import numpy # Say y is y = numpy.array([1,2,3]) Y = numpy.vstack([y,y,y,y]) # Y is array([[1, 2, 3], # [1, 2, 3], # [1, 2, 3], # [1, 2, 3]]) x = numpy.array([[0],[2],[4],[6]]) # a column-vector of your scalars x0, x1... Y - x Hope this is what you meant. cheers, Samuel On 06.10.2011, at 14:08, Neal Becker wrote:

Given a vector y, I want a matrix H whose rows are

y - x0 y - x1 y - x2 ...

where x_i are scalars

On Thu, Oct 6, 2011 at 7:29 AM, Samuel John <scipy@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 <ndbecker2@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

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