Comments (1)
My formula is exactly what your first formula is saying. we sum over each difference and square it. it might be confusing since the points are named x1 and x2. maybe a better naming would be p1 and p2, and then each point p1 has (x1,y1,z1,...).
you can test my formula and compare with this. It's the same result.
def ec(x1,x2):
return np.linalg.norm(x1-x2)
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Related Issues (13)
- Repository has no License HOT 1
- IndexError: index 6 is out of bounds for axis 0 with size 6 HOT 2
- Explanation of `get_hyperplane_value` HOT 1
- Use of log loss function in logistic regression
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- XGBoost Algorithm
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