Dear R-ians, I'm looking for a computational simplified formula to calculate a measure for heterogeneity (let's say H ):
H = sqrt [ (Si (Sj (Xi - Xj)� ) ) /n ] where: sqrt = square root Si = summation over i (= 0 to n) Sj = summation over j (= 0 to n) Xi = element of X with index i Xj = element of X with index j I can simplify the formula to: H = sqrt [ ( 2 * n * Si (Xi) - 2 Si (Sj ( Xi * Xj)) ) / n] Unfortunately this formula stays difficult in iterative programming, because I have to keep every element of X to calculate H. I know a computional simplified formula exists for the standard deviation (sd) that is much easier in iterative programming. Therefore I wondered I anybody knew about analog simplifications to simplify H: sd = sqrt [ ( Si (Xi - mean(X) )� ) /n ] -> simplified computation -> sqrt [ (n * Si( X� ) - ( Si( X ) )� )/ n� ] This simplied formula is much easier in iterative programming, since I don't have to keep every element of X. E.g.: I have a vector X[1:10] and I already have caculated Si( X[1:10]� ) (I will call this A) and Si( X ) (I will call this B). When X gets extendend by 1 element (eg. X[11]) it easy fairly simple to calculate sd(X[1:11]) without having to reuse the elements of X[1:10]. I just have to calculate: sd = sqrt [ (n * (A + X[11]�) - (A + X[11]�)� ) / n� ] This is failry easy in an iterative process, since before we continue with the next step we set: A = (A + X[11]�) B = (B + X[11]) Can anybody help me to do something comparable for H? Any other help to calculate H easily in an iterative process is also welcome! Thanx in advance! Kind regards, Stef ______________________________________________ [email protected] mailing list https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide! http://www.R-project.org/posting-guide.html
