Remove rows from a file that exist in another file with newer timestamp

Question

I have 2 files in the exactly the same format and the same data except for 1 column.
Example row of file 1:

"1/30/2017 11:14:55 AM",Valid customer,jim.smith,NY,1485771295    

Example row of file 2:

"1/26/2017 8:02:01 PM",Valid customer,jim.smith,NY,1485457321  

Of course the files have other rows but what I am interested in is the following:
I want to remove from file 2 all rows that are about the same entity i.e. $3 and exist in file 1 with a later timestamp and keep the rest.
In the example lines you can see that the row in file 1 is newer as we see from the date string of column 1. Now the last integer in the row is the actual epoch of column 1 so this column can be used for e.g. comparing the dates and sorting.
I can accomplish this by scripting e.g in perl something like the following:

#!/usr/bin/perl  

use strict;  
use warnings;  
my $file_a = "file1";
my $file_b = "file2";

open my $file_a_h, $file_a or die "Could not open $file_a";  

sub timestamp_users {  
    my ($fh) = @_;  
    my %recs;   

   while ( my $line =<$fh> ) {    
        my @items = split ",", $line;  
        my $user = $items[3];  
        $recs{$user} = $items[5];    
    }  
    return \%recs;    
}    

my $file_a_recs = timestamp_users($file_a_h);  

close $file_a_h;  

open my $file_b_h, $file_b or die "Could not open $file_b";  

my $file_b_recs = timestamp_users($file_b_h);    
close $file_b_h;  

my $count = 0;
while (my ($user, $last_time) = each %$file_b_recs) {  
    if(exists $file_a_recs->{$user} && $last_time >= $file_a_recs->{$user}) {
        ++$count;  
        `echo $user >> result.txt`;    

    }
}
print "count: $count\n";    

In this case I just output the users and then I would need to do a grep -v on file_b to figure out the rows I need.

But is there a way to do this using command line tools?
This approach seems too complicated for me.

Update:

Example row of file 1:

"1/30/2017 11:14:55 AM",Valid customer,jim.smith,NY,1485771295        
"1/26/2017 5:06:11 AM",New customer,john.doe,CA,1485403571    
"1/30/2017 4:14:30 AM",New customer,tim.jones,CO,1485746070    

Example row of file 2:

"1/26/2017 8:02:01 PM",Valid customer,jim.smith,NY,1485457321    
"1/30/2017 11:09:36 AM",New customer,tim.jones,CO,1485770976   
"1/30/2017 11:14:03 AM",New customer,john.doe,CA,1485771243  
"1/30/2017 11:13:53 AM",New customer,bill.smith,CA,1485771233  

Expected output:

"1/30/2017 11:14:03 AM",New customer,john.doe,CA,1485771243    
"1/30/2017 11:09:36 AM",New customer,tim.jones,CO,1485770976   
"1/30/2017 11:13:53 AM",New customer,bill.smith,CA,1485771233  

Answer 1

To get the newest version of each row in both files:

$  cat file1 file2 | sort -t',' -k3,3 -k5,5nr | sort -t',' -u -k3,3 -o newest

This concatenates the files and sorts the records with the fields from field 3 and 5 as the sorting key. This sorts the concatenated file so that the newest record for each person comes first (thanks to the timestamp in the last column). The last sort uses field 3 as the sorting key and does a unique sort based on this field. This will leave only the newest record for each person in the file newest.

Then we create the complement of the lines in newest, i.e. all records in the two files that are older than the newest record for each person:

$ cat file1 file2 | grep -v -F -x -f newest >older

The grep does a fixed string match (-F) on complete lines (-x) and returns all lines not matching (-v) anything in newest. These lines are stored in older.

The last step is to remove any line from file2 that is present in the older file:

$ grep -v -F -x -f older file2 >new-file2