You might have more luck searching for generalised solutions to the problem. Which in your case is simple counting the occurrences of substrings within a string (doesn't matter what the strings are, or that they are DNA).

This is still a python implementation, but this thread seems to be attempting a similar problem?

Similarly, there's a substring counting loop in C here . You could modify this to count each of your 16 strings easily enough I would think (I know basically 0 C though, so I might not be much use!)

here are a couple of examples of GC calculators that might help (I have no idea how fast they will be - or whether they will help. You would need to modify the output slightly to get your csv-like format.

#!/usr/bin/python

import os
import sys

inFile = open(sys.argv[1],'r')

totalBP = 0
gcBP = 0
headerCount = 0


for line in inFile:
    if line[0] == ">":
        headerCount += 1
    else:
        seqLine = line.strip().lower()
        totalBP += len(seqLine)
        gcBP += seqLine.count('g') # e.g. modify these lines to count the number of occurences of each dinucleotide
        gcBP += seqLine.count('c') # 

print ("Locus %s" % str(sys.argv[1])) + ' GC: ' + str(float(gcBP) / totalBP)

A perl script I stole from seqanswers:

#!/usr/bin/perl -w

use strict;

my $i=1;
my $infasta=$ARGV[0];

open FH1 , "$infasta" or die "$!";

while(<FH1>){
        if ($i==1){
                chomp;
                $_ =~ tr/>//d;
                print "$_\t";
                $i++;}
        elsif ($i==2){
                my $UC = uc($_);
                my $length = ($UC =~ tr/ATGCN//);
                my $numsGCs = ($UC =~ tr/GC/gc/);   # e.g add additional regexes here to find other dinucleotide patterns,
                my $percentGC = ($numsGCs/$length)*100;
                my $rounded = sprintf("%.2f", $percentGC);
                print "$rounded\n";
                $i++;}
        elsif ($i>2){
                $i=1;
                redo;};
}

They might be moderately more performant than your existing script, but without testing it I don't know.