How to sort Alphanumber String in Java

Question

{"SAMPLE:1-1-8-SAM#1", "SAMPLE:1-1-8-SAM#13","DEMO:1-1-4-SAM#13",DEMO:1-1-4-SAM#26,DEMO:1-1-4-SAM#8}

So, the output will be Alphabetical order and numeric also

DEMO:1-1-4-SAM#26
DEMO:1-1-4-SAM#13
DEMO:1-1-4-SAM#8
SAMPLE:1-1-8-SAM#1
SAMPLE:1-1-8-SAM#13

I have tried in ArrayList. The thing is, it is taking entire string as a string. But I want to sort string as well as number.Here what I got the output

DEMO:1-1-4-SAM#8
DEMO:1-1-4-SAM#13
DEMO:1-1-4-SAM#26
SAMPLE:1-1-8-SAM#1
SAMPLE:1-1-8-SAM#13

if the output on the top is desired output, how do you sort ascending and then having #26 in front of #13?? What exactly is the logic of the sort that you need? — display name
– display name, Commented Jan 28, 2015 at 12:29
ho, sorry, i didn't see that. but anyway, I think you got my point — veerendra2
– veerendra2, Commented Jan 28, 2015 at 12:33
Do you wish to sort all numbers or just the final number after the #? — wassgren
– wassgren, Commented Jan 28, 2015 at 12:37

Thomas · Accepted Answer · 2015-01-28 12:28:19Z

4

You will have to write your own Comparator that splits up the strings into lexical parts and numerical parts and then compare them with the appropriate operator.

The following is a pseudo-code implementation of Comparator's only method compareTo to illustrate what I'm suggesting:

public void compare(String s1, String s2) {
  String[] e1 = split(s1);
  String[] e2 = split(s2);

  int n = Math.min(e1.length, e2.length);

  int ret = 0;

  for (int i = 0; ret == 0 && i < n; i++) {
    if (isNumber(e1[i])) {
      ret = compareNumerical(e1[i], e2[i]));
    } else {
      ret = compareLexical(e1[i], e2[i]));
    }
  }

  return ret;
}

answered Jan 28, 2015 at 12:28

Thomas

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2 Comments

Sufiyan Ghori Over a year ago

@Thomas, why not Comparator along with paralletSort from Java 8 API ? like, parallelSort(T[] a,Comparator<?super T> c)... it is much efficient because it assigns sorting tasks to multiple threads available in the thread pool.

Thomas Over a year ago

@SufiyanGhori - This might be a good choice under the right circumstances but the overhead of Thread handling must be taken into account. For short arrays it's probably not worth it (?).

wassgren · Accepted Answer · 2015-01-28 13:28:55Z

If you wish to sort the strings and consider all numbers you can use a combination of regex and Java 8.

Consider a string like the following:

"SAMPLE:1-2-8-SAM#1"

This can be split up using a regular expression to a String[] like this

String s = "SAMPLE:1-2-8-SAM#1";
String[] arr = s.split("((?<=[:\\-#])|(?=[:\\-#]))");
System.out.println(Arrays.stream(arr).collect(Collectors.toList()));

The output would be:

[SAMPLE, :, 1, -, 2, -, 8, -, SAM, #, 1]

This means that we can treat index 0 like a String and index 2, 4, 6 and 10 as Integers.

So, in order to use that for sorting the following code can be used:

String[] strings = {
        "SAMPLE:1-2-8-SAM#1",
        "SAMPLE:1-1-8-SAM#1",
        "SAMPLE:1-1-8-SAM#13",
        "DEMO:1-1-4-SAM#13",
        "DEMO:1-1-4-SAM#26",
        "DEMO:1-1-4-SAM#8"
};

final List<String> sorted = Arrays.stream(strings)
        .map(str -> str.split("((?<=[:\\-#])|(?=[:\\-#]))"))
        .sorted(Comparator
                .<String[], String>comparing((arr) -> arr[0])      // String compare
                .thenComparing(arr -> Integer.valueOf(arr[2]))     // Integer compare
                .thenComparing(arr -> Integer.valueOf(arr[4]))     // Integer compare
                .thenComparing(arr -> Integer.valueOf(arr[6]))     // Integer compare
                .thenComparing(arr -> Integer.valueOf(arr[10])))   // Integer compare
        .map(l -> Arrays.stream(l).collect(Collectors.joining()))  // Map it back to a String
        .collect(Collectors.toList());                             // Collect it to a List for further processing

It is not super-readable but it works as per the OP:s question. If you also use some static imports the code is a bit easier to read.

However, my recommendation is to create a Comparator that does the sorting which will be more readable (and thereby easier to maintain in the long run).

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