how to say, zip code Starts With 2

[polka4ever]

hello,

I am more of an SQL person - so what I am trying to say, I am not sure how to say in Perl. I am sure it is some sort of regular expression. Maybe YOU can help me!

I would like to say,

if ($ZipPlus4 STARTS WITH 29626) then increment n.

any hints?

thank you!

 

[MillerH]

$n++ if $ZipPlus4 =~ /^29626/;

 

[KevinADC]

You don't really need a regexp to search for simple sub-strings within strings.

$n++ if (index $ZipPlus4,'29626') == 0;

'0' (zero) means the location of the sub-string is at the beginning of the string.

- Kevin, perl coder unexceptional!

 

[MillerH]

That's true Kevin. You could also use substr as well:

$n++ if substr($ZipPlus4, 0 , 5) eq '29626';

Overall though, I think the deciding factor should be which code is the most readable. My first example directly translates to "increment $n if $ZipPlus4 begins with '29626'". At least it does if you speak Regular Expression

I went ahead and did a quick benchmark, and these were my results:

Benchmark: timing 5000000 iterations of index, regexp, substr...
     index: 14 wallclock secs (12.51 usr +  0.00 sys = 12.51 CPU) @ 399424.83/s
(n=5000000)
    regexp: 12 wallclock secs (11.03 usr +  0.00 sys = 11.03 CPU) @ 453103.76/s
(n=5000000)
    substr: 16 wallclock secs (14.72 usr +  0.00 sys = 14.72 CPU) @ 339650.84/s
(n=5000000)

It should be noted though that this was with only 9 character strings. If the string were any longer I believe index would end up much slower as it would scan through the entire string for false tests.

Back to real coding fun.

 

[polka4ever]

Thank you much for the tips! This worked great. So, once again coming from the SQL point of view - what if i have SEVERAL zip codes that i'm looking for? what if i want to say, If $ZipPlus4 has a value in ('29626', '29627', '29628', '29788') then increment $n += 1, and if it's NOT in those values, then increment $o += 1.

any help is GREATLY appreciated!

~polka

 

[MillerH]

In that instance, the best method is to turn your $ZipPlus4 to a simple $Zip (as in only 5 characters) and then do a direct search through your array of zips.

my $zip = substr($ZipPlus4, 0, 5); # Only first 5 numbers

if (grep {$zip eq $_} qw(29626 29627 29628 29788)) {
    $n++;
} else {
    $o++;
}

Alternatively, you could also use a %hash lookup table. In fact, that is the fastest method. Especially if you are doing lots of lookups:

my $zip = substr($ZipPlus4, 0, 5); # Only first 5 numbers

my %isSpecialZip = map {$_ => 1} qw(29626 29627 29628 29788);

if (isSpecialZip{$zip}) {
    $n++;
} else {
    $o++;
}

Either method works of course. It's important to know both though since grep works great for one-liners, which are often very useful.

 

[MillerH]

Slight bug in my second bit of code. Forgot the dollar sign:

...
if ([COLOR=red]$[/color]isSpecialZip{$zip}) {
...

 

[KevinADC]

Yep, I have to give you this one Miller. Since the pattern is bound to the beginning of the string the regexp will be faster.

- Kevin, perl coder unexceptional!