can sed search for last two digits of line...

[Ratzie]

I have a huge file with a bunch of data.
2088872167,15 1 01 047
similar
similar

Nothing is consistent except for the spacing.
I need to insert a space before the last two digits of the line...
2088872167,15 1 01 0 47

Can this be done
What would be the syntax...
Thanks

[Alan Meyer_4]

to get the line you specified enter;

grep '47$' filename

Like the $ specifies to find the pattern at the end of a line.

[Ratzie]

I need this done to every line in my file
From this:
2088872167,15 1 01 047
2088872168,15 1 01 123
2088872166,15 1 01 546
2088872164,15 1 01 048
...
...

To this:
2088872167,15 1 01 0 47
2088872168,15 1 01 1 23
2088872166,15 1 01 5 46
2088872164,15 1 01 0 48
...
...

[Alan Meyer_4]

sorry I missread your request...

[Mel Burslan]

while reading the following solution, please bear in mind that my perl programming experience is next to nil. Having said that,

create a little script called L.pl as follows

#!/usr/bin/perl
$a=;
$l=length $a;
printf $l

(if your perl executable is in a different location, modify the first line accordingly)

chmod 755 L.pl

lets say your filename is MYFILE

cat MYFILE | while read line
do

len=`echo $line | ./L.pl`
let m2=$len-3
let m1=$len-2

part1=`echo $line | cut -c 1-${m2}`
part2=`echo $line | cut -c ${m1}-${len}`

echo $part1" "$part2 >> MY_new_FILE
done


again, this is coming from someone who can barely find his way through the infamous perl camelbook. I am sure there are more elaborate solutions using perl to to do the same in one command.

if your data on every line is a fixed length, you do not need the perl portion at all. Just replace that line with

len=your_line_length+1
(+1 because in perl first position is 0 not 1)

then you should be okay

Hope this helps

[Vibhor Kumar Agarwal]

Try this

sed -e 's/\(.*\)../\1 /g' file

will give the output of
2088872167,15 1 01 047
as
2088872167,15 1 01 0 (space)

sed -e 's/\(.*\) \(.*\)/\1/g'
will give the last coloumn
047

Now what you have to do is simply take the last coloumn get the last 2 bits and append it.

sed -e 's/\(.*\) \(.*\)/\1/g' | cut -c 2,3
will give 47

Now just append this to the end of line.
This is general also, and will help with every line.

[curt larson_1]

sed -e 's/\(..\)$/ \1/' file

[Vibhor Kumar Agarwal]

Wow Curt,

I was doing a page full of scripting for that.

[curt larson_1]

just to be sure you know this:

sed -e 's/\(..\)$/ \1/' file
^^^^^^^^^^^^^^^^^^blank/space before \1

[curt larson_1]

vibhor, you had most of it already.
next time will be easier.

and it is a short sed script, but doing it in the shell is probably faster

while read var
do
firstPart=${var%??}
last2=${var#$firstPart}
print $firstPart $last2
done < yourFile

[Vibhor Kumar Agarwal]

Curt,
Can you explain that a bit to me.

I haven't done that type of scripting.
A bit of guidance required.

Thanks

[Muthukumar_5 · Accepted solution]

You can try as,

# cat > testfile
2088872167,15 1 01 047
2088872168,15 1 01 123
2088872166,15 1 01 546
2088872164,15 1 01 048

# perl -pe 's/(..)$/ $1/g' testfile
2088872167,15 1 01 0 47
2088872168,15 1 01 1 23
2088872166,15 1 01 5 46
2088872164,15 1 01 0 48

or

# sed 's/\(..\)$/ \1/g' testfile
2088872167,15 1 01 0 47
2088872168,15 1 01 1 23
2088872166,15 1 01 5 46
2088872164,15 1 01 0 48

hth.

[Ratzie]

That last one was exactly what I was looking for!

Thanks!