similar to: How to stack row vector on top of each other?

I looked at the documentation of source() and summary(), and I could not find the reason why calling something like: > summary(resamps) from the command line, works (it prints the summary) whereas calling summary(resampls) from a file that I source with source("my_file.r") does not print anything. How can I get summary(resamps) to print when I source a file with this command?

Hi All, I am having great trouble doing something pretty simple. Here is what I did: > x <- read.table("clipboard") > dim (x) [1] 126 10 > typeof(x) [1] "list" > w <- array(x) > typeof(w) "list" Q1: How come after constructing an array out of the list, the type of the array is still "list"? > w <- as.array(x) Error in

Hello, How can I apply a function on a vector that refers to actual (n) and previous elements in the vector (e.g. n-1)? For example: I would like to calculate the sum of (n-1) + n for each element of a vector and get a vector as a result. Besides others I tried this: v<-c(3,6,8,1,1,3,9,5,6,3) for (i in 1:NROW(v)){a[i]<-a[i-1]+a[i]} I would like to get this result:

This may be a really obvious question but I just can't figure out how to do it. I have a small dataset that I am trying to compare to some controls. It is essential that the controls are matched on Cancer Stage (a numerical factor between 1 and 4), and then ideally on Age (integer), Gender (factor), Performance Status(factor). I'm using matchit to try and do this, but it seems to give

Hi all, I've simple code to read a file (verify.txt in the same directory as the script file) but when I run this I get "Error in eval(expr, envir, enclos) : object "y" not found". data_model.df = read.table("./verify.txt", header=TRUE, nrows=10); f <- lm(y ~ x) Could someone pls tell me what's wrong with this code? Sincere thanks!

I am trying to concatenate 2 datasets that don't have exactly the same column. In SAS I did: data summary; set agency prop; run; No problem in R I get error message summary <-rbind(agency,prop) Error in match.names(clabs, names(xi)) : names do not match previous names But when I use rbin.fill, that overwrites the second file w/ first one. Is there a way to replicate the sas process

Hi R-Help! I am trying to find a nicer way of extracting all the "complete" diagonals of a matrix. I am working with very large matrices that have many more rows than columns. I want to be able to extract each of the diagonals that are as long as the number of columns in the matrix. I have written a rather ugly function that presently does the job. It illustrates what I am trying to

Dear R helpers, I have a question about drawing random numbers from many categorical distributions. Consider n individuals, each follows a categorical distribution defined over k categories. Consider a simple case in which n=4, k=3 as below catDisMat <- rbind(c(0.1,0.2,0.7),c(0.2,0.2,0.6),c(0.1,0.2,0.7),c(0.1,0.2,0.7)) outVec <- rep(NA,nrow(catDisMat)) for (i in 1:nrow(catDisMat)){

I am very new to Linux so I probably am doing something stupid but I cannot seem to update to R 2.15 Using Ubuntu 12.02 Precise Penguin I realise that debian packages are not updated regularly so I tried to follow the insructions at the R-site So far, I have modified /etc/apt/sources.list to read ## R CRAN added 2012-06-12 deb http://probability.ca/cran/bin/linux/debian squeeze-cran/ Issued

Hello. I am using some dates I read in excel in R. I know the excel origin is supposed to be 1900-1-1. But when I used as.Date with origin=1900-1-1 the dates that R reported me where two days ahead than the ones I read from Excel. I noticed that when I did in R the following: > as.Date("2011-3-4")-as.Date("1900-1-1") Time difference of 40604 days but if I do the same

Hello, I want to use lag on a time variable but I have to take date into consideration ie I don't want days to overlap ie: I don't want my first time of today to match my last time of yeterday. In SAS I would use : data x; set y; by date tim; previous=lag(tim); if first.date then do; previous=.; end; run; How can I do something similar in R? I can't find

#I want to "merge" or "join" 2 dataframes (df1 & df2) into a 3rd (mydf). I want the 3rd dataframe to contain 1 row for each row in df1 & df2, and all the columns in both df1 & df2. The solution should "work" even if the 2 dataframes are identical, and even if the 2 dataframes do not have the same column names. The rbind.fill function seems to work. For

Hi list, I am trying to calculate the sensitivity, specificity, and odds ratio(and confidence interval). Say, my data looks like this 42.53, 37.56, 40.51, 32.67, 38.19, 81.74, 41.55, 68.94, 59, 63, 54.13, 48.85, 50.46, 51.78 Is there any packages in R that does this? Espeically the odds ratio confidence interval. Mike [[alternative HTML version deleted]]

here's my question: suppose I have a matrix: mt<-matrix(1:12,ncol=6) now I have a vector vt<-c(1,2,2,2,1,2) which means I want to get: the 1st row for column1; the 2nd row for column2; the 2nd row for column3; the 2nd row for column4; ... that what I want is this vector: 1,4,6,8,9,12 Does anyone know how to do this fast? I know I can use for-loop to travel all columns,but

Suppose x and y are numeric vectors of the same length. plot(x,y) #scatterplot lmObj1 <- lm(y~x) # best fit line abline(lmObj1) # good lmObj2 <- lm(x~y) #get best fit but with axes interchanged abline(lmObj2) # not what I want. I want the correct line, drawn on the same graph, but with # response and predictor variables interchanged One way to proceed would be to

Hi All I have a vector of about 15 million numbers which I would like to plot. The goal is the see the distribution. I tired the usual steps. 1. Histogram : never gets complete my window freezes w/out log base 10 2. Density : I first calculated the kernel density and then plotted it which worked. It would be nice to superimpose histogram with density but as of now I am not able to get this

Hi, list I am doing 100,000 iterations of Bayesian simulations. What I did is I split it into 4 different R sessions, each one runs 25,000 iteration. But two of the sessions gave the simulation result. I did not use any set.seed(). What is going on here? Thanks, Mike [[alternative HTML version deleted]]

I have some data set which has some values -999.000 & I would like to remove whole row of this kind of values. e.g a<-matrix(c(1,2,3,4,4,5,6,6,-999.99,5,9,-999.00),nrow=4) a<- [,1] [,2] [,3] [1,] 1 4 -999.99 [2,] 2 5 5.00 [3,] 3 6 9.00 [4,] 4 6 -999.00 expected answer [,1] [,2] [,3] [1,] 2 5 5.00 [2,] 3 6 9.00 I

Hello, I'm trying to import a SAS file made using SAS on Unix. Currently I'm using SAS on Windows and I'm trying to import that .ssd04 file to R. The file name of the file is testfile.ssd04 and it is located in 'M:\sasuser'. I'm using Windows XP and R 2.91. Basically what I'm doing is ############ r code ############## > library(foreign) > sashome <-

Hello all, given two vectors X and Y I'd like to receive a vector Z which contains, for each element of X, the index of the corresponding element in Y (or NA if that element isn't in Y). Example: x <- c(4,5,6) y <- c(10,1,5,12,4,13,14) z <- findIndexIn(x, y) z [1] 5 3 NA 1st element of z is 5, because the 1st element of x is at the 5th position in y 2nd element of z is 3,