You can take advantage of the complex type : # build a complex array of your cells z = np.array([complex(c.m_x, c.m_y) for c in cells]) First solution # mesh this array so that you will have all combinations m, n = np.meshgrid(z, z) # get the distance via the norm out = abs(m-n) Second solution … Read more
To get the distance you can use the norm method of the linalg module in numpy: np.linalg.norm(x – y)
One informal but rather intuitive way to think about this is to consider the 2 components of a vector: direction and magnitude. Direction is the “preference”https://stackoverflow.com/”style”https://stackoverflow.com/”sentiment”https://stackoverflow.com/”latent variable” of the vector, while the magnitude is how strong it is towards that direction. When classifying documents we’d like to categorize them by their overall sentiment, so we … Read more
Expanding on my wiki-walk comment in the errata and noting some of the ground-floor literature on the comparability of algorithms that apply to similar problem spaces, let’s explore the applicability of these algorithms before we determine if they’re numerically comparable. From Wikipedia, Jaro-Winkler: In computer science and statistics, the Jaro–Winkler distance (Winkler, 1990) is a … Read more
(Months later) scipy.spatial.distance.cdist( X, Y ) gives all pairs of distances, for X and Y 2 dim, 3 dim … It also does 22 different norms, detailed here . # cdist example: (nx,dim) (ny,dim) -> (nx,ny) from __future__ import division import sys import numpy as np from scipy.spatial.distance import cdist #……………………………………………………………………. dim = 10 nx … Read more
Use numpy.linalg.norm: dist = numpy.linalg.norm(a-b) This works because the Euclidean distance is the l2 norm, and the default value of the ord parameter in numpy.linalg.norm is 2. For more theory, see Introduction to Data Mining: