How to extract numbers from a string and get an ar

Question

I have a String variable (basically an English sentence with an unspecified number of numbers) and I'd like to extract all the numbers into an array of integers. I was wondering whether there was a quick solution with regular expressions?

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I used Sean's solution and changed it slightly:

LinkedList<String> numbers = new LinkedList<String>();

Pattern p = Pattern.compile("\\d+");
Matcher m = p.matcher(line); 
while (m.find()) {
   numbers.add(m.group());
}

Answer 1

The accepted answer detects digits but does not detect formated numbers, e.g. 2,000, nor decimals, e.g. 4.8. For such use -?\\d+(,\\d+)*?\\.?\\d+?:

        Pattern p = Pattern.compile("-?\\d+(,\\d+)*?\\.?\\d+?");
        List<String> numbers = new ArrayList<String>();
        Matcher m = p.matcher("Government has distributed 4.8 million textbooks to 2,000 schools");
        while (m.find()) {  
            numbers.add(m.group());
        }   
        System.out.println(numbers);

Output: [4.8, 2,000]

Answer 2

Fraction and grouping characters for representing real numbers may differ between languages. The same real number could be written in very different ways depending on the language.

The number two million in German

2,000,000.00

and in English

2.000.000,00

A method to fully extract real numbers from a given string in a language agnostic way:

public List<BigDecimal> extractDecimals(final String s, final char fraction, final char grouping) {
    List<BigDecimal> decimals = new ArrayList<BigDecimal>();
    //Remove grouping character for easier regexp extraction
    StringBuilder noGrouping = new StringBuilder();
    int i = 0;
    while(i >= 0 && i < s.length()) {
        char c = s.charAt(i);
        if(c == grouping) {
            int prev = i-1, next = i+1;
            boolean isValidGroupingChar =
                    prev >= 0 && Character.isDigit(s.charAt(prev)) &&
                    next < s.length() && Character.isDigit(s.charAt(next));                 
            if(!isValidGroupingChar)
                noGrouping.append(c);
            i++;
        } else {
            noGrouping.append(c);
            i++;
        }
    }
    //the '.' character has to be escaped in regular expressions
    String fractionRegex = fraction == POINT ? "\\." : String.valueOf(fraction);
    Pattern p = Pattern.compile("-?(\\d+" + fractionRegex + "\\d+|\\d+)");
    Matcher m = p.matcher(noGrouping);
    while (m.find()) {
        String match = m.group().replace(COMMA, POINT);
        decimals.add(new BigDecimal(match));
    }
    return decimals;
}

Answer 3

for rational numbers use this one: (([0-9]+.[0-9]*)|([0-9]*.[0-9]+)|([0-9]+))

Answer 4

I found this expression simplest

String[] extractednums = msg.split("\\\\D++");

Answer 5

I would suggest to check the ASCII values to extract numbers from a String Suppose you have an input String as myname12345 and if you want to just extract the numbers 12345 you can do so by first converting the string to Character Array then use the following psuedocode

for(int i=0;i<CharacterArray.length;i++)
    {
  if(a[i]>=48&&a[i]<=58)
          System.out.print(a[i]);
    }

once the numbers are extracted append them to an array

Hope this helps