update normalisation for average kersize - #14
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please adjust |
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handled comments. Tests run successfully. |
JakobCode
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Jun 8, 2026
| # and divided by num_feas^2 ad we have that many terms | ||
| if num_feas > 0: | ||
| if num_feas > 1: | ||
| # compute 2 times sum over diams to the power p divided by num_feas^2 |
| # get mean over diams, with factor 2 due to symmetry of the norm of the compute vectors in null space of F (norm(x-z)=norm(z-x)) | ||
| # and divided by num_feas^2 ad we have that many terms | ||
| if num_feas > 0: | ||
| if num_feas > 1: |
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What happens for case num_feas == 1?
| if num_feas > 1: | ||
| # compute 2 times sum over diams to the power p divided by num_feas^2 | ||
| diameter_mean_y = 2*np.divide(np.sum(np.power(diam_y,p)), np.power(num_feas,2)) | ||
| diameter_mean_y = 2*np.divide(np.sum(np.power(diam_y,p)), num_feas-1) |
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Please double check that this is correct and align with comment above
| size_feas = len(valid_idx) | ||
| # compute average over sums of differences over twice the feasible sets | ||
| return np.divide(np.nansum(np.power(subdistXX,p_X)), np.power(size_feas,2)) | ||
| return np.divide(np.nansum(np.power(subdistXX,p_X)), size_feas-1), size_feas |
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please double check if correct
nm19000
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Jul 2, 2026
update normalisation for average kersize
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