方法1：http://www.reibang.com/p/824a94f92007
當(dāng)我們運(yùn)行大量數(shù)據(jù)掷酗，必須考慮到計(jì)算機(jī)的內(nèi)存開(kāi)銷時(shí)，我們不得不對(duì)方法進(jìn)行優(yōu)化腹暖。
因此這里在方法1的基礎(chǔ)上介紹另一種方法：
#!/usr/bin/perl -w
# Determining frequency of nucleotides, take 2
# Get the DNA sequence data
print "Please type the filename of the DNA sequence data: ";
$dna_filename = <STDIN>;
chomp $dna_filename;
# Does the file exist?
unless ( -e $dna_filename) {
print "File \"$dna_filename\" doesn\'t seem to exist!!\n";
exit;
}
# Can we open the file?
unless ( open(DNAFILE, $dna_filename) ) {
print "Cannot open file \"$dna_filename\"\n\n";
exit;
}
@DNA = <DNAFILE>;
close DNAFILE;
$DNA = join( '', @DNA);
# Remove whitespace
$DNA =~ s/\s//g;
# Initialize the counts.
# Notice that we can use scalar variables to hold numbers.
$count_of_A = 0;
$count_of_C = 0;
$count_of_G = 0;
$count_of_T = 0;
$errors = 0;
# In a loop, look at each base in turn, determine which of the
# four types of nucleotides it is, and increment the
# appropriate count.
for ( $position = 0 ; $position < length $DNA ; ++$position ) {
$base = substr($DNA, $position, 1);
if ( $base eq 'A' ) {
++$count_of_A;
} elsif ( $base eq 'C' ) {
++$count_of_C;
} elsif ( $base eq 'G' ) {
++$count_of_G;
} elsif ( $base eq 'T' ) {
++$count_of_T;
} else {
print "!!!!!!!! Error - I don\'t recognize this base: $base\n";
++$errors;
}
}
# print the results
print "A = $count_of_A\n";
print "C = $count_of_C\n";
print "G = $count_of_G\n";
print "T = $count_of_T\n";
print "errors = $errors\n";
# exit the program
exit;
