search for: wsum

...in a given interval. Given equal-length vectors of data 'data' and weights 'w', and breaks (intervals) for the histogram, I calculate a weighted histogram as follows (see MASS's 'truehist' for an unweighted histogram): bin <- cut(data, breaks, include.lowest = TRUE) wsums <- sapply(split( w, f = bin), sum) prob <- wsums/(diff(breaks) * sum(w)) where 'bin' is the recoded data vector, 'wsums' is the sum of weights across each factor in 'bin', and 'prob' is the vector of probabilities used to plot the true histogram. But if...

Is there a weighted median function out there similar to weighted.mean() but for medians? If not, I'll try implement or port it myself. The need for a weighted median came from the following optimization problem: x* = arg_x min (a|x| + sum_{k=1}^n |x - b_k|) where a : is a *positive* real scalar x : is a real scalar n : is an integer b_k: are negative and positive scalars

Hi Eik, thank you so much - it works perfectly! Thank you and best wishes Alain   Eik Vettorazzi <E.Vettorazzi@uke.de> hat am 6. Februar 2013 um 17:01 geschrieben: > Hi Alain, > here is a one-liner for a df without the rowSum column > > df<-data.frame(id=c("x01","x02","x03","x04","x05","x06"),a=c(1,2,NA,4,5,6),b=c(2,4,6,8,10,NA),c=c(NA,3,9,12,NA,NA)) > > (df$wSum<-apply(sweep(df[,-1],1,rowSums(df[,-1],na.rm=T),"/"),1,function(x)sum...

...,12,NA,NA),sum=c(3,9,15,24,15,6))    id    a     b     c   sum 1 x01  1     2    NA   3 2 x02  2     4     3      9 3 x03 NA  6     9     15 4 x04  4     8    12    24 5 x05  5    10   NA  15 6 x06  6   NA  NA   6 #my weights w<-c(10.5,9,12.8) #now I want to calculate the proportion of the rowsum = "sum" of every other number in the row, that is df[1,2]/df[1,5], df[1,3]/df[1,5], ... #e.g. df.prop    id         a         b           c          sum 1 x01      0.33     0.66     NA        3 2 x02      0.22     0.44     0.33     9 3 x03     NA       0.4       0.6       15 4 x04...