Remove trailing 0 from the field

Shell Programming and Scripting
1

Hi Freinds,

I have file1.txt as below

file1.txt

1521894~~-0.400~201207
1521794~~-0.486~201207
152494~~-0.490~201207
152154894~~-0.490~201207
1521894354~~-0.489~201207

expected output :

1521894~~-0.4~201207
1521794~~-0.486~201207
152494~~-0.49~201207
152154894~~-0.49~201207
1521894354~~-0.489~201207

Please help. I am very new to unix.

2

Try:

awk -F"~" -vOFS="~" '{sub("0*$","",$3)}1' file
1 Like
3

With sed:

sed 's/0*\(~[0-9]*\)$/\1/' <file

--
Bye

1 Like
4

@Lem ,@bartus11 thanks for reply. it worked good . but if i have a value 9.000 the output which i am getting is 9. but expcted is 9

5

Building on @Lem's and @bartus11's solutions, try

awk -F"~" -vOFS="~" '{sub("[.0]*$","",$3)}1' file
or
sed 's/[\.0]*\(~[0-9]*\)$/\1/' <file

btw, Bartus11's suggestion will work if field4 contains e.g. letters while Lem's will fail on that.

1 Like
6

An awk solution:

 
awk -f b.awk infile

where b.awk:

 
{
 FS=l="~";
 for (i=1; i<=NF; i++) {
    if ($(i) ~ /[.]/) sub("[.0]*$", "", $(i));
    if (i==NF) l="\n";
    printf $i l;
 }
}
7

This should work better:

sed 's/\.*0*\(~[^~]*\)$/\1/' <file

--
Bye

1 Like
8

@All thanks for all the suggestions. Would you mind giving the soultion to the other scenario where the field contains 0

example :

if i use

awk -F"|" -vOFS="|" '{sub("0*$","",$3)}1' file

In the output file it is removing the "0" from third field which should not be removed.

Plz help..

9

try this....

awk -F"|" -vOFS="|" '$3 !~ /^[0-0]+$/{sub("0*$","",$3)}1' file
10 
awk '$3~/\./{sub(/\.?0+$/,x,$3)}1' FS=\| OFS=\| infile