Porting over some code to Python and I don't understand this multiplication syntax

Heya,

So I've been porting over the RAT.m file for my personal use. I'm porting a larger library over to Python that uses some built-in Matlab functions that SciPy and Numpy do not have. RAT.m is one of them. So I've done the bulk of it, except the following line is confusing me.

C = [C*[d;1] C(:,1)];

I know that probably seems basic as hell, but I'm going snowblind looking at it here.

The second bit:

C(:,1)]

I understand that. It translates to Python as:

C[:,0]

Very straight forward.

I even understand this bit:

[d;1]

And the way I'm translating that is like this in Python:

(np.vstack(np.concatenate((d, np.array(1, ndmin=1)))))

It's really this bit that I am confused by:

C*[d;1]

I don't understand what the multiplication is doing. I'm new to MATLAB but not new to Python. I think this is just one of those syntax things that I'm just not seeing. Can anyone help please?

Thank you!

Answer 1

[Mootrax]

Well groovy pants to me! I solved it. In a way. Basically I had some of my matrices in the wrong shape. So I had to work that out and then my multiplier worked. It should be noted that a basic * multiplier in Numpy is not the same as the one in MATLAB.

In Numpy you have to use the dot method.

dstack = np.vstack(np.concatenate((d, np.array(1, ndmin=1)))) # MATLAB equiv: [d;1]
cmult = C.dot(dstack)
cref = np.array(C[:,0], ndmin=2).T
C = np.concatenate((cmult, cref), axis=1)

This is how I arrived at the expected figures. I could lose the two lines in the middle and I could probably simplify the dstack variable, but essentially it's working. I went snowblind last night on this, but managed to figure it out in a short time this morning.

Answer 2

[Walter Roberson]

It is not anything special.

In order for [d;1] to work, d and 1 would have to have the same number of columns. Since 1 has one column, then d must be a column vector. [d;1] then puts 1 at the end of it, for whatever reason.

C*[d;1] can only work if the number of columns in C is the same as the number of rows in [d;1] or [d;1] is a scalar. If d were empty then [d;1] would be the same as [1] and C*[d;1] would be the same as C. But there is not much point in computing that if you know it to be true, so we should assume that [d;1] is a column vector with the same number of rows as C has columns. That would then imply that d is a column vector with one fewer row than C has columns.

This would be like multiplying C by broadcast of [d;1].' and sum the results across the second dimension.

You might do something like this if you were fixing an intercept, or if you were fixing an axis of a rotation matrix.

There is no particular significance to [d;1], just a column of values having a 1 placed at the bottom.

Your expression to do the concatenate looks more complicated than is perhaps needed, but I do not have a lot of experience with python.