How to mimic Python's nice column and row access (i.e matrix[:, 2])?

[henon]

Hey all,

How would you say to do access to a column matrix[:, i] or access to a row matrix[i, :] in NumSharp?

[henon]

I need to be able to get a (n-1)-dimensional slice of a n-dimensional matrix, i.e. a 1d vector out of a 2d matrix.
First step would be to add appropriate accessors to IStorage right?
I am not yet sure how they should look like, please comment if you have ideas.

[henon]

This is not as elegantly possible as in Python, but we can do something like this, given that matrix is an NDArray:

• matrix[Range.All, 1] ... return the first column, Python: matrix[:, 1]
• matrix[new Range(0, 2), 1] ... return the first two entries of the first column, Python: matrix[:2, 1]
• matrix[new Range(1,3), new Range(1,3)] ... return a 2x2 submatrix of matrix starting at 1,1, Python: matrix[1:3, 1:3]

[Oceania2018]

Check indexing to help.

We can override in Slice

Or we can pass string ":1".

[henon]

Yeah, the string idea is great for porting code from Python. I am sure it will be popular.

[Oceania2018]

Seem's like another contributor already implemented the string index. I'll found out later. @PppBr is that true?

[henon]

when you slice a matrix in numpy you essentially get a view of the original data. modifying it will modify the original matrix. Do we have a view concept or implementation already?

[Oceania2018]

Not yet, we havn't have the view class. I've an idea how to implement the new view.

view inherit from ndarray, but keep a filter of data index.
and view will implement a IEnumerate interface.
when interating the view, it will return all the element in the filter.
and view don't need copy the memory. change data will reflect in the original memory.

@henon What do you think of it?

or we need a new NDStorage?

[henon]

@Oceania2018: I think the projection from the view to the original data could be handled entirely by the storage.

This is complicated, I had to think about it for a time to wrap my head around it. If I am correct, what you need is a function that projects a view index onto an index on the original data (which could be another view that does further projection - think about that).

I am not yet sure how that function looks like if you i.e. for instance get a 1D slice out of a 17-dimensional matrix but once we have that generic formula we can easily project view_index to underlying data_index and view_index+1, view_index+2, ... etc. to enumerate it.

[Oceania2018]

Actually, we have the formula to convert n-d index to 1d index and wise-verse.
chech the function Shape.GetDimIndexOutShape
So don't worry about the dimension. the storage is persistent data in 1d array.

I still think the view should inherit from ndarray with a filters in view instance, not in ndstorage level. Let's keep this talk open.

[henon]

Nice we have the projection and the inverse projection already, so the difficult part is actually already solved. The rest is just software design.

One thing I missed, that the view also needs to enumerate on the underlying data (or view), is a stride. If you take a row out of a matrix the stride is 1, but if you take a column, the stride is the width of the matrix.

ok, you may be right about inheriting from ndarray. I have only limited knowledge about the design at this point, so I trust your judgement.

[Oceania2018]

I will write the first version of view, then we can discuss further.

[henon]

Don't forget to make the design recursive, so that you can have a view of a view of a view of a 'storage'.

[Oceania2018]

Sure

[henon]

Update: Python's slice notation for accessing slices of NDArray has been implemented. However, the implementation of a view that serves that slice of the original data still needs to be done.

Here is a very simple test case that demonstrates the problem to be solved:

    public void GetAColumnOf2dMatrix()
    {
        var x = np.arange(9).reshape(3, 3);
        var v = x[":, 1"];
        Assert.IsTrue(Enumerable.SequenceEqual( v.Data<double>(), new double[]{ 1, 4, 7 }));
        v[1] = 99;
        Assert.AreEqual(99, x[1,1]);
    }