I am a bit new to Objective C and was wondering if there is a better way to count words in a string.
ie:
NSString *str = @"this is a string";
// return should be 4 words ..
The way I now how to do it is by breaking the string into an array of words space (' ') character and count the array.
Any advise will be appreciated! Thanks!! :)
EDIT: For those of you who came here looking for answer; I found a similar post with an excellent reply.
There are two ways that don't involve collecting an array of words, and should be smarter than just breaking on spaces:
enumerateSubstringsInRange:options:usingBlock:
method. Introduced to Cocoa Touch in iOS 4.0.I would use one of these, even if I did want to collect or otherwise use the words.
Are you sure you have a bottleneck in that part of code? If not (which is quite probable), then splitting on spaces seems perfectly acceptable to me. You could create a C string and count the spaces instead, but a lot of times such an “optimized” version is actually slower than the original one. That is, assuming that your current code looks like this:
This is not exactly correct (see
@"___"
where underscore is a space), but maybe you really use a regex and split on\s+
?In this situation, I'd use an NSScanner like so:
This only creates two additional objects, no matter how long the string is.
This code will count the number of words (i.e., non-empty substrings) contained in a string that are separated by any number of space or line break characters:
It's a slight improvement to zoul's answer without recurring to regexes.
Unless you're going to be doing it hundreds of times a second, I would just opt for the readable solution, something like the following pseudocode:
It seems a bit of a waste to create a whole array of other strings just so you can count them and throw them away.
And if, for some reason, it becomes an issue, you can just replace the function body with a faster version. Make sure it is a problem first however. Optimisation of code that's fast enough already is wasted effort.
One liner accurate solution: