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 <[EMAIL PROTECTED]> 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 > > _______________________________________________ > Numpy-discussion mailing list > Numpy-discussion@scipy.org > http://projects.scipy.org/mailman/listinfo/numpy-discussion -- Christopher Barker, Ph.D. Oceanographer Emergency Response Division NOAA/NOS/OR&R (206) 526-6959 voice 7600 Sand Point Way NE (206) 526-6329 fax Seattle, WA 98115 (206) 526-6317 main reception [EMAIL PROTECTED] _______________________________________________ Numpy-discussion mailing list Numpy-discussion@scipy.org http://projects.scipy.org/mailman/listinfo/numpy-discussion
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