Sorting list of Strings having integer values- Java

Question

I have this list of Strings

Car1

Car2

Car3

......


Carn

I want to sort this list according to numerics in it.
for instance if I have Car3, Car1, Car12, Car45, I want it to be sorted like Car1, Car3, Car12, Car45.
I have used Collections.sort(), but it returns something like Car1, car12, car3, Car45.
what should I do to put it in correct order?

Have you seen [this question :)](http://stackoverflow.com/questions/14337682/sorting-alphanumeric-strings-java) — Freak, Jun 26 '13 at 09:44
Check this http://stackoverflow.com/questions/104599/sort-on-a-string-that-may-contain-a-number — Rohan, Jun 26 '13 at 09:46

score 6 · Answer 1 · answered Jun 26 '13 at 09:46

6

You need a custom Comparator, something like

    Collections.sort(list, new Comparator<String>() {
        public int compare(String s1, String s2) {
            int i1 = Integer.parseInt(s1.replaceAll("\\D", ""));
            int i2 = Integer.parseInt(s2.replaceAll("\\D", ""));
            return Integer.compare(i1, i2);
        }
    });

answered Jun 26 '13 at 09:46

Evgeniy Dorofeev

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+1, but for performance I'd compile the regex and use a positive match to extract the numbers. Comparators usually are performance-critical. – Marko Topolnik Jun 26 '13 at 09:54

score 4 · Answer 2 · answered Jun 26 '13 at 09:45

4

Use this:

class ComparatorOfNumericString implements Comparator<String>{

    public int compare(String string1, String string2) {
        // TODO Auto-generated method stub
        String a = string1.substring(3);
        String b = string2.split(3);
        return Integer.parseInt(a)-Integer.parseInt(b);
    }
}

Now when sorting pass this comparator as follows:

Collections.sort(stringList,new ComparatorOfNumericString ());

answered Jun 26 '13 at 09:45

ajay.patel

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2

Be careful when comparing integer using minus - it might overflow. – Howard Jun 26 '13 at 09:53

score 3 · Answer 3 · answered Jun 26 '13 at 09:48

3

The sort is lexicographical and you get this order because 1 is smaller than 3 in the alphabet and the comparer disregards what comes next. What you need to do is to either write your own comparer, that will cut the "Car" leaving only the number in the string, then you need to parse that string representation of the number to an int and compare the ints ( as one of the solutions )

answered Jun 26 '13 at 09:48

user2511124

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+1 for explaining lexicographical ordering of strings – selig Jun 26 '13 at 09:53
Thank you for your response I already did this but I want to be sure that is it the only way to do it or I am just re-inventing the wheel :) – Just_another_developer Jun 26 '13 at 10:00

herman · Answer 4 · 2013-06-26T13:05:23.583

You need to specify a custom Comparator<String> to the Collections.sort() method that defines the ordering that you want for the strings.

This works for a number at any position in the string, and also handles strings that have no numbers (falls back to ordinary string comparison with natural ordering):

final Pattern intsOnly = Pattern.compile("\\d+");

Comparator<String> comparator = new Comparator<String>() {
    @Override
    public int compare(final String string1, final String string2) {
        String int1 = null;
        String int2 = null;
        Matcher matcher1 = intsOnly.matcher(string1);
        if (matcher1.find()) {
            int1 = matcher1.group();
            Matcher matcher2 = intsOnly.matcher(string2);
            if (matcher2.find()) {
                int2 = matcher2.group();
            }
        }
        if (int1 == null || int2 == null) { return string1.compareTo(string2); }
        return Integer.valueOf(int1).compareTo(Integer.valueOf(int2));
    }
};
Collections.sort(strings, comparator);

Of course this means that Truck1 will come before Car2, but that is how you wanted it according to your question.

UPDATE: The above solution does not make sure that Car6 will come before Truck6. If you also want to fall back to natural string ordering in case of equal numbers, use this:

final Pattern intsOnly = Pattern.compile("\\d+");

Comparator<String> comparator = new Comparator<String>() {
    @Override
    public int compare(final String string1, final String string2) {
        String int1 = null;
        String int2 = null;
        Matcher matcher1 = intsOnly.matcher(string1);
        if (matcher1.find()) {
            int1 = matcher1.group();
            Matcher matcher2 = intsOnly.matcher(string2);
            if (matcher2.find()) {
                int2 = matcher2.group();
            }
        }
        int result = 0;
        if (int1 != null && int2 != null) {
            result = Integer.valueOf(int1).compareTo(Integer.valueOf(int2));
        }
        if (result == 0) {
            return string1.compareTo(string2);
        }
        else {
            return result;
        }
    }
};
Collections.sort(strings, comparator);

score 1 · Answer 5 · edited May 23 '17 at 11:45

1

You will need to create a custom Comparator that does the sorting how you want it. You can then call

Collections.sort(myList, myComparator)

See Collections.sort(List, Comparator).

You can read up on comparators here: Implementing Java Comparator

edited May 23 '17 at 11:45

Community

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answered Jun 26 '13 at 09:43

cowls

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score 1 · Answer 6 · answered Jun 26 '13 at 09:47

1

You need to use the overloading of sort with the comparator. The following code is for your specific strings (CarXXX):

Collections.sort(carList, new Comparator<String>() {
    public int compare(String s1, String s2) {
        int t1 = Integer.parseInt(s1.substring(3));
        int t2 = Integer.parseInt(s2.substring(3));
        return t1 - t2;
    }
});

answered Jun 26 '13 at 09:47

Aleks G

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Be careful when comparing integer using minus - it might overflow. – Howard Jun 26 '13 at 09:54
@Howard if an overflow is of concern, then even parsing the values into integers may be a problem. – Aleks G Jun 26 '13 at 09:58

score 1 · Answer 7 · answered Jun 26 '13 at 09:49

1

Collections.sort() is correct, sou just have to write the comparator:

public class testComp implements Comparator<String> {

@Override
public int compare(String o1, String o2) {
    int number1 = Integer.parseInt(o1.substring(3, o1.length()));
    int number2 = Integer.parseInt(o2.substring(3, o2.length()));
    if (number1 > number2) {
        return -1;
    } else {
        return 1;
    }
}

answered Jun 26 '13 at 09:49

Steffen Nieuwenhoven

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Pardon my ignorance, what is this with -1 and 1 .... what they indicate? – Just_another_developer Jun 26 '13 at 10:10
This works only for numbers at index 3 in the string. I'm sure "Car1", "Car2" etc. was just an example. See my answer for a complete solution. – herman Jun 26 '13 at 10:14
Return value of "-1" indicates that "o2" should be above "o1" in the list. Return value of "1" puts "o1" above "o2". A return value of 0 indicates that o1 & o2 are equal. You are basically indicating which way to sort the list. – Ian Dec 12 '18 at 13:38

score 0 · Answer 8 · answered Jun 26 '13 at 09:50

Car1 is object so you have to create a custom Comparator for that. and specify the ordering nature of it.

Collection.Sort() is Sorts the specified list into ascending order, according to the natural ordering of its elements. just example: string will do like a-z, number is 0-9. but objects does not have any natural order.

score 0 · Answer 9 · answered Jun 26 '13 at 11:01

This will work for you. But you can't have duplicate

    String text = "Car3, Car1, Car12, Car45";
    String[] str = text.split(", ");
    Map<Integer,String> myMap=new HashMap<>();
    int[] numOnly=new int[str.length];
    for (int i = 0; i < str.length; i++) {
          numOnly[i] = Integer.parseInt(str[i].replaceAll("\\D", ""));
          myMap.put(numOnly[i],str[i]);
    }       
    System.out.println(myMap.values());

Sorting list of Strings having integer values- Java

9 Answers9