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I need to sort an array and get indexes linked to an unsorted array. The problem is that if the unsorted array contains duplicate entries, i.e. [1,1,1,2,2,], then indexes are the same for these entries. For the example [3,5,5,3,3,] the indexes would be [0,1,1,0,0]. But I would need to get the following indexes [0,3,4,1,2]. How to do this?

ArrayList<Double> nfit = new ArrayList<Double>();
ArrayList<Double> nfit_copy = new ArrayList<Double>(nfit);

// Fill nfit

Collections.sort(nfit);

int[] ind = new int[nfit.size()];

for (int n = 0; n < nfit.size(); n++){
    ind[n] = nfit_copy.indexOf(nfit.get(n));
}
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3 Answers 3

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1

This is because you use nfit_copy.indexOf(nfit.get(n)); which will gave you the same index if you have duplicated number

For Example

[3,5,5,3,3,]----> every time you will use :nfit.indexOf(3) ,will give you index 0

so , may be you need to change this value or set it null (Don't reomve it because it will change the index) , to allow you to get the next duplicate number

Try this:

ArrayList<Integer> nfit = new ArrayList<Integer>();
ArrayList<Integer> nfit_copy = new ArrayList<Integer>(nfit);

// Fill nfit 

  nfit.add(3);
  nfit.add(5);
  nfit.add(5);
  nfit.add(3);
  nfit.add(3);


  nfit_copy = (ArrayList<Integer>) nfit.clone();


  Collections.sort(nfit);

  int[] ind = new int[nfit.size()];

  for (int n = 0; n < nfit.size(); n++) {
         ind[n] = nfit_copy.indexOf(nfit.get(n));
         nfit_copy.set(nfit_copy.indexOf(nfit.get(n)), null);
   }
   for (int i = 0; i < ind.length; i++) {
         int j = ind[i];
         System.out.println(j);
    }
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5 Comments

It provides an incorrect result for nfit.add(3); nfit.add(3); nfit.add(5); nfit.add(3); nfit.add(3);. The result is 0 1 3 4 2, instead of 0 1 4 2 3.
How 0 1 4 2 3 ??you have number 3 in index (0 ,1 ,3 ,4 ) and one 5 in index 2 , so please check it again :)
Probably I misunderstood your idea. I checked the code of Tee Gee (see below), and it provides 0 1 4 2 3 for the array [3,3,5,3,3], which seems to be correct to me.
0 1 4 2 3 will mean your array is 3 3 3 5 3 after sorted , so this is not sorted index you want , and seem incorrect to me :)
We were talking about different things. In your case you assign indexes to unsorted array (and sorry, this is exactly what I need), but Tee Gee provides sorted indexes.
1 
public static void main(String[] args) {
        Integer[] input = new Integer[]{1,2,2,1,5,6,2,3,2,3,4,5,6,1};
        List<Integer> mappedIndexes = new ArrayList<Integer>();

        List<Integer> sorted = new ArrayList<Integer>();
        for (int num : input) {
            sorted.add(num);
        }       
        Collections.sort(sorted);

        for (int number : input) {
            System.out.println("finding for input -> " + number);
            for (int i = 0; i < sorted.size(); i++) {
                int sortedNumber = sorted.get(i);
                if (number == sortedNumber) {
                    if (mappedIndexes.contains(i)) {
                        continue;
                    } else {
                        System.out.println("setting index as -> " + i);
                        mappedIndexes.add(i);
                        break;
                    }
                }
            }
        }
        System.out.println(mappedIndexes);
    }
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Comments

1

Here is my solution with Set and HashMap structures. Read the code comments for more info, please.

int[] input = { 45, 3, 4, 9, 2, 1, 45, 3 };
// Backup the initial array for later use.
int[] originalArray = Arrays.copyOf(input, input.length);    
Arrays.sort(input);

// Determine all occurences with a set.
Set<Integer> set = new HashSet<Integer>();
for (int i : input) {
    set.add(i);
}

// Populate a hashmap for keeping <value,index> pairs.
HashMap<Integer, Integer> map = new HashMap<Integer, Integer>();
int counter = 0;
for (int i : set) {
    map.put(i, counter++);
}

// Populate the output array.
int[] output = new int[input.length];
for (int i = 0; i < output.length; i++) {
    output[i] = map.get(originalArray[i]);
}

That's all. If you print the contents of output to console, you can see the results.

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