Understanding numpy.linalg.norm() in IPython

Question

I'm creating a linear regression model for supervised learning.

I have a bunch of data points plotted on a graph (x1, y1), (x2, y2), (x3, y3), etc, where the x's are the real data and the y values are the training data values.

As part of the next step in writing a basic nearest neighbor algorithm, I want to create a distance metric to measure the distance (and similarity) between two instances.

If I wanted to write a generic function to compute the L-Norm distance in ipython, I know that a lot of people use numpy.linalg.norm(arr, ord = , axis=). What I'm confused about is how to format my array of data points so that it properly calculates the L-norm values.

If I had just two data points, say (3, 4) and (5, 9), would my array need to look like this with each data point's values in one row?

arry = ([[3,4] 
         [5,9]])

or would it need to look like this where all the x-axis values are in one row and y in another?

arry = ([[3,5]
         [4,9]])

Answer 1

numpy.linalg.norm(x) == numpy.linalg.norm(x.T) where .T denotes the transpose. So it doesn't matter.

For example:

>>> import numpy as np
>>> x = np.random.rand(5000, 2)
>>> x.shape
(5000, 2)
>>> x.T.shape
(2, 5000)
>>> np.linalg.norm(x)
57.82467111195578
>>> np.linalg.norm(x.T)
57.82467111195578

Edit:

Given that your vector is basically

x = [[real_1, training_1],
     [real_2, training_2],
      ...
     [real_n, training_n]]

then the Frobenius norm is basically computing

np.sqrt(np.sum(x**2))

Are you sure this is the right metric. There are a whole bunch of other norms. Here are 3

np.sum((x[:,0]**2 - x[:,1]**2) # N-dimensional euclidean norm
np.sqrt(np.sum(x[:,0]**2) + np.sum(x[:,1]**2)) # L^2 norm
np.sqrt(x[:,0].dot(x[:,1])) # sqrt dot product