Extract number and operators from a String in Objective-C - regex

Question

I have this string (2.5+1)*5
I want to make a NSMutableArray containing each of the elements:
[(, 2.5, +, 1, ), *, 5]
This is the code I wrote:
+ (NSMutableArray*) convertToList: (NSString *) exp {
NSRegularExpression * reg = [NSRegularExpression regularExpressionWithPattern: #"(\\d+\\.\\d+)|(\\d+)|([+-/*///^])|([/(/)])" options:0 error:nil];
NSArray *matches = [ reg matchesInString: exp
options: 0
range:NSMakeRange(0, [exp length])];
return [matches mutableCopy];
}
int main (int argc, char *argv[]) {
NSString * exp = #"(2+5)";
NSMutableArray * array = [ExpressionTexting convertToList:exp];
NSLog(#"%#", array);
}
I get this output:
(
"<NSExtendedRegularExpressionCheckingResult: 0x10010d2e0>{0, 1}{<NSRegularExpression: 0x10010abb0> (\\d+\\.\\d+)|(\\d+)|([+-/*///^])|([/(/)]) 0x0}",
"<NSExtendedRegularExpressionCheckingResult: 0x10010d3f0>{1, 1}{<NSRegularExpression: 0x10010abb0> (\\d+\\.\\d+)|(\\d+)|([+-/*///^])|([/(/)]) 0x0}",
"<NSExtendedRegularExpressionCheckingResult: 0x10010d470>{2, 1}{<NSRegularExpression: 0x10010abb0> (\\d+\\.\\d+)|(\\d+)|([+-/*///^])|([/(/)]) 0x0}",
"<NSExtendedRegularExpressionCheckingResult: 0x10010d4f0>{3, 1}{<NSRegularExpression: 0x10010abb0> (\\d+\\.\\d+)|(\\d+)|([+-/*///^])|([/(/)]) 0x0}",
"<NSExtendedRegularExpressionCheckingResult: 0x10010d570>{4, 1}{<NSRegularExpression: 0x10010abb0> (\\d+\\.\\d+)|(\\d+)|([+-/*///^])|([/(/)]) 0x0}"
)
In Java I got the answer:
public static ArrayList<String> convertToList(String exp){
String regex = "(\\d+\\.\\d+)|(\\d+)|([+-/*///^])|([/(/)])";
Matcher m3 = Pattern.compile(regex).matcher(exp);
ArrayList<String> list = new ArrayList<String>(exp.length());
while (m3.find()) {
list.add(m3.group());
}
return list;
}
Which works, but I can't translate it in Objective-c

Answer 1

You should use regex pattern
(\\d+\\.?\\d*|\\.\\d+|[^\\d.])
Java code:
String str = "(2.5+1)*5";
Pattern p = Pattern.compile("(\\d+\\.?\\d*|\\.\\d+|[^\\d.])");
Matcher m = p.matcher(str);
while (m.find()) {
String match = m.group();
System.out.println(match);
}
Output:
(
2.5
+
1
)
*
5
Check this code here.

Answer 2

Maybe you need a math parser.
check DDMathParser

Answer 3

I found the solution, the error was due to the fact that I didn't really know how to handle the result, the NSArray returned is made of NSTextCheckingResult, which I have to use to extract the numbers and the operators.
+ (NSMutableArray*) convertToList: (NSString *) exp {
NSRegularExpression * reg = [NSRegularExpression regularExpressionWithPattern: #"(\\d+\\.\\d+)|(\\d+)|([+-/*///^])|([/(/)])" options:0 error:nil];
NSArray *matches = [ reg matchesInString: exp
options: 0
range:NSMakeRange(0, [exp length])];
NSMutableArray *mArray = [[NSMutableArray alloc] init];
for (NSTextCheckingResult *match in matches) {
NSString* matchText = [exp substringWithRange:[match range]];
[mArray addObject:matchText];
//NSLog(#"%#", matchText);
}
return mArray;
}

Answer 4

As was alluded to in another answer, DDMathParser can do this quite easily. It looks like you don't care about the hierarchy of the actual terms, so you could get away with just using the resulting tokens. That's easy:
NSError *error = nil;
DDMathStringTokenizer *tokenizer = [[DDMathStringTokenizer alloc] initWithString:#"(2.5+1)*5" error:&error];
NSArray *tokens = [tokenizer tokens];
tokens = [tokens valueForKey:#"token"];
NSLog(#"%#", tokens);
The -tokens method on DDMathStringTokenizer returns an array of DDMathStringToken objects. Those objects have an underlying NSString of the actual token (accessed via the token property), so you'll need the valueForKey: call on the array to turn it into an array of strings. But if you do this, then you'll get this logged:
(
"(",
"2.5",
"+",
1,
")",
"*",
5
)