similar to: replacing double for loops with apply's

Dear R-users, I don't know if there is a problem in wtd.quantile (from library "Hmisc"): -------------------------------- x <- c(1,2,3,4,5) w <- c(0.5,0.4,0.3,0.2,0.1) wtd.quantile(x,weights=w) ------------------------------- The output is: 0% 25% 50% 75% 100% 3.00 3.25 3.50 3.75 4.00 The version of R I am using is: 2.1.0 Best,Jing

The function wtd.var(x,w) in Hmisc calculates the weighted variance of x where w are the weights. It appears to me that wtd.var(x,w) = var(x) if all of the weights are equal, but this does not appear to be the case. Can someone point out to me where I am going wrong here? Thanks. Tom La Bone [[alternative HTML version deleted]]

Hi everyone, now I am trying to finish writing the code (I had asked for assistance on subtracting arrays) This is what I what I am running in R: > source("/home/ie/Documents/TTU/GA_Research/GLUE/R-Project/R_GLUE_Example/NSEr.R") NSEr <- function (obs, sim) { {jjh <- (as.vector(obs) - sim)^2 Xjjhs <- apply(Xjjh, 2, sum) Yii <- (obs - mean(obs))^2 Yiis <- apply(Yii, 2,

Hello list, I have the following function which, as you can see, uses mean: meanratings <- round(apply(stack03[,c(102:121)],2,function(x) (tapply(x ,actcode, mean, na.rm=T))), digits=1) The above function yields the following output: q27a q27b q27c q27d q27e q27f q27g q27h q27i q27j q27k q27l q27m q27o q27p 1 7.8 8.1 7.7 7.9 7.9 NaN NaN 8.4 7.8 7.0 7.6 NaN NaN 7.1 6.0 2

Hello, Using this data: http://n4.nabble.com/file/n1676330/US_Final_Values.txt US_Final_Values.txt and the following code i got the image at the end of this message: US.final.values<-read.table("c:/tmp/US_Final_Values.txt",header=T,sep=" ") US.nls.1<-nls(US.final.values$ECe~a*US.final.values$WTD^b+c,data=US.final.values,start=list(a=2.75,b=-0.95,c=0.731),trace=TRUE)

Hello, I probably have a syntax error in trying to generate an expected survival curve from a weighted cox model, but I can't see it. I used the help sample code to generate a weighted model, with the addition of a "weights=albumin" argument (I only chose albumin because it had no missing values, not because of any real relevance). Below are my code with the resulting error

Hi all, I'm having some difficulty passing arguments into lm() from within a function, and I was hoping someone wiser in the ways of R could tell me what I'm doing wrong. I have the following: lmwrap <- function(...) { wts <- somefunction() print(wts) # This works, wts has the values I expect fit <- lm(weights=wts,...) return(fit) } If I call my function lmwrap, I get

Hello, all. I'm new to R and just beginning to learn to write functions. I know I'm out of my depth posting here, and I'm sure my issue is mundane. But here goes. I'm analyzing the American National Election Study (nes), looking at mean values of a numeric dep_var (environ.therm) across values of a factor (partyid3). I use ddply from plyr and wtd.mean from Hmisc. The nes requires a

Hello, I'm looking to calculate a 95% confidence interval about my estimate for a sample's weighted mean, where the calculated confidence interval would equal the t-test confidence interval of the sample in the case when all of the weights are equal. My initial thought was to simply implement a modified version of the t-test function but substituting the weighted variance and mean for the

Hello, I've been working with "nnet" and now I'd like to use the weigths, from the fitted model, to iterpret some of variables impornatce. I used the following command: mts <- nnet(y=Y,x=X,size =4, rang = 0.1, decay = 5e-4, maxit = 5000,linout=TRUE) X is (m x n) Y is (m x 1) And then I get the coeficients by: Wts<-coef(mts) b->h1 i1->h1

Hello everyone, I have tried several ways of doing this and searched the documentation and help lists and I have been unable to find an answer or even whether it is possible to do it. I am pasting together a formula and I need to insert double quotes around the strings. Here's an example: location <- c("AL", "AK", "MA", "PA") v=2 test <-

Hi, I have a data frame with two columns: dt and tf. The dt column is datetime and the tf column is a temperature. dt tf 1 2009-06-20 00:53:00 73 2 2009-06-20 01:08:00 73 3 2009-06-20 01:44:00 72 4 2009-06-20 01:53:00 71 5 2009-06-20 02:07:00 72 ... I need a subset of the rows where the minutes are 53. The hour is immaterial. I can not find a wildcard

Hello! I have data frame with 'weights' in one of the columns. I need to compute weighted mean on another column other factor variable and i am trying to: res<-tapply(data$k,list(data$model),weighted.mean,w=data$w,na.rm=T) and i get: Warning messages: 1: longer object length is not a multiple of shorter object length in: x * w 2: longer object length is not a multiple of shorter

Hi Duncan I tried Ecdf and/or wtd.quantile from Hmisc and it is working (probably). Ecdf(x, q=.5) Ecdf(x, weights=xw,col=2, add=T, q=.5) wtd.quantile(x) 0% 25% 50% 75% 100% 10 10 10 100 300 wtd.quantile(x, weights=xw, type="i/n") 0% 25% 50% 75% 100% 10.0000 138.8667 192.5778 246.2889 300.0000 But could you please be more specific in this? >

Hi Petr, I think that Duncan suggests something like this: x<- c(rep(10,20), rep(300,5), rep(100, 10)) tx <- table(x) prop.x <- tx / sum(tx) vx <- as.integer(names(tx)) prop.wx <- tx * vx / sum(tx * vx) plot(ecdf(x)) plot(vx, cumsum(prop.x), ylim = 0:1) plot(vx, cumsum(prop.wx), ylim = 0:1) Best regards, Thierry ir. Thierry Onkelinx Statisticus / Statistician Vlaamse

Hello, I was unable to find clues to my problem in ?optim. Using the data and code below, I get an error ("system is exactly singular") when a particular line of code is left in, but have found that 'optim' works when I comment it out. The line of code in question is after the closeAllConnections() line of code and contains a call to "na.approx" from the zoo package.

I could use some help understanding how nls parses the formula argument to a model.frame and estimates the model. I am trying to utilize the functionality of the nls formula argument to modify garchFit() to handle other variables in the mean equation besides just an arma(u,v) specification. My nonlinear model is y<-nls(t~a*sin(w*2*pi/365*id+p)+b*id+int,data=t1,

Dear R users, I have a question on using weighted.mean() while aggregating a data frame. I have a data frame with columns Sub, Length and Slope: > x[1:5,] Sub Length Slope 1 2 351.547 0.0025284969 2 2 343.738 0.0025859390 3 1 696.659 0.0015948968 4 2 5442.338 0.0026132544 5 1 209.483 0.0005304225 and I would like to calculate the weighted.mean of Slope, using Length

Dear All, I am economist and working on poverty / income inequality. I need descriptive statitics like the ratio of education expentitures between different income quintiles where each household has a different weight. After a bit of google search I found 'Hmisc' and 'quantreg' libraries for weighted quantiles. The problem is that these packages give me only weighted quintiles;

Dear all, just a quick question regarding weights in logistic regression. I do results <- lrm(y.js ~ h.hhsize + h.death1 + h.ill1 + h.ljob1 + h.fin1 + h.div1 + h.fail1 + h.sex + h.ch.1