How to extract numbers from a string and get an array of ints?

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 wond

Answer 1

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

Answer 2

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 pseudocode

    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

Answer 3

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 4

I found this expression simplest

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

Answer 5

Pattern p = Pattern.compile("-?\\d+");
Matcher m = p.matcher("There are more than -2 and less than 12 numbers here");
while (m.find()) {
  System.out.println(m.group());
}

... prints -2 and 12.

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-? matches a leading negative sign -- optionally. \d matches a digit, and we need to write \ as \\ in a Java String though. So, \d+ matches 1 or more digits.

Answer 6

What about to use replaceAll java.lang.String method:

    String str = "qwerty-1qwerty-2 455 f0gfg 4";      
    str = str.replaceAll("[^-?0-9]+", " "); 
    System.out.println(Arrays.asList(str.trim().split(" ")));

Output:

[-1, -2, 455, 0, 4]

---

Description

[^-?0-9]+

• [ and ] delimites a set of characters to be single matched, i.e., only one time in any order
• ^ Special identifier used in the beginning of the set, used to indicate to match all characters not present in the delimited set, instead of all characters present in the set.
• + Between one and unlimited times, as many times as possible, giving back as needed
• -? One of the characters “-” and “?”
• 0-9 A character in the range between “0” and “9”