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Re: Approaches for this Matching Problem?
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Re: Approaches for this Matching Problem?


  • Subject: Re: Approaches for this Matching Problem?
  • From: Steve Cronin <email@hidden>
  • Date: Mon, 22 Jun 2009 13:25:39 -0500

Ken;

Thanks as always for a thought-provoking and thorough response!

1) I haven't given up on those Tiger users yet...

2) I don't see how looping on the 'words' in thePhrase would change the 'ham' issue you raise - it would still match either on the word or the phrase.
Am I missing something?

3) The actual tags in this particular context are extensive enough that this 'ham-confusion' is pretty unlikely and could be easily remedied by the user's management of the tag list... [e.g. service, management, institute] -- I know there remains the root forms issue but let's not drag that issue in here


Steve


On Jun 22, 2009, at 12:52 PM, Ken Thomases wrote:

On Jun 22, 2009, at 3:10 AM, Steve Cronin wrote:

So I've been pondering and testing...

1) Why would I have to bust up/loop on thePhrase?

BUT in my case I don't care where, what word or how often - I just want to know IF a tag occurs in thePhrase!!

Then you shouldn't refer to "words". If you just care if any of a set of strings appear in a given string, that's a different question.


If one of your tags is "ham", do you want it to match a phrase containing "shame"?


- (BOOL) containsTag:(NSString *)thePhrase {
BOOL tagFound = NO;
NSString *tag;
NSEnumerator *tagEnum = [tags objectEnumerator]; //tags is an NSSet instance variable built from Core Data
while ((tag=[tagEnum nextObject])!=nil) {
if ([thePhrase rangeOfString:tag options:NSCaseInsensitiveSearch].location!=NSNotFound) {
tagFound = YES; break;
}
}
return tagFound;
}

If you can target Leopard or later, you can simplify this even more by using "for (NSString* tag in tags) ...". It's more efficient, too.


Regards,
Ken


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References: 
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 >Re: Approaches for this Matching Problem? (From: Steve Cronin <email@hidden>)
 >Re: Approaches for this Matching Problem? (From: Ken Thomases <email@hidden>)

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