Hi Experts,
How to sepearate the list digit with letters : with a space from where the letters begins, or other words from where the digits ended.
file
52087mo(enbatl)
52049mo(enbatl)
52085mo(enbatl)
25051mo(enbatl)
The output should be looks like:
52087 mo(enbatl)
52049 mo(enbatl)
52085 mo(enbatl)
25051 mo(enbatl)
Thanks a lot,
Try
sed 's/^[0-9]*/& /' file
Another approach:
awk '{ match($0,/^[0-9]*/); print substr($0, RSTART, RLENGTH) " " substr($0, RSTART+RLENGTH) } ' file
@mirni. That would work in this case, but if if there is also text that begins with a non-digit then as space would be inserted at the start of the line.. To avoid that one could use:
sed 's/^[0-9][0-9]*/& /' file
@bipinajiith, the same applies to your awk approach..
Yes, I was matching nothing also. Thank you.
I have a feeling lines may start with letters too, re-reading the OP's post:
In which case I would omit the anchor, and let it match the first group of digits:
sed 's/[0-9][0-9]*/& /' file
Ah yes..that's a consequence of the * trying to match as much as possible but also as early as possible. "Being early" wins over "being greedy", if it can.
Thanks all, it worked. Scrutinizer thanks for explaining 'text that begins with non-digit' as well.
I am wondering what would be the code if the digits stays in the middle:
Example text:
abce1234jklm
Want the output as :
abce 1234 jklm
echo "abce1234jklm" | sed 's/[0-9]*/ & /'
Doesn't works ,
Thanks a lot.
---------- Post updated at 01:47 PM ---------- Previous update was at 01:39 PM ----------
ok I am able to figure it out:
echo "abce1234jklm" | sed 's/^[a-z][a-z]*/& /;s/[0-9][0-9]*/& /'
abce 1234 jklm
Thanks for all the helps all.
It's because you missed scrutinizer's point.
[0-9]* matches zero or more digits, so it matches the beginning of the line (containing zero digits)
[0-9][0-9]* matches one or more digits
echo "abce1234jklm" | sed 's/[0-9][0-9]*/ & /'
works as you want
Bipinajith ,
this is nice coding btw, and it too worked for me: Thanks a lot.
awk '{ match($0,/^[0-9]*/); print substr($0, RSTART, RLENGTH) " " substr($0, RSTART+RLENGTH) } ' file
---------- Post updated at 02:31 PM ---------- Previous update was at 02:27 PM ----------
Mirini,
Thanks for poinitng that out & explaining that earlier mentioned by Scrutinizer, and I got it fixed...wroks like charm.
echo "abce1234jklm" | sed 's/[0-9][0-9]*/ & /'
abce 1234 jklm
Thanks much...
Can this be used?
sed 's/[0-9]+/ & /'
This should match minimum one digit?
[0-9]+ gives the same [0-9][0-9]* ?
You can with GNU sed en BSD sed -E (extended regex) option...
sed -E 's/[0-9]+/ & /' file
with regular sed an alternative to the earlier solution would be:
sed 's/[0-9]\{1,\}/ & /' file