similar to: Problem with Median

Hello. I'm a newbie here. In my script (I name it readData.R), I wrote the followings: readData <-function(){ med = read.csv("medicalData.csv");} Then I tested the script by 'Source R Code' then on the command I typed 'readData()' then I typed 'med' to check if the variable contains the medical data but it returned 'Object med is not found'. What

Dear R-users, I have the following 2 files; A V1 V2 A 1 A 2 A 3 A 4 B 1 B 4 C 1 C 3 D 4 B V1 V2 process1 1 process2 2 process3 3 process4 4 I want to get this output C V1 V2 V3 A 1 process1 A 2 process2 A 3 process3 A 4 process4 B 1 process1 B

Greetings R-users, I have been using the fpc package in R to cluster my data. Speficically I am using kmeansruns clustering. I would like to know how I use R to partition data into clusters. What I am doing is as follows. # Use csv file as input ##################### wholeset = read.csv("Spellman800genesImputed.csv") # exclude first col (gene names) ##########################

Hi! Recently I try to find the method maximum likelihood for gamma,weibull,Pearson type III,Kappa Distribution, mixed exponential distribution, skew distribution. I have tried function ms() for gamma two parameters and weibull two parameters.It works but not for Pearson type III. I have problem to find the likelihood function for mixed exponential distribution and kappa distribution. So can

Hi, 1.) The "median" function does not work well. Please refer to the data below (same data is attached as txt-delimited). This is what I try to do in R: median ( dataf [2:9] ) I get warning: "needs numeric data" 2.) BUT if apply the median to a single vector: median ( dataf [,2]] ) then it works: 3.) How come the "median"

Hello. I need calculate the median of several column of a data.frame, in a new column of this data frame, but the median operator only calculate from a vector. I have made a functionc that calculate the median but it is very slow. Are there any method in any package to calculate this? Best regards. Jose Sierra. A B C -0.01678042

Hi, set.seed(45) test1<-data.frame(columnA=rnorm(7,45),columnB=rnorm(7,10)) #used an example probably similar to your actual data apply(test1,2,function(x) sprintf("%.1f",median(x))) #columnA columnB # "44.5"? "10.2" par(mfrow=c(1,2)) lapply(test1,function(x) {b<-

Hello, I am trying to calculate the median of numbers across each row for the data shown below , with the condition that if the number is negative, that it should be ignored and the median should be taken of only the positive numbers. For eg: data is in Column A,B,C. Column D and E demonstrates what I want to get as answer A B C Median median value -13.6688115 -32.50914055

Hi all, I have found the following function online median.test<-function(y1,y2){ z<-c(y1,y2) g <- rep(1:2, c(length(y1),length(y2))) m<-median(z) fisher.test(z<m,g)$p.value } in http://www.mail-archive.com/r-help@r-project.org/msg95278.html I have the following data > group1 <- c(2, 2, 2, 1, 4, 3, 1, 1) > group2 <- c(3, 1, 3, 1, 4, 1, 1, 1, 7, 1, 1, 1, 1,

Dear friends, I'm probably wrong but is there anything better than bootstrap to get a confidence interval of the median from a population with unspecified distribution ? Best wishes Troels Ring, Aalborg, Denmark

Hi, I tried looking through google search on whether there's a way to computer the median for each row of a nxn matrix and return the medians for each row for further computation. And also if the number of columns in the matrix are even, how could I specify which median to use? Thank you very much! -- Edward Chen [[alternative HTML version deleted]]

Doese anyone know a R function to find the median of a gamma distribution? [[alternative HTML version deleted]]

I am studying "Using R for Introductory Statistics" and find it in general very useful. At present, I am stumbling over the function simple.median.test. > x [1] 12.8 3.5 2.9 9.4 8.7 0.7 0.2 2.8 1.9 2.8 3.1 15.8 > simple.median.test (x,median=5) [1] 0.3876953 > simple.median.test (x,median=10) [1] 0.03857422 until here it is identical to the document. Then I

Hi there, I got a data frame with more than 200k columns. How could I get median of each column fast? mapply is the fastest function I know for that, it's not yet satisfied though. It seems function "median" in R calculates median by "sort" and "mean". I am wondering if there is another function with better algorithm. Any hint? Thanks, Xin Zheng

Okay, so I have a homework projecr for R, and we had to input the following link as some sort of data: nb10 <- read.table("http://www.adjoint-functors.net/su/web/314/R/NB10"). Afterwards, we have to use fivenum(nb10) to find max, min, quantiles, and sd, but I'm okay with this. The next question is where I'm stuck. The question is as follows; Compare the median (use the

Dear R-users, Does anyone knows how I can order my serie of boxplots from lowest to highest median (which is much better for visualization purposes). thanks in advance, willem [[alternative HTML version deleted]]

Dear useRs, 1. how to calculate single median value for two columns of a matrix? 2. i have a matrix of 16 columns and 365 rows, how to calculate median between columns 1 and 16, 2 and 16, 3 and 16, 4 and 16, 5 and 16 till 15th column. is there a loop command to do the said operation? regards eliza [[alternative HTML version deleted]]

Hi, Can we get the code for calculating Oja median for multivariate data Thanks and Regards Rahul Agarwal Analyst Equities Quantitative Research UBS_ISC, Hyderabad On Net: 19 533 6363 [[alternative HTML version deleted]]

By considering this reproducible example #start code library(lattice) dotplot(variety ~ yield | site, data = barley, layout = c(1,6), index.cond= function(x,y){median(x)}, panel = function(x,y,...) { panel.dotplot(x,y,...) median.values <- median(x) panel.abline(v=median.values, col.line="red")

The generic stats::median method is defined as median <- function (x, na.rm = FALSE) {UseMethod("median")} I suggest that this should become median <- function (x, na.rm = FALSE, ...) {UseMethod("median")} This would allow additional S3 methods to be developed with additional arguments. Currently I have to over-ride this generic definition in the demography