Sorting array of strings by value within the string

Question

I've been stumped by this for a while now. I have an array of strings all of which contain some numeric value that I need to extract and use for comparison in order to sort the array. I have tried making use of the built in method sort with a comparator but have not been successful with it.

If anyone could give me a little insight or point me in the right direction that would be greatly appreciated.

The array contains strings in the format: "NAME has worked x hours"

My intent is to pull the number out and sort based off the value of the integer while still keeping the relationship to the name.

public static <T> void sort(T[] a, Comparator<? super T> c)

can you post the array of string? and the result you want it to be — Rod_Algonquin
– Rod_Algonquin, Commented Sep 4, 2014 at 4:45
There are tons of answers on implementing "natural sort" in all languages... — Alexei Levenkov
– Alexei Levenkov, Commented Sep 4, 2014 at 4:47
possible duplicate of Natural sort order string comparison in Java - is one built in? — Alexei Levenkov
– Alexei Levenkov, Commented Sep 4, 2014 at 4:48
@Rod_Algonquin - He has an array of Strings like {"10","2","1"}. The problem with natural sorting of Strings (default Arrays.sort(stringArray) is that he will get the answer as {"1","10","2"}. He needs {"1","2","10"} (integer sort). — TheLostMind
– TheLostMind, Commented Sep 4, 2014 at 4:50
@TheLostMind - I agree, I think this is the problem. However, I think the answer is to parse the strings into custom objects, and then sort those instead. I can't think of any benefit to dealing directly with Strings, especially if your objects have an equivalent toString method. — DaoWen
– DaoWen, Commented Sep 4, 2014 at 4:53

Rod_Algonquin · Accepted Answer · 2014-09-04 04:56:30Z

2

You can use the Collections.sort to sort your string using the comparator and parsing the String to Integer to enable you to sort it by integer.

sample:

    String s[] = { "10", "2", "1" };
    Collections.sort(Arrays.asList(s), new Comparator<String>() {

        @Override
        public int compare(String o1, String o2) {
            int i = Integer.parseInt(o1);
            int i2 = Integer.parseInt(o2);
            if (i > i2)
                return 1;
            else if (i < i2)
                return -1;
            else
                return 0;
        }
    });
    System.out.println(Arrays.toString(s));

result:

[1, 2, 10]

answered Sep 4, 2014 at 4:56

Rod_Algonquin

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1 Comment

TheLostMind Over a year ago

The array contains strings in the format: "NAME has worked x hours" - So, he needs to change his compare() logic somewhat (based on the edited question). But more or less, this is the correct answer. +1.

Upio · Accepted Answer · 2014-09-04 08:00:41Z

2

Make sure you handle all the edge cases but the below code should work for you.

public class NumericStringComparator implements Comparator<String> {
    @Override
    public int compare(String str1, String str2) {
        Integer num1 = extractNumericValue(str1);
        Integer num2 = extractNumericValue(str2);

        // handle null/error cases

        return num1.compareTo(num2);
    }

    private static Integer extractNumericValue(String str) {
        // extract numeric value however but as an example
        return Integer.parseInt(str);
    }
}

With this comparator you can use the Arrays.sort(..) method to sort your list

String[] array = ...;
Arrays.sort(array, new NumericStringComparator());

edited Sep 4, 2014 at 8:00

answered Sep 4, 2014 at 4:59

Upio

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1 Comment

Jan Groth Over a year ago

Best of many answers IMHO. BTW, if extractNumericValue() returned an Integer, the whole comparison part would shrink to extractValue(str1).compareTo(extractValue(str2))

Kevin · Accepted Answer · 2014-09-04 04:57:50Z

0

if each string only contains numbers, then you can just use ParseInt to get integer values for each string and then sort them using a regular comparator. If the strings contain non-numeric characters and integers, then this complicates things, and you might want to make a helper method to do the dirty work for you. A possible (though inefficient) implementation might look like...

public static int getNumbersFromString(String string)
{
    String numbers = "";
    for(int i = 0; i < string.length; i++)
    {
        if((int)(string.charAt(i) >= 48 && (int)(string.charAt(i) <= 57)
            numbers += string.charAt(i);
    }
    int foo = Integer.parseInt(numbers);
    return foo;
}

answered Sep 4, 2014 at 4:57

Kevin

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