Euclidean distance with weights

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I am currently using SciPy to calculate the euclidean distance

dis = scipy.spatial.distance.euclidean(A,B)

where; A, B are 5-dimension bit vectors. It works fine now, but if I add weights for each dimension then, is it still possible to use scipy?

What I have now: sqrt((a1-b1)^2 + (a2-b2)^2 +...+ (a5-b5)^2)

What I want: sqrt(w1(a1-b1)^2 + w2(a2-b2)^2 +...+ w5(a5-b5)^2) using scipy or numpy or any other efficient way to do this.

Thanks

The suggestion of writing your own weighted L2 norm is a good one, but the calculation provided in this answer is incorrect. If the intention is to calculate

then this should do the job:

def weightedL2(a,b,w):
    q = a-b
    return np.sqrt((w*q*q).sum())

When to use weighted Euclidean distance and how to determine the , Weights for standardisation. The setup you have is a variant of Mahalanobis distance. So when w is the reciprocal of each measurement's� dis = scipy.spatial.distance.euclidean(A,B) where; A, B are 5-dimension bit vectors. It works fine now, but if I add weights for each dimension then, is it still possible to use scipy?

Simply define it yourself. Something like this should do the trick:

def mynorm(A, B, w):
    import numpy as np
    q = np.matrix(w * (A - B))
    return np.sqrt((q * q.T).sum())

Euclidean Distance, The weighted Euclidean distance metric incorporates the feature weights w1, w2,…, wn on each dimension: w 1 2 ( x 1 − y 1 ) 2 + w 2 2 ( x 2 − y 2 ) 2 + ⋯ + w n 2 ( x n − y n ) 2 . Illustration for n=3, repeated application of the Pythagorean theorem yields the formula In mathematics, the Euclidean distance or Euclidean metric is the "ordinary" straight-line distance between two points in Euclidean space. With this distance, Euclidean space becomes a metric space. The associated norm is called the Euclidean norm. Older literature refers to the metric as the Pythagorean

If you want to keep using scipy function you could pre-process the vector like this.

def weighted_euclidean(a, b, w):
    A = a*np.sqrt(w)
    B = b*np.sqrt(w)
    return scipy.spatial.distance.euclidean(A, B)

However it's look slower than

def weightedL2(a, b, w):
    q = a-b
    return np.sqrt((w*q*q).sum())

[PDF] Chapter on Euclidean distance, Euclidean distances, which coincide with our most basic physical idea of distance, but generalized The weights of the three variables in our example are (to 4. I thought of using the Weighted Euclidean Distance like this: d x, b = ( ∑ i = 1 n w i ( x i − b i) 2)) 1 / 2. where. x i is the value of the i-th measure for the particular data. b i is the corresponding benchmark value for that measure.

Euclidean distance, The suggestion of writing your own weighted L2 norm is a good one, but the calculation provided in this answer is incorrect. If the intention is to� Euclidean distance min-max problem in Rn. This paper extends their work to include a weight corresponding to each point in R3. 1.2 Literature Sylvester [10] rst explored the minimum covering circle problem in 1857. Sylvester [11] utilized a geometric, primal approach to solving the problem. In other words, he started with a

Euclidean distance with weights, Weight functions apply weights to an input to get weighted inputs. Z = dist(W,P,FP ) takes these inputs,. W. I am trying to create my own weights for relative work task importance, or weight. For every task, I have a value of importance, frequency, and relevance. I was recommended to use Euclidean distance from zero to create the weight per task. However, I can't seem to find anything on the internet, why to use such measure at all, and also why it

Euclidean distance weight function - MATLAB dist, I want to compute a distance matrix (between all pairs of individuals) using euclidean distance. But for two variables, I would like to give twice as many weight� Okay, so this is how we think about writing Euclidean distance in terms of an inner product, which we see right here. And then if we looked at our scale to Euclidean distance, it's a simple modification where we add this matrix a, where this matrix is this diagonal matrix with our weights a1 to ad.

Comments
  • That isn't the norm contained in the question - you have squared the weights. Also the .sum() is completely redundant, q*q.T is the inner product of the vector with itself, ie. it is the sum.
  • You are correct about the weights, I should have been more careful, however your criticism about the .sum() being completely redundant is misguided. The result of q * q.T would be a 1x1 matrix, which would be an unexpected return type for a norm function, the sum will turn it into a scalar.
  • But why use sum() to cast to a scalar? np.asscalar will be several times faster`?
  • I don't know the reason, but that is how it is implemented in scipy.spatial.distance.euclidean .. I just assume the authors of scipy know what's best