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Number of numeric commas inside a string, excluding commas between double quotes

I have the following function to count the number of commas (or any other character) in a String without taking into account internal double quotes. I want to know if there is a better way to achieve this, or even if you find some case where this function might work.

public int countCharOfString(char c, String s) {
    int numberOfC = 0;
    boolean doubleQuotesFound = false;
    for(int i = 0; i < s.length(); i++){
        if(s.charAt(i) == c && !doubleQuotesFound){
            numberOfC++;
        }else if(s.charAt(i) == c && doubleQuotesFound){
            continue;
        }else if(s.charAt(i) == '\"'){
            doubleQuotesFound = !doubleQuotesFound;
        }
    }
    return numberOfC;
}

Thanks for any advice.

+3
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7 answers

This implementation has two differences:

  • Use CharSequenceinstead of string
  • There is no need booleanfor tracking value if we are inside the specified subsequence.

Function:

public static int countCharOfString(char quote, CharSequence sequence) {

    int total = 0, length = sequence.length();

    for(int i = 0; i < length; i++){
        char c = sequence.charAt(i);
        if (c == '"') {
            // Skip quoted sequence
            for (i++; i < length && sequence.charAt(i)!='"'; i++) {}
        } else if (c == quote) {
            total++;
        }
    }

    return total;
 }
+3
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public static int countCharOfString(char c, String s)
{
    int numberOfC = 0;
    int innerC = 0;
    boolean holdDoubleQuotes = false;
    for(int i = 0; i < s.length(); i++)
    {
        char r = s.charAt(i);
        if(i == s.length() - 1 && r != '\"')
        {
            numberOfC += innerC;
            if(r == c) numberOfC++;
        }
        else if(r == c && !holdDoubleQuotes) numberOfC++;
        else if(r == c && holdDoubleQuotes) innerC++;
        else if(r == '\"' && holdDoubleQuotes)
        {
            holdDoubleQuotes = false;
            innerC = 0;
        }
        else if(r == '\"' && !holdDoubleQuotes) holdDoubleQuotes = true;
    }
    return numberOfC;
}


System.out.println(countCharOfString(',', "Hello, BRabbit27, how\",,,\" are, you?"));

CONCLUSION:

3

An alternative would be to use a regular expression:

public static int countCharOfString(char c, String s)
{
   s = " " + s + " "; // To make the first and last commas to be counted
   return s.split("[^\"" + c + "*\"][" + c + "]").length - 1;
}
+2
  • charAt() . char.
  • length() . int.
  • c - if/else.
+1

, ...

public int countCharOfString(char c, String s) {
    final String removedQuoted = s.replaceAll("\".*?\"", "");
    int total = 0;
    for(int i = 0; i < removedQuoted.length(); ++i)
        if(removedQuoted.charAt(i) == c)
            ++total;
    return total;
}
+1

.

, , , 1,5 3 . , , .

public static void main(String... args) {
    String s = generateString(20 * 1024 * 1024);
    for (int i = 0; i < 15; i++) {
        long start = System.nanoTime();
        countCharOfString(',', s);
        long mid = System.nanoTime();
        countCharOfString2(',', s);
        long end = System.nanoTime();
        System.out.printf("countCharOfString() took %.3f ms, countCharOfString2() took %.3f ms%n",
                (mid - start) / 1e6, (end - mid) / 1e6);
    }
}

private static String generateString(int length) {
    StringBuilder sb = new StringBuilder(length);
    Random rand = new Random(1);
    while (sb.length() < length)
        sb.append((char) (rand.nextInt(96) + 32)); // includes , and "
    return sb.toString();
}

public static int countCharOfString2(char c, String s) {
    int numberOfC = 0, i = 0;
    while (i < s.length()) {
        // not quoted
        while (i < s.length()) {
            char ch = s.charAt(i++);
            if (ch == c)
                numberOfC++;
            else if (ch == '"')
                break;
        }
        // quoted
        while (i < s.length()) {
            char ch = s.charAt(i++);
            if (ch == '"')
                break;
        }
    }
    return numberOfC;
}


public static int countCharOfString(char c, String s) {
    int numberOfC = 0;
    boolean doubleQuotesFound = false;
    for (int i = 0; i < s.length(); i++) {
        if (s.charAt(i) == c && !doubleQuotesFound) {
            numberOfC++;
        } else if (s.charAt(i) == c && doubleQuotesFound) {
            continue;
        } else if (s.charAt(i) == '\"') {
            doubleQuotesFound = !doubleQuotesFound;
        }
    }
    return numberOfC;
}


countCharOfString() took 33.348 ms, countCharOfString2() took 31.381 ms
countCharOfString() took 28.265 ms, countCharOfString2() took 25.801 ms
countCharOfString() took 28.142 ms, countCharOfString2() took 14.576 ms
countCharOfString() took 28.372 ms, countCharOfString2() took 14.540 ms
countCharOfString() took 28.191 ms, countCharOfString2() took 14.616 ms
+1

Simplified, less error prone (and yes, less efficient than walking the char string to char and tracking everything manually):

public static int countCharOfString(char c, String s) {
  s = s.replaceAll("\".*?\"", "");
  int cnt = 0;
  for (int foundAt = s.indexOf(c); foundAt > -1; foundAt = s.indexOf(c, foundAt+1)) 
    cnt++;
  return cnt;
}
+1
source

You can also use regex and String.split ()

It might look something like this:

public int countNonQuotedOccurrences(String inputstring, char searchChar)
{
  String regexPattern = "[^\"]" + searchChar + "[^\"]";
  return inputString.split(regexPattern).length - 1;
}

Denial of responsibility:

It just shows a basic approach.

The above code will not check for searchChar at the beginning or end of a line.

You can either check this manually or add to regexPattern.

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