add a new column with sum of columns in perl

This is fairly easy to do in Perl, even as a one-liner:

perl -MList::Util=sum -anE 'if (1 == $.) { say join(q{ }, @F, q{FullYear}) } else { say join(q{ }, @F, sum(@F[2..13])) }' «YOUR-FILE»

Explanation:

-MList::Util=sum loads the List::Util module and imports the sum function. This is the same as use List::Util qw(sum).

-n tells Perl to process the input file line-by-line, running the script for each line. (Actually redundant, as the next option implicitly turns this on). -a turns on autosplit mode, so we get an array @F with one entry per field. -E means we're going to provide a script as a command-line argument, using current Perl features (for "say" in this case).

Full details for those options can be found in the perlrun manpage/podfile.

Then, here is the script, with spacing added, and comments explaining:

if (1 == $.) {                        # $. is the line number. Line 1 is header line.
    say join(' ', @F, q{FullYear});   # print out the heder + FullYear
}
else {
    # print out rows + sum of columns 2..13. Remember Perl counts from 0 in arrays,
    # so column 2 is the 3rd column (the number for January).
    say join(' ', @F, sum(@F[2..13]));
}

BTW: You can ask Perl to help understand one-liners (at least ones you trust — this is not safe with untrusted scripts) with -MO=Deparse, which gives output like this:

command:

perl -MO=Deparse -MList::Util=sum -anE 'if (1 == $.) { say join(q{ }, @F, q{FullYear}) } else { say join(q{ }, @F, sum(@F[2..13])) }' t-file 

output:

use List::Util (split(/,/, 'sum', 0));
use feature 'current_sub', 'bitwise', 'evalbytes', 'fc', 'postderef_qq', 'say', 'state', 'switch', 'unicode_strings', 'unicode_eval';
LINE: while (defined($_ = readline ARGV)) {
    our @F = split(' ', $_, 0);
    if (1 == $.) {
        say join(' ', @F, 'FullYear');
    }
    else {
        say join(' ', @F, &sum(@F[2..13]));
    }
}
-e syntax OK

So you can see the List::Util load, the -n going line-by-line, and -a adding the split.

Would Math::BigFloat do for your "huge precision"?

perl -MMath::BigFloat -ape 'my $s=0; $s += new Math::BigFloat($_) for @F[2..$#F]; s/$/ $s/'

abc dec 7.5 8.5
abc dec 7.5 8.5 16

You could also use List::Util::sum with Math::BigFloat; it's quite pointless, though:

perl -MMath::BigFloat -MList::Util=sum -ape 's/$/" ".sum map new Math::BigFloat($_), @F[2..$#F]/e'

It's not perl, but this seems to get the job done:

awk 'NR==1 {$(NF+1) = "FullYear"; print} NR>1 {subtotal=0; for(f=0;f<=NF; f++) {subtotal+=$f}; $(NF+1)=subtotal; printf( "%s %s %5.10f %5.10f %5.10f %5.10f %5.10f %5.10f %5.10f %5.10f %5.10f %5.10f %5.10f %5.10f %5.10f\n", $1, $2, $3, $4, $5, $6, $7, $8, $9, $10, $11, $12, $13, $14, $15 ) }' inputfile

Just a variant of @derobert:

perl -MList::Util=sum -nlE 'say "$_ ", sum((split)[2..13])||"FullYear"' input

or using -a

perl -MList::Util=sum -nalE 'say "$_ ", sum(@F[2..13])||"FullYear"' input