similar to: "unique" rows in data frame

Dear R users, i have this problem with matrices i want to check between two matrices if they are isomorphic i will give an example for what excactly i want   1 -1  1                                 -1  1   1 -1   1  -1                                1  -1  -1   1  1   -1                                1   1  -1 this two matrices are isomorphic beacause if i change the first 2 columns the matrices

Hi, all, I have a little problem to solve. I'd like to collapse rows which are next to each other but have same value to one row. The following is an example. say x is a data frame like: X1 X2 X3 X4 X5 a 1 0 0 0 1 b 1 0 1 0 1 c 1 0 0 0 1 d 1 0 0 0 1 e 1 0 0 0 1 f 1 1 0 0 1 g 1 1 0 0 1 notice that a, c,d,e are the same. since c,d,e are next to each

Dear R-users, I would like to create a new data frame composed of 2 columns of another data frame. But it does not give me what I want... > casesCNST[1:10,] case X1 X2 X3 X4 expected 1 A1 0 0 0 0 E 2 A2 0 0 0 1 C 3 A3 0 0 0 2 C 4 A4 0 0 0 3 C 5 A5 0 0 0 4 C 6 A6 0 0 1 0 C 7 A7 0 0 1 1 C 8

A user sent me an example where coxph fails, and the root of the failure is a case where names(mf) is not equal to the term.labels attribute of the formula -- the latter has an extraneous newline. Here is an example that does not use the survival library. # first create a data set with many long names n <- 30? # number of rows for the dummy data set vname <- vector("character",

Hi everyone, my question might be very trivial, but I could not come up with an answer... I want to find out how often a matrix contains a certain vector as row: x1<-c(1,2,3) x2<-c(1,5,6) x3<-c(7,8,9) A<-matrix(c(rep(x1,5),rep(x2,5),rep(x3,5),rep(x1,5)),nrow=20,ncol=3,byrow=T) How can I find out, how many times x1 is a row of A? Thanks in advance and best regards, Sebastian

Dear All, I have got the limits for removing extreme values for each variables using following function . f=function(x){quantile(x, c(0.25, 0.75),na.rm = TRUE) - matrix(IQR(x,na.rm = TRUE) * c(1.5), nrow = 1) %*% c(-1, 1)} #Example: n <- 100 x1 <- runif(n) x2 <- runif(n) x3 <- x1 + x2 + runif(n)/10 x4 <- x1 + x2 + x3 + runif(n)/10 x5 <-

Hello, I found a potential problem in R 2.1.0 (and R 2.0.1) I expect that > tmp <- FUN(x1, x2, x3, x4) > as.data.frame(tmp) is the same as > as.data.frame(FUN(x1, x2, x3, x4)) since the tmp variable in this case is unnecessary. However, below I will demonstrate that under an odd set of conditions, I can correctly perform as.data.frame(tmp), but not as.data.frame(FUN(x1, x2, x3,

Dear all, I'm looking to create a formula within a function to pass to glmer() and I'm having a problem that the following example will illustrate: library(lme4) y1 = rnorm(10) x1 = data.frame(x11=rnorm(10), x12=rnorm(10), x13=rnorm(10)) x1 = data.matrix(x1) w1 = data.frame(w11=sample(1:3,10, replace=TRUE), w12=sample(1:3,10, replace=TRUE), w13=sample(1:3,10, replace=TRUE)) test1 <-

Dear R users, I have a data with 1000 variables named "x1", "x2", ..., "x1000", and I want to construct a formula like this format: ~x1+x2+...+x1000+x1:x2+x1:x3+x999:x1000+log(x1)+...+log(x1000) That is: the base variables followed by all interaction terms and all base feature log-transformations. I know I can use several paste functions to construct it. But is

Hello, How does one specify a formula to lm inside a function (with variable names not known in advance) and have the formula appear explicitly in the output? For example, f <- function(d) { in.model <- sample(c(0,1), ncol(d)-1, replace=T) current.model <- lm(paste(names(d)[1], "~", paste(names(d[2:ncol(d)])[which(in.model == 1)], collapse= "+")), data=d) #***

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: I try to revise the maximum likelihood function below using something constrains. But it seems something wrong with it. Becasue R would not allow me to edit the function like this. It is very appreciate if you can help. function (parameters,y,x1,x2) { p<-parameters[1] alpha1<-parameters[2] beta1<-parameters[3)] delta1<-parameters[4] alpha2<-parameters[5]

Suppose I have the following: x1<-as.vector(rnorm(10)) x2<-as.vector(rnorm(10)) x3<-as.vector(rnorm(10)) x4<-as.vector(rnorm(10)) x5<-as.vector(rnorm(10)) x6<-as.vector(rnorm(10)) x7<-as.vector(rnorm(10)) x8<-as.vector(rnorm(10)) x9<-as.vector(rnorm(10)) x10<-as.vector(rnorm(10)) I would like the mean of x1 to x6 for each vector position. I would do something else

Hi suppose I have two arrays x1,x2 of dimensions a1,b1,c1 and a2,b2,c2 respectively. I want x = x1 "+" x2 with dimensions c(max(a1,a2), max(b1,b2),max (c1,c2)) with x[a,b,c] = x1[a1,b1,c1] + x2[a2,b2,c2] if a <=min(a1,a2) , b<=min (b1,b2), c<=min(c1,c2) and the other bits either x1 or x2 or zero according to whether the coordinates are "in range" for

Hello, I am trying to make a data frame from many elements after running a function which creates many elements, some of which may not end up being real elements due to errors or missing data. For example, I have the following three elements p1s, p2s, and p3s. p9s did not generate the same data as there was an error in the function for some reason. I currently have to delete p9s from the

Is it possible to use "paste" to write out an expression and evaluate it? Suppose I want to add two vectors X1 and X2, defined as follows: X1 <- 1:6 X2 <- 6:1 If I write the following it looks like what I want but is a character: noquote(paste(paste("X", 1, sep = ""), paste("X", 2, sep = ""), sep = "+")) Is there a way to tell R

Dear All, I have a data frame with n columns: X1, X2, ., Xn. Now I want to create a new column: if X1 = X2 = . = Xn, the value is 1; Otherwise, the value is 0. How to do that in a quick way instead of doing (n choose 2) comparisons? Thank you, Frank [[alternative HTML version deleted]]

Hello folks, Any ideas how to do this? data.frame is a data frame with column names "x1",...,"xn" y is a response variable of length dim(data.frame)[1] I want to write a function function(y, data.frame){ lm(y~x1+...+xn) } This would be easy if n was always the same. If n is arbitrary how could I feed the x1+...+xn terms into lm(response~terms)? Thanks Richard -- Dr.

Failed? What was the error message? Cheers, Bert Bert Gunter "The trouble with having an open mind is that people keep coming along and sticking things into it." -- Opus (aka Berkeley Breathed in his "Bloom County" comic strip ) On Tue, Aug 22, 2017 at 8:17 AM, Stephen O'hagan <SOhagan at manchester.ac.uk> wrote: > I'm trying to use boot.stepAIC for

The error is "the model fit failed in 50 bootstrap samples Error: non-character argument" Cheers, SOH. On 22/08/2017 17:52, Bert Gunter wrote: > Failed? What was the error message? > > Cheers, > > Bert > > > Bert Gunter > > "The trouble with having an open mind is that people keep coming along > and sticking things into it." > -- Opus (aka