Computing F-measure for clustering

StackOverflow https://stackoverflow.com/questions/12725263

Question

Can anyone help me to calculate F-measure collectively ? I know how to calculate recall and precision, but don't know for a given algorithm how to calculate one F-measure value.

As an exemple, suppose my algorithm creates m clusters, but I know there are n clusters for the same data (as created by another benchmark algorithm).

I found one pdf but it is not useful since the collective value I got is greater than 1. Reference of pdf is F Measure explained. Specifically I have read some research paper, in which the author compares two algorithms on the basis of F-measure, they got collectively values between 0 and 1. if you read the pdf mentioned above carefully, the formula is F(C,K) = ∑ | ci | / N * max {F(ci,kj)}
where ci is reference cluster & kj is cluster created by other algorithm, here i is running from 1 to n & j is running from 1 to m.Let say |c1|=218 here as per pdf N=m*n let say m=12 and n=10, and we got max F(c1,kj) for j=2. Definitely F(c1,k2) is between 0 and 1. but the resultant value calculated by above formula we will get value above 1.

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Solution 2

The paper Characterization and evaluation of similarity measures for pairs of clusterings by Darius Pfitzner, Richard Leibbrandt and David Powers contains a lot of useful information regarding this subject, including the following example:

Given the set,

           D = {1, 2, 3, 4, 5, 6}

and the partitions,

           P = {1, 2, 3}, {4, 5}, {6}, and
           Q = {1, 2, 4}, {3, 5, 6}

where P is set created by our algorithm and Q is set created by standard algorithm we known

           PairsP = {(1, 2), (1, 3), (2, 3), (4, 5)},
           PairsQ = {(1, 2), (1, 4), (2, 4), (3, 5), (3, 6), (5, 6)}, and
           PairsD = {(1, 2), (1, 3), (1, 4), (1, 5), (1, 6), (2, 3), (2, 4),
                      (2, 5), (2, 6), (3, 4), (3, 5), (3, 6), (4, 5), (4, 6), (5, 6)}

so,

           a = | PairsP intersection PairsQ | = |(1, 2)| = 1
           b = | PairsP- PairsQ | = |(1, 3)(2, 3)(4, 5)| = 3
           c = | PairsQ- PairsP  | = |(1, 4)(2, 4)(3, 5)(3, 6)(5, 6)| = 5
         
     F-measure= 2a/(2a+b+c)

Note: There is an error in the publication on page 364 where a, b, c, and d are computed and the result of b and c are actually switched incorrectly. This switch would throw off the results of some other measures. Obviously, the F-measure is unaffected.

OTHER TIPS

The term f-measure itself is underspecified. It's the harmonic mean, usually of precision and recall. Actually you should even say F1-score if you mean the unweighted version, because you can put different weight on the two input values. But without saying which two values are averaged (not in the sense of the arithmetic mean!) this doesn't say much.

https://en.wikipedia.org/wiki/F1_score

Note that the values must be in the 0-1 value range. Otherwise, you have an error earlier on.

In cluster analysis, the common approach is to apply the F1-Measure to the precision and recall of pairs, often referred to as "pair counting f-measure". But you could compute the same mean on other values, too.

Pair-counting has the nice property that it doesn't directly compare clusters, so the result is well defined when one result has m cluster, the other has n clusters. However, pair counting needs strict partitions. When elements are not clustered or assigned to more than one cluster, the pair-counting measures can easily go out of the range 0-1.

Discusses some of these metrics (including Rand index and such) and gives a simple explanation of the "pair counting F-measure".

The N in your formula, F(C,K) = ∑ | ci | / N * max {F(ci,kj)}, is the sum of the |ci| over all i i.e. it is the total number of elements. You are perhaps mistaking it to be the number of clusters and therefore are getting an answer greater than one. If you make the change, your answer will be between 1 and 0.

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