I did the substitution you're after like this: s{ \s+ \S+ \s* \z }{}msx

Here's the test code:

#!perl

use 5.12.0;
use warnings;

my $sought = q{sb1 1.0 142r};
my $last;

while (my $line = <DATA>) {
    chomp $line;
    if ($line eq $sought) {
        $last =~ s{ \s+ \S+ \s* \z }{}msx;
    }
    say $last if $. > 1;
    $last = $line;
}
say $last;

__DATA__
pre 2
pre 1
4.62399E+09 2.90871E+09 7.17338E+14 
sb1 1.0 142r
post 1
post 2

[download]

Here's the output:

$ match_exp_prob.pl
pre 2
pre 1
4.62399E+09 2.90871E+09
sb1 1.0 142r
post 1
post 2

[download]

For your code, where I have DATA, you'll want IN and where I have say $last, you'll want print OUT $last, "\n".

-- Ken

In reply to Re: how to read in a line containing exponentials by kcott
in thread how to read in a line containing exponentials by pattobw

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