Cutting/Replacing characters at the end of a field 2

[johngiggs]

How do I go about replacing/deleting the characters "2003" if they exist in field 1? The following cut command works assuming that all lines have 2003 as the 9th - 12th characters.

cut -c1-8,13-70 file

and

sed 's/2003 / /g'

Also works because there are 5 spaces between fields 1 and 2.

What is the best sed solution for performing this task?

Sample Input

Name Description
========================================================
tuesdayb2003 Tuesday wk1
tuesdays2003 2003 Tuesdays: Tuesdays w/oholiday
tuesfrid2003 2003 Tuesfrid: Tuesdays thru Fridays w/o

Desired Output

Name Description
========================================================
tuesdayb Tuesday wk1
tuesdays 2003 Tuesdays: Tuesdays w/oholiday
tuesfrid 2003 Tuesfrid: Tuesdays thru Fridays w/o

Any help would be greatly appreciated.

Thanks,

John

 

[johngiggs]

Actually, I only want to delete/cut the string "2003" if it exists and they are the last characters in the first field.

Thanks,

John

 

[CaKiwi]

Assuming field 1 ends in a space and 2003 is always at the end of it (untested)

sed 's/^\(.*\)2003 /\1 /'

CaKiwi

"I love mankind, it's people I can't stand" - Linus Van Pelt

 

[vgersh99]

sed -e '/2003/s/^\([^ ][^ ]*\)2003\(.*\)/\1\2/g' file.txt

vlad
+----------------------------+
| #include<disclaimer.h> |
+----------------------------+

 

[PHV]

If you the number of spaces between fields does'nt matter:

awk '$1~/2003$/{$1=substr($1,1,length($1)-4)}
{print}' /path/to/inputfile >/path/to/outputfile

 

[johngiggs]

Thanks, vlad!! That did the trick. Can you please explain this portion of the command:

/^\([^ ][^ ]*\)2003\(.*\)/\1\2/


Thanks,

John

 

[vgersh99]

/^\([^ ][^ ]*\)2003\(.*\)/\1\2/

/ / <-- pattern to match for substitution.

^ <-- beginning of the line

\( \) <-- FIRST referencable expression

2003 <-- string '2003'

[^ ][^ ]* <-- ANY character BUT space
repeated more than ONCE
\( \) <-- SECOND referencable expression
.* <-- any number of ANY character


/ / <-- replacment for the
preceeding 'match'
\1\2 <-- subst with back-reference
expression ONE folowed by the
back-reference expression TWO


I'm not sure if that's the best way of explaining it, but...... What's taking place here is 'tagging' regexp matching and BACK-REFERENCING them in the 'replacement portion of the 'substitute command'. You can think of it as:

ANYTHING_but_space followed_by_2003 the_rest_of_the_line

what you need to get is:

ANYTHING_but_space the_rest_of_the_line

hope it helps



vlad
+----------------------------+
| #include<disclaimer.h> |
+----------------------------+

 

[johngiggs]

Thanks, Vlad!! That was a good explanation. I understood what was happening, but I didn't fully understand the syntax that was used.

PHV, your solution works well also. Thanks for your help.

John