(unix shell scripting) Trim last 3 characters of a line WITHOUT using sed, or perl, etc.

Question

Hi,

I've got a shell script outputting data like this:

1234567890  *
1234567891  *

I need to remove JUST the last three characters " *". I know I can do it via

(whatever) | sed 's/\(.*\).../\1/'

But I DON'T want to use sed for speed purposes. It will always be the same last 3 characters.

Qny quick way of cleaning up the output?

Answer 1

You can use awk just to print the first 'field' if there won't be any spaces (or if there will be, change the separator'.

I put the fields you had above into a file and did this

awk '{ print $1 }' < test.txt 
1234567890
1234567891

I don't know if that's any better.

Answer 2

Both awk and sed are plenty fast, but if you think it matters feel free to use one of the following:

If the characters that you want to delete are always at the end of the string

echo '1234567890  *' | tr -d ' *'

If they can appear anywhere within the string and you only want to delete those at the end

echo '1234567890  *' | rev | cut -c 4- | rev

The man pages of all the commands will explain what's going on.

I think you should use sed, though.

Answer 3

what do you mean don't want to use sed/awk for speed purposes? sed/awk are faster than the shell's while read loop for processing files.

$ sed 's/[ \t]*\*$//' file
1234567890
1234567891

$ sed 's/..\*$//' file
1234567890
1234567891

with bash shell

while read -r a b
do
 echo $a
done <file

Answer 4

I can guarantee you that bash alone won't be any faster than sed for this task. Starting up external processes in bash is a generally bad idea but only if you do it a lot.

So, if you're starting a sed process for every line in your input, I'd be concerned. But you're not. You only need to start one sed which will do all the work for you.

You may however find that the following sed will be a bit faster than your version:

(whatever) | sed 's/...$//'

All this does is remove the last three characters on each line, rather than substituting the whole line with a shorter version of itself. Now maybe more modern RE engines can optimise your command but why take the risk.

To be honest, about the only way I can think of that would be faster would be to hand-craft your own C-based filter program. And the only reason that may be faster than sed is because you can take advantage of the extra knowledge you have on your processing needs (sed has to allow for generalised procession so may be slower because of that).

Don't forget the optimisation mantra: "Measure, don't guess!"

---

If you really want to do this one line at a time in bash (and I still maintain that it's a bad idea), you can use:

pax> line=123456789abc
pax> line2=${line%%???}
pax> echo ${line2}
123456789
pax> _

---

You may also want to investigate whether you actually need a speed improvement. If you process the lines as one big chunk, you'll see that sed is plenty fast. Type in the following:

#!/usr/bin/bash

echo This is a pretty chunky line with three bad characters at the end.XXX >qq1
for i in 4 16 64 256 1024 4096 16384 65536 ; do
    cat qq1 qq1 >qq2
    cat qq2 qq2 >qq1
done

head -20000l qq1 >qq2
wc -l qq2

date
time sed 's/...$//' qq2 >qq1
date
head -3l qq1

and run it. Here's the output on my (not very fast at all) R40 laptop:

pax> ./chk.sh
20000 qq2
Sat Jul 24 13:09:15 WAST 2010

real    0m0.851s
user    0m0.781s
sys     0m0.050s
Sat Jul 24 13:09:16 WAST 2010
This is a pretty chunky line with three bad characters at the end.
This is a pretty chunky line with three bad characters at the end.
This is a pretty chunky line with three bad characters at the end.

That's 20,000 lines in under a second, pretty good for something that's only done every hour.

Answer 5

Note: This answer is somewhat intended to be a joke, but it actually does work...

#!/bin/bash
outfile="/tmp/$RANDOM"
cfile="$outfile.c"
echo '#include <stdio.h>
int main(void){int e=1;char c;while((c=getc(stdin))!=-1){if(c==10)e=1;if(c==32)e=0;if(e)putc(c,stdout);}}' >> "$cfile"
gcc -o "$outfile" "$cfile"
rm "$cfile"
cat somedata.txt | "$outfile"
rm "$outfile"

You can replace cat somedata.txt with a different command.

Answer 6

Assuming all data is formatted like your example, use 'cut' to get the first column only.

cat $file | cut -d ' ' -f 1  

or to get the first 10 chars.

cat $file | cut -c 1-10

Answer 7

If the script always outputs lines of 10 characters followed by 3 extra (in other words, you just want the first 10 characters), you can use

script | cut -c 1-10

If it outputs an uncertain number of non-space characters, followed by a space and then 2 other extra characters (in other words, you just want the first field), you can use

script | cut -d ' ' -f 1

... as in majhool's comment earlier. Depending on your platform, you may also have colrm, which, again, would work if the lines are a fixed length:

script | colrm 11

Answer 8

Another answer relies on the third-to-last character being a space. This will work with (almost) any character in that position and does it "WITHOUT using sed, or perl, etc.":

while read -r line
do
    echo ${line:0:${#line}-3}
done

If your lines are fixed length change the echo to:

echo ${line:0:9}

or

printf "%.10s\n" "$line"

but each of these is definitely much slower than sed.