Computing F-measure for clustering

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.

• E. Achtert, S. Goldhofer, H.-P. Kriegel, E. Schubert, A. Zimek
  Evaluation of Clusterings Metrics and Visual Support
  Int. Conf. Data Engineering (ICDE 2012)
  http://www.computer.org/portal/web/csdl/doi/10.1109/ICDE.2012.128

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

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.

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.