Replace Text with Part of Text Using Regex with Bash Perl

Question

For example I have this output:

string1 anynameveryveryverylong string2
string1 othernameveryveryverylong   string2

I want truncate the name to the first ten characters:

string1 anynamever  string2
string1 othernamev  string2

a pseudo regex can be:

perl -pe "s/([^\t]+\t)([^\t]+)\t/\1\2{10}\t/g"

How do i get this?

Accepted Answer

perl -pe 's/^(\S+\s+)(\S{10})\S*/$1$2/'

^ matches at the start of the string
\S means non-whitespace
+ means repeated at least once
\s means whitespace
{10} means repeated 10 times

I.e. Keep the first word and the first 10 characters of the following word while forgetting the remaining characters of the second word.

Your pseudoregex has one substantial problem: the {10} is placed in the replacement part, but the replacement is just a string. The regex happens in the pattern part only.

Answer 2

Some more choices:

Perl with autosplitting on tabs:

$ perl -F"\t" -lae '$F[1]=substr($F[1],0,10); print join "\t",@F' file 
string1  anynamever  string2
string1  othernamev  string2

awk

$ awk -F"\t" -vOFS="\t" '{$2=substr($2,1,10)}1' file 
string1  anynamever  string2
string1  othernamev  string2

sed

$ sed -E 's/(\S+\t\S{10})[^\t]+/\1/' file 
string1  anynamever  string2
string1  othernamev  string2

One more Perl

$ perl -pe 's/(\S+\t\S{10})[^\t]+/\1/' file 
string1  anynamever  string2
string1  othernamev  string2

Replace Text with Part of Text Using Regex with Bash Perl

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Replace Text with Part of Text Using Regex with Bash Perl

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Alexander Ross