Matching Values in Hashes - regex

Question

I have two arrays of hashes. I want to narrow down the second one according to variables in the first.
The first array contains hashes with keys seqname, source, feature, start, end, score, strand, frame, geneID and transcriptID.
The second array contains hashes with keys
organism, geneID, number, motifnumber, position, strand and sequence.
What I want to do, is remove from the first array of hashes, all the hashes which have a variable geneID which is not found in any of the hashes of the second array. - Note both types of hash have the geneID key. Simply put, I want to keep those hashes in the first array, which have geneID values which are found in the hashes of the second array.
My attempt at this so far was with two loops:
my #subset # define a new array for the wanted hashes to go into.
for my $i (0 .. $#first_hash_array){ # Begin loop to go through the hashes of the first array.
for my $j (0 .. $#second_hash_array){ # Begin loop through the hashes of the 2nd array.
if ($second_hash_array[$j]{geneID} =~ m/$first_hash_array[$i]{geneID}/)
{
push #subset, $second_hash_array[$j];
}
}
}
However I'm not sure that this is the right way to go about this.

Answer 1

For starters, $a =~ /$b/ doesn't check for equality. You'd need
$second_hash_array[$j]{geneID} =~ m/^\Q$first_hash_array[$i]{geneID}\E\z/
or simply
$second_hash_array[$j]{geneID} eq $first_hash_array[$i]{geneID}
for that.
Secondly,
for my $i (0 .. $#first_hash_array) {
... $first_hash_array[$i] ...
}
can be written more succinctly as
for my $first (#first_hash_array) {
... $first ...
}
Next on the list is that
for my $second (#second_hash_array) {
if (...) {
push #subset, $second;
}
}
can add $second to #subset more than once. You either need to add a last
# Perform the push if the condition is true for any element.
for my $second (#second_hash_array) {
if (...) {
push #subset, $second;
last;
}
}
or move the push out of the loop
# Perform the push if the condition is true for all elements.
my $flag = 1;
for my $second (#second_hash_array) {
if (!...) {
$flag = 0;
last;
}
}
if ($flag) {
push #subset, $second;
}
depending on what you want to do.
To remove from an array, one would use splice. But removing from an array messes up all the indexes, so it's better to iterate the array backwards (from last to first index).
Not only is it complicated, it's also expensive. Every time you splice, all subsequent elements in the array need to moved.
A better approach is to filter the elements and assign the resulting element to the array.
my #new_first_hash_array;
for my $first (#first_hash_array) {
my $found = 0;
for my $second (#second_hash_array) {
if ($first->{geneID} eq $second->{geneID}) {
$found = 1;
last;
}
}
if ($found) {
push #new_first_hash_array, $first;
}
}
#first_hash_array = #new_first_hash_array;
Iterating through #second_hash_array repeatedly is needlessly expensive.
my %geneIDs_to_keep;
for (#second_hash_array) {
++$geneIDs_to_keep{ $_->{geneID} };
}
my #new_first_hash_array;
for (#first_hash_array) {
if ($geneIDs_to_keep{ $_->{geneID} }) {
push #new_first_hash_array, $_;
}
}
#first_hash_array = #new_first_hash_array;
Finally, we can replace that for with a grep to give the following simple and efficient answer:
my %geneIDs_to_keep;
++$geneIDs_to_keep{ $_->{geneID} } for #second_hash_array;
#first_hash_array = grep $geneIDs_to_keep{ $_->{geneID} }, #first_hash_array;

Answer 2

This is how I would do it.
Create an array req_geneID for geneIDs required and put all geneIds of the second hash in it.
Traverse the first hash and check if the geneId is contained in the req_geneID array.(its easy in ruby using "include?" but you may try this in perl)
and,
Finally delete the hash that doesnot match any geneID in req_geneID using this in perl
for (keys %hash)
{
delete $hash{$_};
}
Hope this helps.. :)