That makes sense. I'll add an optional eps value, and handle the case of 0
when it comes up.
Thanks,
Robert
On 14 October 2011 14:23, Skipper Seabold <[email protected]> wrote:
> On Thu, Oct 13, 2011 at 11:10 PM, Robert Layton <[email protected]>
> wrote:
> > I'm working on adding Adjusted Mutual Information, and need to calculate
> the
> > Mutual Information.
> > I think I have the algorithm itself correct, except for the fact that
> > whenever the contingency matrix is 0, a nan happens and propogates
> through
> > the code.
> >
>
> FWIW, scipy.stats defines entropy of p(x) = 0 to be 0, and I think it
> is so by definition. The other option I've seen in software is to let
> the user define the eps.
>
>
> https://github.com/scipy/scipy/blob/master/scipy/stats/distributions.py#L5284
>
> > Sample code on the net [1] uses an eps=np.finfo(float).eps. Should I do
> > this, adding eps to anything that is a denominator or parameter to log?
> > Is there a better way?
> > [1]
> http://blog.sun.tc/2010/10/mutual-informationmi-and-normalized-mutual-informationnmi-for-numpy.html
> > FYI: My current code:
> > def mutual_information(labels_true, labels_pred, contingency=None):
> > if contingency is None:
> > labels_true, labels_pred = check_clusterings(labels_true,
> > labels_pred)
> > contingency = contingency_matrix(labels_true, labels_pred)
> > # Calculate P(i) for all i and P'(j) for all j
> > pi = np.sum(contingency, axis=1)
> > pi /= float(np.sum(pi))
> > pj = np.sum(contingency, axis=0)
> > pj /= float(np.sum(pj))
> > # Compute log for all values
> > log_pij = np.log(contingency)
> > # Product of pi and pj for denominator
> > pi_pj = np.outer(pi, pj)
> > # Remembering that log(x/y) = log(x) - log(y)
> > mi = np.sum(contingency * (log_pij - pi_pj))
> > return mi
> > --
> >
> >
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threats, fraudulent activity and more. Splunk takes this data and makes
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