CODEWITHSUNDEEP
iphoneobjective-ciosipad
innesngunn
innesngunn

Reputation: 323

Fastest way to compare/set strings

I have an array of Place objects. Each Place object has a name and code property, both strings. Each Place object already has a code, but I need to look up the name property from a server. I get back 2 arrays: one contains name, the other codes. These arrays are ordered so that the name at some index in the nameArray corresponds exactly with the code at the same index of the codeArray.

I have been looping through the array of Place objects, then checking to see if the code property for that Place is the same as the current index in the codeArray. If it is, I set the name of that Place by using the same index in the nameArray:

for(int i = 0; i < [placesArray count]; i++)
{
    for(int j = 0; j < [codeArray count]; j++) {

       if([[[placesArray objectAtIndex:i] code] isEqualToString:[codeArray objectAtIndex:j]]) {   
            [[placesArray objectAtIndex:i] setName:[nameArray objectAtIndex:j]];
       }
    }
}

This works but isn't terribly fast - it can take 30 seconds with 1000+ Places to loop through.

Is there a faster way?

Upvotes: 0

Views: 281

Answers (4)

Raymond Wang
Raymond Wang

Reputation: 1484

The problem was not the counting for the array, it's the embedded for loop which will take O(n*n) and in Andrew's solution, it's only O(n)+O(n)+O(n)+whatever take to find a object of the key in the dictionary, which i guess would be in a hash table lookup and that's really fast.

Colby, you probably will be ok with Andrew's solution. If you still wanna improve the performance, then a good idea would be sort the array's first then do lookup.

Hope this helps.

Upvotes: 0

Joel Kravets
Joel Kravets

Reputation: 2471

Try using a break statement in the inner loop. This way you don't need to loop through the entire loop each time.

for(int i = 0; i < [placesArray count]; i++)
{
    for(int j = 0; j < [codeArray count]; j++) {

       if([[[placesArray objectAtIndex:i] code] isEqualToString:[codeArray objectAtIndex:j]]) {   
            [[placesArray objectAtIndex:i] setName:[nameArray objectAtIndex:j]];
            break;
       }
    }
}

You could also make the second array become smaller as you find more results. It will cost you more memory but 1000 strings isn't much anyway.

NSMutableArray * tempCodeArray = [NSMutableArray arrayWithArray:codeArray];
for(int i = 0; i < [placesArray count]; i++)
{
    for(int j = 0; j < [tempCodeArray count]; j++) {

       if([[[placesArray objectAtIndex:i] code] isEqualToString:[tempCodeArray objectAtIndex:j]]) {   
            [[placesArray objectAtIndex:i] setName:[nameArray objectAtIndex:j]];
            [tempCodeArray removeObjectAtIndex:j];
            break;
       }
    }
}

Upvotes: 0

Andrew Madsen
Andrew Madsen

Reputation: 21373

As with anytime you're trying to optimize performance, you should profile the code using Instruments to find out where the bottleneck actually is. That said, looping through the placesArray for each name in the nameArray and doing a string comparison is pretty inefficient.

How about something like this?

NSMutableDictionary *placesByCode = [NSMutableDictionary dictionaryWithCapacity:[placesArray count]];
for (Place *aPlace in placesArray) {
    [dictionary setObject:aPlace forKey:aPlace.code];
}

NSMutableDictionary *namesByCode = [NSMutableDictionary dictionaryWithCapacity:[namesArray count]];
for (int i=0; i<[namesArray count]; i++) {
    NSString *name = [namesArray objectAtIndex:i];
    NSString *code = [codeArray objectAtIndex:i];
    [namesByCode setObject:name forKey:code];
}

for (NSString *code in namesByCode) {
    Place *place = [placesByCode objectForKey:code];
    place.name = [namesByCode objectForKey:namesByCode];
}

Looking up each place by its code in the dictionary should be quite a bit faster than manually looping through the whole place array for each name.

Upvotes: 1

Jaume
Jaume

Reputation: 923

You can use for NSArray -containsObject

if ([myarray containsObject:myObject]) {
    // ...
}

Upvotes: 0

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