# Ask a Jedi: Pair Counting

I hate to bug you with a question, but I am at a loss. I have a database with 200K records and one of the columns contains a string of 0-30 characters that represent 0-15 pairs. I need to find a way to separate these into some form of list and then count the number of instances of a given pair. I am "sure" there is a way to dump this into an Array or Struct, but I still don't have a grasp of how to use these. Can you think of an "easy" way to do this with CF?

Hey, everything is easy in ColdFusion! (Ok, so maybe I’m a little bit biased.) This almost seems like a perfect Friday Puzzler, but I already have one in mind, so let’s look at a simple solution for this.

His database contains a column that looks like this:

01333910394828013392948281

Each two characters represents one pair. What he wanted then was a structure containing each pair along with the number of times it shows up in the string. Here is a simple UDF, and test code, to demonstrate this:

``` <cfset str = "01333910394828013392948281">```

<cffunction name=”getPairStats” returnType=”struct” output=”true”> <cfargument name=”pairStr” type=”string” required=”true”> <cfset var result = structNew()> <cfset var x = ““> <cfset var pair = ““>

``````&lt;!--- must be pairs ---&gt;
&lt;cfif len(arguments.pairStr) mod 2 is not 0&gt;
&lt;cfthrow message="Avast ye matey! This string is not an even set of pairs!"&gt;
&lt;/cfif&gt;

&lt;cfloop index="x" from="1" to="#len(arguments.pairStr)#" step="2"&gt;
&lt;cfset pair = mid(arguments.pairStr, x, 2)&gt;
&lt;cfif not structKeyExists(result, pair)&gt;
&lt;cfset result[pair] = 0&gt;
&lt;/cfif&gt;
&lt;cfset result[pair] = result[pair] + 1&gt;
&lt;/cfloop&gt;

&lt;cfreturn result&gt; &lt;/cffunction&gt;
``````

<cfdump var=”#getPairStats(str)#”> </code>

The UDF, getPairStats, accepts a string and returns a struct. It first does a sanity check to ensure that there is an even number of characters. If not, it throws an error using my special “Pirate-Mode(tm)” brand error handling. (Coming soon to BlogCFC5.) After that it is nothing more than a simple loop using step=”2” to tell ColdFusion to skip over every other letter. I use Mid() to grab the pairs and then simply update a structure.

Raymond Camden's Picture