Some facts:
- Small pieces of code like this one can be pasted directly in your message, they are easier to consult. Use the Insert Code (n-2 button) command to format it after pasting. Pastebin is good for long listing, though.
- Each time I see delay(...) at the end of draw(), it rings an alarm bell... You have to wait one minute before seeing the result of the request... Check millis() against the delay before doing an update.
- Are you sure you mean "regular expressions"? You can do simple searches simply, no need for regexes to search "colour". If you search both color and colour, two searches will probably be faster than compiling the RE.
- You already iterate on the list, so just do your search on the getText() result. Look at String's indexOf() or contains() for simple searches.

The reason I thought regular expressions were the way to go (and this is probably my bad for not explain properly) is because I don't want to search for the terms colour or color but instead the actual names of colours (specifically this list  http://en.wikipedia.org/wiki/List_of_colors) 

THen maybe you do want a regex, so that you can distinguish "red" from "scared"

Use the Java Pattern and Matcher classes.

To get the color values quickly, put them in a Java Hashmap, with the color names as keys -- e.g., in your setup include

1. HashMap colorMap;
2. 
   
3. /// inside setup()....
4. colorMap = new HashMap();
5. colorMap.put("red",color(255,0,0)); // etc......

and then when you get a color name, grab it using get

1. // for some string "colorName"
2. if (colorMap.containsKey(colorName)) {
3.     int newCol = colorMap.get(colorName);
   
4. } else {
5.    // do nothing, or something -- no such color
   
6. }
   

kb, http://www.riftgame.com/

Could someone please go into a little more detail about how the 2nd part would work..

My code has changed a fair bit since earlier.  http://pastebin.com/xmAf1CTm and I've run into a new issue..

Just declare:
HashMap<String, Integer> colorMap = new HashMap<String, Integer>();
(thus don't init in setup()) and get() will return an Integer without needing further cast (by default, it fetches Object, the super-generic object type of Java, hence the message).
Note that this means you compare (with match() which is, I persist, not the best fit here) the color name with the color value, which isn't what you want.
Beside, you completely miss the usefulness of HashMap: if coloursString is a single word that can happen in colorMap (I suggest to lowercase the names, and coloursString), you can do simply:
Integer col = colorMap.get(coloursString);
if (col == null) println("no colour");
else println("Found colour " + (color) col);

colourString is the latest tweet downloaded which could be 1 word it could be a sentence and it will always start with @username as the tweets are @replies from other users. So the keys in colorMap are names of colours which could occur in coloursString i.e "@twitteraccount I like the colour Amber" It needs to be able to pick up the work Amber, which is a Key in my colorMap, but not require the whole string to match just the key word.

" didn't work, it will now returns the error"
Yes, that's what I wrote; get() returns a color, not a color name. So it works, but it is still not what you want.
Although, no, there is something I overlooked: get(i) just won't work, as you must do get() with color names, not indexes.
I fear that in your case, HashMap is next to useless. Just put an array list of couples of color names and color values, and iterate over it.
Something like:

1. class ColorInfo {
2.   String name;
3.   color value;
4.   ColorInfo(String n, color v) { name = n; value = v; }
5. }
6. 
   
7. ArrayList<ColorInfo> colorMap = new ArrayList<ColorInfo>();
8. colorMap.add(new ColorInfo("Air Force Blue", color(93,138,168)));
9. colorMap.add(new ColorInfo("Alice Blue", color(240,248,255)));
10. // etc.
11. 
    
12.     while (checkcolour.hasNext()) {
13.       Status element = checkcolour.next();
14.       String coloursString = element.getText();
15.       
16.       for (int i = 0; i < colorMap.size(); i++) { // Initialize i! Don't hard code size!
17.         String ci = colorMap.get(i);
18.         if (coloursString.contains(ci.name)) {
19.           println("found colour " + ci.name + " -> " + hex(ci.value));
20.         }
21.       }
22.     }

Untested!

This one won't differentiate "I like red" from "I am scared" -- which is why you should use match()



kb, http://www.riftgame.com/

Given the list of colors (mostly composed), the risk here is minor. And match() won't work better if you don't tell how to ensure whole word matching (using \\b).

Yes, using \\b is exactly the point :)

kb, http://www.riftgame.com/

So I've tried changing it to an array 

The incorrect line (you should indicate it when you report errors) is probably String ci = colorMap.get(i);
You defined colorMap as an array list of ColorInfo, you cannot get a String out of it!
What you want to do is:

1.       for (ColorInfo ci : colorMap) {
2.         if (match(coloursString, "\\b" + ci.name + "\\b") != null) { // Lot of stuff was wrong there too!
   
3.              println("found colour " + ci.name + " -> " + hex(ci.value));
4.         } else {
5.              println("GTFO");
6.         }
7.       }

Or with your numerical for loop: ColorInfo ci = colorMap.get(i);

Thanks, that works!