]]/chr&/'
command
awk '{if(!/^#/ && !/^$/){$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
awk -F'\t'since there can be spaces in the INFO fields.