similar to: sequence of vectors

Given an x-vector with, say, 3 elements, I would like to compute the following vector of permutated products (1-x1)*(1-x2)*(1-x3) (1-x1)*(1-x2)*x3 (1-x1)*x2*(1-x3) x1*(1-x2)*(1-x3) (1-x1)*x2*x3 x1*(1-x2)*x3 x1*x2*(1-x3) x1*x2*x3 Now, I already have the correctly sorted matrix of permutations! So, the input looks something like: #input x<-c(0.3,0.1,0.2) Nx<-length(x) Ncomb<-2^Nx

Dear All, I have a vector x = (1,1,1,2,2,2,3,3,3,3) I am looking for a function to return a vector containing the distinct elements of x i,e y = (1,2,3) The following code gives the desired results: as.numeric(levels(as.factor(x))) Is there any other elegant way? Thanks, B --------------------------------- [[alternative HTML version deleted]]

Dear All, I have written the following programs to find a non-singular (10*10) covariance matrix. Here is the program: nitems <- 10 x <- array(rnorm(5*nitems,3,3), c(5,nitems)) sigma <- t(x)%*%x inverse <- try(solve(sigma), TRUE) while(inherits(inverse, "try-error")) { x <- array(rnorm(5*nitems,3,3), c(5,nitems)) sigma <- t(x)%*%x inverse <-

Dear All, How can I sort a data frame with respect to more than one variable? I know that for one variable X one may use: df[order(df$X), ] where df is the data frame containing X. Many thanks, Bernard --------------------------------- [[alternative HTML version deleted]]

Dear All, I would be grateful if you can help me. My problem is the following: I have a data set like: ID time X1 X2 1 1 x111 x211 1 2 x112 x212 2 1 x121 x221 2 2 x122 x222 2 3 x123 x223 where X1 and X2 are 2 covariates and "time" is the time of observation and ID indicates the

Dear all, How can I plot the number of observations per box in a boxplot. Many thanks, Bernard --------------------------------- [[alternative HTML version deleted]]

Daer All, I have a factor variable, X with 5 levels. When I type tables(X) it gives me: table(X) 1 2 3 4 5 10 5 0 0 0 How to drop the levels with zeros such that when I will type: table(X) it will give me: table(X) 1 2 10 5 Thank a lot, Bernard --------------------------------- [[alternative HTML version deleted]]

Dear All, I wonder how to compute the age from the date of birth and the date of examination. Suppose that have the following data: df <- as.data.frame(rbind(c(1,"10/08/1950","15/03/1998"), c(1,"10/08/1950","20/07/1998"), c(1,"10/08/1950","23/10/1998"))) names(df) <- c("ID", "date_birth",

Dear All, I am wondering if there is an R function to convert a covariance matrix to a correlation matrix. I have a covariance matrix sigma and I want to compute the corresponding correlation matrix R from sigma. Thank you very much, Bernard --------------------------------- [[alternative HTML version deleted]]

Dear All, I am looking for an R function or any other reference to generate a series of correlated Binomial (not a Bernoulli) data. The "bindata" library can do this for the binary not the binomial case. Thank you, Bernard --------------------------------- [[alternative HTML version deleted]]

Dear all, mydata<-data.frame(x1=c(1,4,6),x2=c(3,1,2),x3=c(2,1,3)) cor(mydata) x1 x2 x3 x1 1.0000000 -0.5960396 0.3973597 x2 -0.5960396 1.0000000 0.5000000 x3 0.3973597 0.5000000 1.0000000 I wonder if it is possible to fill only lower triangle of this correlation matrix? Using 'dist' doesn't seem to be useful as it doesnt allow to convert this table

Hi, I am tring to write a loop to compute this, ========================== x1=c( rep(-1,4), rep(1,4) ) x2=c( rep(c(-1,-1,1,1),2) ) x3=c( rep(c(-1,1),4) ) x1*x2 x1*x3 x2*x3 ======================== suppose i have x1,x2,x3 i want to compute their ' two factor interactions', x1x2,x1x3 and x2x3, I wrote ======================== for(i in 1:2){ for( j in i+1:3){ xij=c()

Dear All, I wonder how to re-code a categorical variable without using the ifelse loops. For example: Let X be a categorical variable with 6 levels, levels(X) = c(1,2,3,4,5,6) How to create a new categorical variable Y with levels, 10,11,12, say , such that: Y = 10 if X =1 or X=2 Y = 11 if X =3 or 4 Y =11 if X = 5 or 6 many thanks, Bernard

Hello I have data like this x1 x2 x3 x4 x5 I want to create a matrix similar to a correlation matrix, but with the difference between the two values, like this x1 x2 x3 x4 x5 x1 x2-x1 x3-x1 x4-x1 x5-x1 x2 x3-x2 x4-x2 x5-x2 x3 x4-x3 x5-x3 x4 x5-x4 x5 Then I

Dear All, Is there any R function to test if two data frames, say df1 and df2, having the same row names and the same column names are identiques? Thanks in advance, Bernard --------------------------------- [[alternative HTML version deleted]]

Dear All, I am looking how to specify the values on the y axis and x axis for a curve function. example: curve(2*x + 3, 0,10) gives me a range of x axis in (0,2,4,6,8,10) amd for yaxis (5,10,15,20). How to do in order to have the x axis in (0,1,2,3,4,5,6,7,8,9,10) and y axis in (5,6,7,8,......, 10). Thanks a lot, Bernard ---------------------------------

Dear All, I would like to use a nonparametric bootstrap to calculate the confidence intervals for the 5% and 95% quantiles using boot.ci. As you know, boot.ci requires the use of boot to generate bootstrap replicates for my statistic. However this last function doesn't work in my case because I am missing something. Here is an example y <- rnorm(100) Quantile <-

Hello all, I am brand new to R and doing an exercise for a class. I need to find the nearest neighbour for points in the following matrix: > DistanceMatrix x1 x2 x3 x4 x5 [1,] 0.000000 2.828427 1.581139 2.236068 2.000000 [2,] 2.828427 0.000000 1.581139 4.123106 2.000000 [3,] 1.581139 1.581139 0.000000 2.549510 2.121320 [4,] 2.236068 4.123106 2.549510

Dear all, I have created the following data (that you can run) in order to explain my problem: y <- rep(c(1,2), 8) id <- rep(1:8,each=2) x1 <- rep(c("-a","+a"), each = 8) x2 <- rep(c("-b","+b"), each = 2, times = 4) x3 <- rep(c("-c", "+c"), each = 4,2) df <- data.frame(cbind(id,y,x1,x2,x3)) If I do:

Dear All, I wonder if there is any implemented statistical test in R to test the equality between many correlations. As an example, let X1, X2, X3 X4 be four random variables. let Phi(X1,X2) , Phi(X1,X3) and Phi(X1,X4) be the corresponding correlations. How to test Phi(X1,X2) = Phi(X1,X3) = P(X1,X4)? Many thanks in advance, Bernard