Re: [Numpy-discussion] beginner question: rank-1 arrays
On Nov 16, 2007 5:55 PM, Emanuel Woiski [EMAIL PROTECTED] wrote: Sorry for coming very late to the thread, but you mean something like: for i in range(len(cols): [sending too soon...] On Oct 9, 2007 4:36 AM, Sven Schreiber [EMAIL PROTECTED] wrote: Alan G Isaac schrieb: On Mon, 8 Oct 2007, Robin apparently wrote: However in my code (I am converting from MATLAB) it is important to maintain 2d arrays, and keep the difference between row and column vectors. Well, I have noticed that numpy doesn't care very much about rows and cols. Mind you, if you slice along a col, you end up with a row - just try it and see. But how can you evaluate an expression such as a[i] - b[j], for all (i,j), with i for rows and j for cols? The trick here is 'newaxis' . With 'newaxis' you have a temporary dimension for a or b, without actually changing a or b shapes. Following the usual meaning, the expression becomes: c = a[:,newaxis] - b See: a = arange(3.) a array([ 0., 1., 2.]) b = a # just an example... Now I want: c1 = zeros((3,3)) for i in range(3): for j in range (3): c1[i,j] = a[i] - b[j] c1 array([[ 0., -1., -2.], [ 1., 0., -1.], [ 2., 1., 0.]]) That's exactly the same as the one-liner: c2 = a[:,newaxis] - b c2 array([[ 0., -1., -2.], [ 1., 0., -1.], [ 2., 1., 0.]]) Nice isn't it? Hope that help you somehow:) cheers woiski ___ Numpy-discussion mailing list Numpy-discussion@scipy.org http://projects.scipy.org/mailman/listinfo/numpy-discussion
Re: [Numpy-discussion] beginner question: rank-1 arrays
Sorry for coming very late to the thread, but you mean something like: for i in range(len(cols): On Oct 9, 2007 4:36 AM, Sven Schreiber [EMAIL PROTECTED] wrote: Alan G Isaac schrieb: On Mon, 8 Oct 2007, Robin apparently wrote: However in my code (I am converting from MATLAB) it is important to maintain 2d arrays, and keep the difference between row and column vectors. How about using matrices? help(numpy.mat) hth, Alan Isaac Robin, Alan is right, you want numpy matrices which are always 2d. Check out numpy.matlib; if you replace from numpy import [whatever] by from numpy.matlib import [whatever] you get everything there is in numpy, and things like ones() zeros() empty() etc. will always be 2d matrices. -sven ___ Numpy-discussion mailing list Numpy-discussion@scipy.org http://projects.scipy.org/mailman/listinfo/numpy-discussion ___ Numpy-discussion mailing list Numpy-discussion@scipy.org http://projects.scipy.org/mailman/listinfo/numpy-discussion
Re: [Numpy-discussion] beginner question: rank-1 arrays
On Mon, Oct 08, 2007 at 11:00:39PM +0100, Robin wrote: Coming from matlab and being use to 0:10 for row or (0:10)' for column this seems a bit messy. Is there a better way of constructing row/column 2d arrays from a slice type range? r_[0:10] and c_[0:10]. Does that suit you ? The first one is indeed only 1D, but I don't see the problem with that. If you really want 2D you can use c_[0:10].T . Gaël ___ Numpy-discussion mailing list Numpy-discussion@scipy.org http://projects.scipy.org/mailman/listinfo/numpy-discussion
Re: [Numpy-discussion] beginner question: rank-1 arrays
On Mon, Oct 08, 2007 at 11:00:39PM +0100, Robin wrote: Hi, I am trying to implement a project in scipy. I think I am getting somewhere finally. However in my code (I am converting from MATLAB) it is important to maintain 2d arrays, and keep the difference between row and column vectors. After working through some initial problems I think I am getting more of a picture of how things work in numpy. However I am finding my code littered with things like: np.array(np.r_[0:nterms],ndmin=2)(for a row vector) or np.array(np.r_[0:nterms],ndmin=2).T (for a column vector) You can use N.arange(10).reshape(-1,1) or N.c_[:10,] But if you use this sort of thing often, just write your own factory method: def col(n): return N.arange(n).reshape(-1,1) Cheers Stéfan ___ Numpy-discussion mailing list Numpy-discussion@scipy.org http://projects.scipy.org/mailman/listinfo/numpy-discussion
Re: [Numpy-discussion] beginner question: rank-1 arrays
On Mon, Oct 08, 2007 at 11:12:07PM +0100, Robin wrote: On 10/8/07, Gael Varoquaux [EMAIL PROTECTED] wrote: r_[0:10] and c_[0:10]. Does that suit you ? The first one is indeed only 1D, but I don't see the problem with that. If you really want 2D you can use c_[0:10].T . Thanks, but not really :) Firstly - for me I don' see any difference between the two, they both give a numpy rank-1 array which doesn't have a row/column characteristic. x.shape for both of the above is (10,) I need (10,1) or (1,10) for my code. Damn it. Shame on me. I meant c_[0:10,]. If you really need a shape of (1,10) (I have never had such a need) you can use c_[0:10,].T. HTH, Gaël ___ Numpy-discussion mailing list Numpy-discussion@scipy.org http://projects.scipy.org/mailman/listinfo/numpy-discussion
Re: [Numpy-discussion] beginner question: rank-1 arrays
On Mon, 8 Oct 2007, Robin apparently wrote: However in my code (I am converting from MATLAB) it is important to maintain 2d arrays, and keep the difference between row and column vectors. How about using matrices? help(numpy.mat) hth, Alan Isaac ___ Numpy-discussion mailing list Numpy-discussion@scipy.org http://projects.scipy.org/mailman/listinfo/numpy-discussion
Re: [Numpy-discussion] beginner question: rank-1 arrays
On 10/8/07, Gael Varoquaux [EMAIL PROTECTED] wrote: Damn it. Shame on me. I meant c_[0:10,]. If you really need a shape of (1,10) (I have never had such a need) you can use c_[0:10,].T. Thanks! - the trick with the , is just the sort of thing I was looking for - I knew there must be an easy way... Cheers Robin ___ Numpy-discussion mailing list Numpy-discussion@scipy.org http://projects.scipy.org/mailman/listinfo/numpy-discussion
