Re: Getting equivalent elements in a range/array

[Ali Çehreli]

On 05/07/2011 09:07 PM, Andrej M. wrote:

I want to turn this:
auto arr = [1, 1, 2, 3, 4, 4];

into this:
auto arr2 = [[1, 1], [2], [3], [4, 4]];

I want an array of arrays of the same elements. Lazy or not, I don't care.

I thought I could get away with this inside some while loop:
auto equals = array(filter!"a == b"(arr));
arr = arr[equals.length-1..$];

Nope.

I need this for some buffered output, where the requirement is the elements of 
the buffer all need to have the same properties so a function can output a 
buffer of elements in one call instead of calling the function for each element 
(the function call is expensive).

This seems to work, but needs more work. :)

import std.stdio;
import std.array;

struct EquivalentElements
{
    int[] range;
    int[] front_;

    this(int[] range)
    {
        this.range = range;
        this.front_ = popEqualFront(this.range);
    }

    bool empty()
    {
        return front_.empty;
    }

    int[] front()
    {
        return front_;
    }

    void popFront()
    {
        this.front_ = popEqualFront(this.range);
    }

    private int[] popEqualFront(ref int[] range)
    {
        int[] front;

        if (!range.empty) {
            do {
                front ~= range[0];
                range = range[1..$];

            } while (!range.empty &&
                     (front[$-1] == range[0]));
        }

        return front;
    }
}

void main()
{
    writeln(EquivalentElements([1, 1, 2, 3, 4, 4]));
}