问题描述:

I have a two-dimension ArrayList that contains double values:

ArrayList<ArrayList<Double>> data = new ArrayList<ArrayList<Double>>();

In analogy with classic arrays , I would like to sort the "cols" of this matrix :I want to take the items having the same index in the sub ArrayLists, and then sort them. Like calling Collections.sort() for every column...

By rows I mean the outer level and inner level are columns.

What is the proper way to do this?

I thought about iterating over the matrix to invert it and then sort each row with Collections.sort() ? but maybe it's not the best solution because the matrix is about 400*7000 .

I can't use classic arrays since the size of the matrix is unknown.

Thanks for help.

网友答案:

Do something like this:

    final int COLUMN = 5;
    Comparator<ArrayList<Double>> myComparator = new Comparator<ArrayList<Double>>() {
        @Override
        public int compare(ArrayList<Double> o1, ArrayList<Double> o2) {
            return o1.get(COLUMN).compareTo(o2.get(COLUMN));
        }
    };
    Collections.sort(list, myComparator);

Set COLUMN to whatever column you're sorting on.

Update:

Yes, this won't work at all.

I like ahanin's second suggestion to make your own List which wraps your original List. You would have to wrap the object returned by get() as well so that the variable wrappedList contains the column values and wrappedList.get(0) also returns a column of values. Then the sorting can work. I wonder what the minimum methods are that you have to implement for Collections.sort() to work on your List.

It would probably be easiest to just take someone else's quicksort and make it work with your list.

Here is one implementation: http://www.vogella.de/articles/JavaAlgorithmsQuicksort/article.html

网友答案:

So here's an option that is like the inversion idea but instead of inverting the entire thing you are building one column at a time, sorting it, and discarding it.

ArrayList<ArrayList<Double>> data = new ArrayList<ArrayList<Double>>(); 

for(int c=0; c<data.get(0).size(); c++) {
    List<Double> col = new ArrayList<Double>();
    for( int r=0; r<data.size(); r++ )
        col.add( data.get(r).get(c) );

    Collections.sort(col);

    for( int r=0; r<col.size(); r++ )
        data.get(r).set( c, col.get(r) );
}

I doubt that you will be able to get anything more efficient, besides perhaps the option of creating your own class that provides a view of a column of the table as a list.

网友答案:

I got two crazy ideas: either to implement your own sorting algorithm that would be aware of your inverted matrix structure, or write an implementation of Collection that would wrap your structure and represent the column, which could later be used as an argument to Collections.sort(). With ArrayList this should be rather fast.

网友答案:

Here is an approach to this situation of converting the data into an object array. Change column to whatever you like.

final int column = 3;
ArrayList<ArrayList<Double>> data = new ArrayList<ArrayList<Double>>(); 

// Convert data into an object array
Object [] temp = data.toArray();

Arrays.sort(temp, new Comparator<Object>() {
    @Override
    public int compare(Object o1, Object o2) {
        // Get inner arrays to be compared by type-casting
        ArrayList<Double> temp1 = (ArrayList<Double>) o1;
        ArrayList<Double> temp2 = (ArrayList<Double>) o2;

        // Then compare those inner arrays' indexes
        return temp1.get(column).compareTo(temp2.get(column));
    }
});

// Clear the old one and add the ordered list into
data.clear();

for(Object o: temp)
{
    data.add((ArrayList<Double>) o);
}
网友答案:

I guess if storage is not a problem you can use a SortedMap in your ArrayList. In that way you don't have to worry for sorting the elements.

ArrayList<SortedMap<Double,byte>> data;
网友答案:

Not sure if i understood your Q correctly, but this will sort all "columns" (assuming outer level are rows and inner level are columns):

    final ArrayList<ArrayList<Double>> data = new ArrayList<ArrayList<Double>>(); 
    //...

    final Integer[] rows = new Integer[data.size()];
    for(int i=0;i<rows.length;i++)
        rows[i]=i;

    int cols = data.get(0).size();
    for(int c=0;c<cols;c++)
    {
        final int colidx = c;
        Arrays.sort(rows,new Comparator<Integer>(){
            @Override
            public int compare(Integer arg0,
                    Integer arg1) {
                return data.get(arg0).get(colidx).compareTo(data.get(arg1).get(colidx));        
            }});    

        for(int i=0;i<rows.length;i++)
            data.get(i).add(colidx+1,data.get(rows[i]).get(colidx));
        for(int i=0;i<rows.length;i++)
            data.get(i).remove(colidx);
    }
网友答案:

You will have to visit all the elements in order to sort them in any case. So what you could do is this.

int temp[] = new temp[col.length];
int x = 0;
while(x < row.length)
{
    for(int i = 0; i<col.length; i++)
    {
        temp[i] = mat[x][i];
    }
    sort(temp);
    for(int i = 0; i<col.length; i++)
    {
        mat[x][i] = temp[i];
    }
x++;
}

The running time here would be O(n^2logn) but since you are sorting just 400 doubles that would not take much time. and since you have to visit each element in the matrix atleast once, you can't go below O(n^2)

相关阅读:
Top