On Thu, Oct 6, 2011 at 7:29 AM, Samuel John <sc...@samueljohn.de> wrote:
> I just learned two things: > > 1. np.newaxis > 2. Array dimension broadcasting rocks more than you think. > > Yup. :) > > The x[:, np.newaxis] might not be the most intuitive solution but it's > great and powerful. > Intuitive would be to have x.T to transform [0,1,2,4] into > [[0],[1],[2],[4]]. > I agree, creating a new dimension by indexing with np.newaxis isn't the first thing I would guess if I didn't already know about it. An alternative is x.reshape(4,1) (or even better, x.reshape(-1,1) so it doesn't explicitly refer to the length of x). (Also, you probably noticed that transposing won't work, because x is one-dimensional. The transpose operation simply swaps dimensions, and with just one dimension there is nothing to swap; x.T is the same as x.) Warren > Thanks Warren :-) > Samuel > > On 06.10.2011, at 14:18, Warren Weckesser wrote: > > > > > > > On Thu, Oct 6, 2011 at 7:08 AM, Neal Becker <ndbeck...@gmail.com> wrote: > > Given a vector y, I want a matrix H whose rows are > > > > y - x0 > > y - x1 > > y - x2 > > ... > > > > > > where x_i are scalars > > > > Suggestion? > > > > > > > > In [15]: import numpy as np > > > > In [16]: y = np.array([10.0, 20.0, 30.0]) > > > > In [17]: x = np.array([0, 1, 2, 4]) > > > > In [18]: H = y - x[:, np.newaxis] > > > > In [19]: H > > Out[19]: > > array([[ 10., 20., 30.], > > [ 9., 19., 29.], > > [ 8., 18., 28.], > > [ 6., 16., 26.]]) > > > > > > Warren > > > > > > _______________________________________________ > > NumPy-Discussion mailing list > > NumPy-Discussion@scipy.org > > http://mail.scipy.org/mailman/listinfo/numpy-discussion > > > > _______________________________________________ > > NumPy-Discussion mailing list > > NumPy-Discussion@scipy.org > > http://mail.scipy.org/mailman/listinfo/numpy-discussion > > _______________________________________________ > NumPy-Discussion mailing list > NumPy-Discussion@scipy.org > http://mail.scipy.org/mailman/listinfo/numpy-discussion >
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