How to get multiple values by index in an ojAlgo-array?

Question

Let's say I have an ojAlgo-array defined like this

ArrayAnyD<Double> regularArray = ArrayAnyD.PRIMITIVE64.make(10);
regularArray.loopAll((final long[] ref) -> regularArray.set(ref, ref[0]*ref[0]));

which just contains the squares of the numbers from 0 to 9, and some indexes:

long[] idxs = {1, 3, 4, 7, 9}

Now, I'd like to do something akin to

slicedArray = regularArray[idxs]

which should give me another Array containing 1.0, 9.0, 16.0, 49.0, 81.0. How would I do that with ojAlgo?

regularArray.get(idxs) only gives me the first value.
regularArray.sliceSet(idxs, 0) returns all values from the first one onwards and ignores the following indexes.

I suspect I'd need to use some form of regularArray.loop or .loopAll, but I'm not sure how to make it work

There's no way to do that directly. You have to, manually, create a copy. You can't slice using arbitrary indices. Slicing needs to be structured somehow (uniform stride between the indices). — apete
– apete, Commented Apr 27, 2022 at 14:15
You created a 1-dimensional ArrayAnyD? How many dimension do you need? If you use matrices instead of arrays, you can create them as row or column vectors. These then have methods like rows(int[]) and columns(int[]) that works the way you want. — apete
– apete, Commented Apr 27, 2022 at 14:35
@apete This was just meant as a sized down example, maybe I shouldn't have minimized it that much. In reality I'd need up to three dimensions (nx3x3), where n can be as much as a million or more. I might be able to rewrite my routines to flatten the three dimensions into two so that a matrix would be possible in theory (although I'd really prefer to avoid that), but I'm sceptical about the size. In the introduction of ojAlgo, it says that Arrays are advantageous for larger sizes. — PebNischl
– PebNischl, Commented Apr 28, 2022 at 7:36
Don't understand your use case. Maybe start a discussion over at GitHub: github.com/optimatika/ojAlgo/discussions — apete
– apete, Commented Apr 28, 2022 at 7:51
If that huge array is actually a "collection" of 3x3 matrices, then it is better to define the array as 3x3xn rather than nx3x3. That way the matrices are contiguous, and can be a iterated over using array.matrices(); — apete
– apete, Commented Apr 28, 2022 at 7:57

apete · Accepted Answer · 2022-05-06 10:43:27Z

1

Here's some of what you can do:

ArrayAnyD<Double> array = ArrayAnyD.PRIMITIVE64.make(3, 3, n);

MatrixView<Double> matrices = array.matrices();

// A MatrixView is Iterable,
// but you can also go directly to the matrix you want

matrices.goToMatrix(78);

// Specific rows and columns
Access2D<Double> select = matrices.select(new int[] { 1, 2 }, new int[] { 0 });

Access2D<Double> rows = matrices.rows(1, 2); // All columns

Access2D<Double> columns = matrices.columns(0); // All rows

double value10 = matrices.doubleValue(1, 0);
double value20 = matrices.doubleValue(2, 0);

matrices.goToMatrix(99); // ...

To do all of that (exactly that way) you need v51.3.0-SNAPSHOT or later.

answered May 6, 2022 at 10:43

apete

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