I would like to hear your opinion on developing an explicit sparse/dense 2D matrix class with indexing similar to Matlab, and without significant differences between sparse and dense matrices from the user's perspective...
I know that it's not one of Numpy/Scipy's goals to clone Matlab, but I think I represent a potentially large scientific Python user base, who would find Python matrices that feel a bit more like Matlab/Octave etc. extremely useful. I have a great respect for all the work in Numpy and Scipy,
but at the same time I feel that numerical linear algebra in Python would benefit from a more dedicated matrix library, and I think that Numpy (or another single package) should provide that
in a homogeneous way - without ties to how Numpy array works.
- Joachim
On 3/27/07, Christopher Barker <Chris.Barker@noaa.gov> wrote:
Zachary Pincus wrote:
> rest of linear algebra -- e.g. that m[0] yields a matrix if m is a
> matrix-- it almost certainly would violate the principle of least
> surprise for iteration over m (intuitively understood to be choosing m
> [0] then m[1] and so forth) to yield anything other than a matrix.
I don't think the OP was suggesting that. Rather, I think the suggestion
was that for a mXn matrix, M:
M[i].shape == (n,)
M[i:i+1].shape == (1,n)
that is, indexing (or iterating returns a vector, and slicing returns a
matrix). This is, indeed exactly how numpy arrays behave!
The problem with this is:
numpy matrices were created specifically to support linear algebra
calculations. For linear algebra, the distinction between row vectors
and column vectors is critical. By definition, a row vector has shape:
(1,n), and a column vector has shape (m,1).
In this case, perhaps the OP is thinking that a shape (n,) array could
be considered a row vector, but in that case:
M[1,:].shape == (n,)
M[:,1].shape == (m,)
which of these is the row and which the column? This is why matrices
index this way instead:
M[1,:].shape == (1, n)
M[:,1].shape == (m, 1)
now we know exactly what is a row and what is a column.
By the way, I think with the way numpy works: M[i] == M[i,:] by
definition, so those couldn't yield different shaped results. Is that right?
I think we got a bit sidetracked by the example given. If I have a bunch
of points I want to store, I'm going to use an (m,2) *array*, not a
matrix, then then A[i] will yield a (2,) array, which makes sense for
(2-d) points. In fact, I do this a LOT.
If I happen to need to do some linear algebra on that array of points,
I'll convert to a matrix, do the linear algebra, then convert back to an
a array (or just use the linear algebra functions on the array).
I hope this helps
-Chris
> This can't possibly be what you're asking for, right? You aren't
> suggesting that m[0] and list(iter(m))[0] should be different types?
>
> There are many clear and definite use cases for m[0] being a matrix,
> by the way, in addition to the theoretical arguments -- these aren't
> hard to come by. Whether or nor there are actual good use-cases for
> iterating over a matrix, if you believe that m[0] should be a matrix,
> it's madness to suggest that list(iter(m))[0] should be otherwise.
>
> My opinion? Don't iterate over matrices. Use matrices for linear
> algebra. There's no "iteration" construct in linear algebra. The
> behavior you find so puzzling is a necessary by-product of the
> fundamental behavior of the matrix class -- which has been explained
> and which you offered no resistance to.
>
> Zach
>
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--
Christopher Barker, Ph.D.
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Chris.Barker@noaa.gov
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