Java regular expression to identify strings with more digits than non-digits

Question

How can I identify strings containing more digits than non-digits using regular expression (Pattern) in Java? Thank you.

Answer 1

That's not a regular language, and thus it cannot be captured by a vanilla regex. It may be possible anyway, but it will almost certainly be easier not to use a regex:

public static boolean moreDigitsThanNonDigits(String s) {
    int diff = 0;
    for(int i = 0; i < s.length(); ++i) {
        if(Character.isDigit(s.charAt(i))) ++diff;
        else --diff;
    }
    return diff > 0;
}

Answer 2

I'm not sure that using regular expressions would be the best solution here.

Answer 3

You won't be able to write a regexp that does this. But you already said you're using Java, why not mix in a little code?

public boolean moreDigitsThanNonDigits(String input) {
    String nonDigits = input.replace("[0-9]","");
    return input.length() > (nonDigits.length * 2);
}

Answer 4

regex alone can't (since they don't count anything); but if you want to use them then just use two replacements: one that strips out all the digits and one that only keeps them. then compare string lengths of the results.

of course, i'd rather use Dave's answer.

Answer 5

Regular expressions are conceptually not able to preform such a task. They are equivalent to formal languages or (regular) automatons. They have no notion of memory (or a stack), so they cannot count the occurences of symbols. The next extension in terms of expressiveness are push-down automatons (or stack machines), which correspond to context free grammars. Before writing such a grammer for this task, using a method like the moreDigitsThanNonDigits above would be appropriate.