integer based sort in 2-d string array

Question

I have this 2 dimensional string array.

2 10 BakerSarah D 
2 11 SmothersSally A 
2 12 SillySall C 
2 13 Viper B 
2 5 LouieChef B 
2 6 Lawson C

Each column is string . Now I want to sort this on second column . i have tried this code

void sortarray(final int index){
        Arrays.sort(data, new Comparator<Object[]>(){
            @Override
            public int compare(Object[] o1,  Object[] o2) {
                String[] a = (String[])o1;
                String[] b = (String[])o1;
                return a[index].compareTo(b[index]);
            }
        });

    }

but this is giving sort in

    2 10 BakerSarah D 
    2 11 SmothersSally A 
    2 12 SillySall C 
    2 13 Viper B 
    2 5 LouieChef B 
    2 6 Lawson C

in order . Why so ?? How to change that to sort

2 5 LouieChef B 
2 6 Lawson C   
2 10 BakerSarah D 
2 11 SmothersSally A 
2 12 SillySall C 
2 13 Viper B

you want integer sorting, but your value is still a string, you need to convert it to integer, and compare the integers — x4rf41
– x4rf41, Commented Aug 20, 2013 at 12:09

RNJ · Accepted Answer · 2013-08-20 12:13:31Z

1

When the compare returns 0 (ie they are equal) then you need compare on another index. I have updated your code to have this new index - index2.

void sortarray(final int index, final int index2){
    Arrays.sort(data, new Comparator<Object[]>(){
        @Override
        public int compare(Object[] o1,  Object[] o2) {
            String[] a = (String[])o1;
            String[] b = (String[])o1;
            Integer i = a[index].compareTo(b[index]);
            if (i == 0) {
               return a[index2].compareTo(b[index2]);
            } 
            return i;
        }
    });

}

I am assuming (perhaps incorrectly) that you want to sort my the first column and then by the second. If it is just the second then try what @x4rf41 says and do an Integer.valueOf to convert the string to an integer

Personally though I would create an object and implement Comparator on this so that you can sort in a more OO way.

edited Aug 20, 2013 at 12:13

answered Aug 20, 2013 at 12:08

RNJ

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4 Comments

x4rf41 Over a year ago

you need to convert the string to an integer, then the sorting will work

RNJ Over a year ago

index2 is the second index you want to sort on. In this example it is the column containing 5,6,10,....

user2699538 Over a year ago

but that's index parameter itself

JB Nizet Over a year ago

That really doesn't answer the question.

JB Nizet · Accepted Answer · 2013-08-20 12:15:01Z

Strings have a natural, lexicographic order. Which means that "10" comes before "5". Integers have a natural, numeric order. So you should transform your strings into numbers and compare the numbers:

Arrays.sort(data, new Comparator<Object[]>(){
    @Override
    public int compare(Object[] o1,  Object[] o2) {
        String[] a = (String[])o1;
        String[] b = (String[])o1;
        if (index == 2) { // lexicographic order
            return a[index].compareTo(b[index]);
        }
        else { // numeric order
            int left = Integer.parseInt(a[index]);
            int right = Integer.parseInt(b[index]);
            return Integer.compare(left, right);
        }
    }
});

Note that this wouldn't happen if, instead of using a String[] to hold your information, you used a proper class, with fields of the appropriate type:

public class Row { // choose a better name
    private int field1; // choose a better name
    private int field1; // choose a better name
    private String name;

    // constructor and getters omitted
}

Java is an OO language. Use objects.

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