similar to: keep the row indexes/names when do aggregate

Dear all, If I have a data frame x<-data.frame(x1=c(1,7),x2=c(4,6),x3=c(8,2)): x1 x2 x3 1 4 8 7 6 2 I want to sort the whole data and get this: x1 1 x3 2 x2 4 x2 6 x1 7 x3 8 If I do sort(X), R reports: Error in order(list(x1 = c(1, 7), x2 = c(4, 6), x3 = c(8, 2)), decreasing = FALSE) : unimplemented type 'list' in 'orderVector1' The only way

Hi, netters, First of all, thanks a lot for all the prompt replies to my earlier question about "merging" data frames in R. Actually that's an equivalence to the "join" clause in mysql. Now I have another question. Suppose I have a data frame X with lots of columns/variables: Name, Age,Group, Type, Salary. I wanna do a subtotal of salaries: aggregate(X$Salary,

Dear R users, I am trying to have an xyplot of a data set which has the following variables: case (n=10,20,30) parameter (parm=a,b) group (grp=g1,g2) y (y values) x (x=2,4,8) My plot should be parameter by case such that I have 2 rows (each row= each parameter) and 3 columns (each column=each case). My R-code is as follows but I am not able to get what I want to: tp1.sim <- xyplot(y~ x | case

I'm trying to collapse the following table along the sub-group factor. In this case, collapsing means taking the average ages within a subgroup and creating a new table. I seem to be running into trouble trying to create this new data frame. I can use the ave() function to find averages within a subgroup but how do I maintain the Group/Gender factors after collapsing? (see bottom table) Can

Dear Sirs, I had the following (reproducible) problem: g1 <- gl(3,4) g2 <- gl(2,2,12) y <- rnorm(12) AA <- aggregate(y,list(g1,g2),mean) print(AA,2) # I should've said 'digits=2' but ... Segmentation fault (core dumped) # print(AA) and print(AA,digits=2) work as expected Many thanks, -Rashid Nassar --please do not edit the information below--

Hi R-users, Since R.2.11 aggregate can now deal with non-scalar functions, which is very useful to me. However, I have a question about how best to process the output. test <- data.frame(a = rep(c("g1", "g2"), each = 50), b = runif(100)) res <- aggregate(test$b, list(group = test$a), function(x) quantile(x, probs = c(0.05, 0.95))) > res group x.5%

Dear all, I''m trying to write a function, that will take as an argument, some aligned genome sequences, and using a sliding window, do pairwise comparisons of sequence similarity. Coding the sliding window I think I can manage but what I''m trying to get to grips with is getting it so as every pairwise comparison is made, no matter how many genomes are added, from 3 to N. So if

Hello, I have the following data.frame structure(list(Study = structure(c(1L, 2L, 3L, 4L, 5L, 6L, 7L, 8L, 9L, 10L, 11L, 12L, 13L, 14L, 15L, 16L, 17L, 18L, 19L, 1L, 2L, 3L, 4L, 5L, 6L, 7L, 8L, 9L, 10L, 11L, 12L, 13L, 14L, 15L, 16L, 17L, 18L, 19L, 1L, 2L, 3L, 4L, 5L, 6L, 7L, 8L, 9L, 10L, 11L, 12L, 13L, 14L, 15L, 16L, 17L, 18L, 19L, 1L, 2L, 3L, 4L, 5L, 6L, 7L, 8L, 9L, 10L, 11L, 12L, 13L, 14L,

Hi I am using the following script to average a set of data 0f 62 columns into 31 colums. The data consists of values of ln(0.01) or -4.60517 instead of NA's. These need to be averaged for each row (i.e 2 values being averaged). What I would I need to change for me to meet the conditions: 1. If each run of the sample has a value, the average is given 2. If only one run of the sample has a

Hi, I wonder how to pass several functions and their arguments as arguments to a function. For example, the main function is f = function(X ) { process1(X) ... process2(X) } I have a few functions that operate on X, e.g. g1(X, par1), g2(X, par2), g3(X, par3). par1, par2 and par3 are parameters and of different types. I would like to pass g1, g2, g3 and their arguments to f and g1,

I'm solving 4 complex equations simultaneously. Code is below. The code returns only zero's for the solution though there should also be a non-zero result. I'm pretty confident that the equations are correct because they are straight from a published paper and I checked them pretty thoroughly. The parameter values I used are from the published paper as well. Any suggestions for how

Dear R users, I used to "OPTIM" to minimize the obj. function below. Even though I used the true parameter values as initial values, the results are not very good. How could I improve my results? Any suggestion will be greatly appreciated. Regards, Kathryn Lord #------------------------------------------------------------------------------------------ x = c(0.35938587,

Dear R users, I used to "OPTIM" to minimize the obj. function below. Even though I used the true parameter values as initial values, the results are not very good. How could I improve my results? Any suggestion will be greatly appreciated. Regards, Kathryn Lord #------------------------------------------------------------------------------------------ x = c(0.35938587,

Hi, I run this code to get the power of the test for modified bartlett's test..but I'm not really sure that my coding is right.. #normal distribution unequal variance asim<-5000 pv<-rep(NA,asim) for(i in 1:asim) {print(i) set.seed(i) n1<-20 n2<-20 n3<-20 mu<-0 sd1<-sqrt(25) sd2<-sqrt(50) sd3<-sqrt(100) g1<-rnorm(n1,mu,sd1) g2<-rnorm(n2,mu,sd2)

?s 17:39 de 25/07/2024, Yuan Chun Ding via R-help escreveu: > Hi R users, > > I generated a square correlation matrix for the dat dataframe below; > dat<-data.frame(g1=c(1,0,0,1,1,1,0,0,0), > g2=c(0,1,0,1,0,1,1,0,0), > g3=c(1,1,0,0,0,1,0,0,0), > g4=c(0,1,0,1,1,1,1,1,0)) > library("Hmisc") > dat.rcorr =

Greetings All. I want to do the following simple thing. I have defined a function med3x3() such that, given vectors X,Y, med3x3(X,Y) returns a 3x3 table where: Row 1: X > median(X) Row 2: X = median(X) Row 3: X < median(X) Col 1: Y < median(Y) Col 2: Y = median(Y) Col 3: Y > median(Y) (with intersections of these conditions for the individual cells). I can easily define fixed

Dear R members I have a problem regarding storing the lists. Let L=number of distinct values of any predictor (say L=5) P=number of predictors (say P=20) g1 <- c() for(i in 1:P){ if(L > 1){ for(j in 1:(L-1)){ g <- .... g1[j] <- g } } g2[]=sort.list(g1) } Now the question is: What should we use inside brackets of g2[....], whether

Hi, I want to compute a distribution of the intersection of a graph and 'randomized' graphs induced by the permutations of node labels (to preserve the graph topology). Since I ll have many permutations to perform, I was thinking of using the snow package and in particular "parSapply" to divide the work between my 4 CPUs. But I get the following error message : Error in

After reading the help page on set.seed, I am unsure about how to solve the following problem. I need to call function f a thousand times. The list of values returned by f, should be as random as possible. f calls g twice: f <- function(){g1 <- g(1); g2 <- g(2); c(g1; g2)} The function g in turn calls sample and returns a number, but also depends on its argument, so, starting from the

-----BEGIN PGP SIGNED MESSAGE----- Hash: SHA1 Hello, I like to know what will happend if: - - one Domain with 2 DCs - - two Admins working on their Windows-ws with RSAT. - - Admin1 is connected to DC1 - - Admin2 is connected to DC2 - - Admin1 creats a user "u1" and a group "g1" - - then the two DCs will lose connection, the user and group is replicated to DC2 - - now Admin1