search for: x3

hello I make a subroutine that give-me a (mathematical) function in string format. I would like transform this string into function ( R function ). thanks for any tips. cleber #e.g. fun_String = "-100*x1 + 0*x2 + 100*x3" fun <- function(x1,x2,x3){ return( ############ evaluation( fun_String ) ############ ) True String mathematical function :-( :-( > nomes [1] "8.49*x1*z1 + 6.13*x1*z2 + 6.4*x1*z3 + 6.9*x2*z1 + 4.54*x2*z2 + 3.99*x2*z3 + 19.31*x3*z1 + 12.49*x3*z2 + 3.86*x3*z3 + 5.25*x1*z1*z...

I have a sparse contingency table (most cells are 0): > xtabs(~.,data[,idx:(idx+4)]) , , x3 = 1, x4 = 1, x5 = 1 x2 x1 1 2 3 1 0 0 31 2 0 0 112 3 0 0 94 , , x3 = 2, x4 = 1, x5 = 1 x2 x1 1 2 3 1 0 0 0 2 0 0 0 3 0 0 0 , , x3 = 3, x4 = 1, x5 = 1 x2 x1 1 2 3 1 0 0 0 2 0 0 0 3 0 0 0 , , x3 =...

Hi All, I'm trying to do the following constrained optimization example. Maximize x1*(1-x1) + x2*(1-x2) + x3*(1-x3) s.t. x1 + x2 + x3 = 1 x1 >= 0 and x1 <= 1 x2 >= 0 and x2 <= 1 x3 >= 0 and x3 <= 1 which are the constraints. I'm expecting the answer x1=x2=x3 = 1/3. I tried the "constrOptim" function in R and I'm running into some issues. I first start off by setting my...

...rmula contained within quotation marks. Here's what it looks like: > dd [1] "lm(y ~ 1,data=Cement)" "lm(y ~ X,data=Cement)" "lm(y ~ X1,data=Cement)" [4] "lm(y ~ X2,data=Cement)" "lm(y ~ X3,data=Cement)" "lm(y ~ X4,data=Cement)" [7] "lm(y ~ X + X1,data=Cement)" "lm(y ~ X + X2,data=Cement)" "lm(y ~ X + X3,data=Cement)" [10] "lm(y ~ X + X4,data=Cement)" "lm(y ~ X1...

Dette er en melding med flere deler i MIME-format. --=_alternative 004C4E4A00257091_= Content-Type: text/plain; charset="US-ASCII" Yes. so (x1*x2*x3*x4*x5*x6*x7*x8)^2 = (x1+x2+x3+x4+x5+x6+x7+x8)^8 ? and there is a difference in (x1*x2*x3*x4*x5*x6*x7*x8)^2 and (x1*x2*x3*x4*x5*x6*x7*x8) althoug the resulting formulas are the same, or? This fikses my problem, but R still crashes for the large formula. It may be due to stack owerflow, but i guess...

...le. For example, if I were measuring weight, I might want to have each individual's weight, and also the group mean by, say, race, sex, and geographic region. The following code works: > x1<-rep(c("A","B","C"),3) > x2<-c(rep(1,3),rep(2,3),1,2,1) > x3<-c(1,2,3,4,5,6,2,6,4) > x<-as.data.frame(cbind(x1,x2,x3)) > x3.mean<-rep(0,nrow(x)) > for (i in 1:nrow(x)){ + x3.mean[i]<-mean(as.numeric(x[,3][x[,1]==x[,1][i]&x[,2]==x[,2][i]])) + } > cbind(x,x3.mean) x1 x2 x3 x3.mean 1 A 1...

On Wed, 5 Oct 2005 Hallgeir.Grinde at elkem.no wrote: > And some more informastion I forgot. > R does not crash if I write out the formula: > > set.seed(123) > x1 <- runif(1000) > x2 <- runif(1000) > x3 <- runif(1000) > x4 <- runif(1000) > x5 <- runif(1000) > x6 <- runif(1000) > x7 <- runif(1000) > x8 <- runif(1000) > y <- rnorm(1000) > fit <- lm(y~(x1*x2*x3*x4*x5*x6*x7*x8)^2) > -> R crashes > > fit <- lm(y~x1+x2+x3+x4+x5+x6+x7+x8 >...

On Wed, 5 Oct 2005 Hallgeir.Grinde at elkem.no wrote: > Yes. > so (x1*x2*x3*x4*x5*x6*x7*x8)^2 = (x1+x2+x3+x4+x5+x6+x7+x8)^8 ? Yes in the sense that the simplified formula given by terms() is the same. > and there is a difference in > (x1*x2*x3*x4*x5*x6*x7*x8)^2 > and > (x1*x2*x3*x4*x5*x6*x7*x8) > althoug the resulting formulas are the same, or? The first...

Dear list-subscriber, in the process of writing a general code snippet to extract coefficients in an expression (in the example below: 0.5 and -0.7), I stumbled over the following peculiar (at least peculiar to me:-) ) sorting behaviour of the function all.names(): > expr1 <- expression(x3 = 0.5 * x1 - 0.7 * x2) > all.names(expr1) [1] "-" "*" "x1" "*" "x2" > all.vars(expr1) [1] "x1" "x2" > expr2 <- expression(x3 <- 0.5 * x1 - 0.7 * x2) > all.names(expr2) [1] "<-" "x3" &q...

Hello, I am trying to reformat some data so that it is organized by group in the columns. The data currently looks like this: group X3.Hydroxybutyrate X3.Hydroxyisovalerate ADP 347 4 4e-04 3e-04 5e-04 353 3 5e-04 3e-04 6e-04 359 4 4e-04 3e-04 6e-04 365 4 6e-04...

Hi R users, Is there an easy way in R to generate the results table below using table 1 and the formula (simplified version of the real problem)? It would be easy if I knew the R equivalent of SAS's retain function, but could not find one. Thanks in Advance for any help! table1: ID X2 X3 1.00 1.00 0 2.00 0.00 3.00 1.00 4.00 3058 5.00 0.00 6.00 6.00 Formula: X3 = x2 + (.24 * x3) where the values in the x3 column of the result table are retained from previous x3 rows.. Also the first x3 value is initialized to 0 to start e.g. for ID=1 we have 1 + .24(0) = 1.0...

Hello, 1. After I generate a 100x50 matrix by x3<-matrix(0,100,50);for (i in 1:100) {x1<-rpois(50, mu[i]);x2<-x1; x2[runif(50)<.01]<-0; x3[i,]<-x2}, 2. I want to calculate means and sample variances of each row and create a new matrix 100x2; 3. I then want to repeat above procedure 500 times so that eventually I will have 500...

hi Using a ts or tprs basis, I expected gcv to decrease when increasing the basis dimension, as I thought this would minimise gcv over a larger subspace. But gcv increased. Here's an example. thanks for any comments. greg #simulate some data set.seed(0) x1<-runif(500) x2<-rnorm(500) x3<-rpois(500,3) d<-runif(500) linp<--1+x1+0.5*x2+0.3*exp(-2*d)*sin(10*d)*x3 y<-rpois(500,exp(linp)) sum(y) library(mgcv) #basis dimension k=5 m1<-gam(y~x1+x2+te(d,bs="ts")+te(x3,bs="ts")+te(d,x3,bs="ts"),family="poisson") #basis dimension k=10...

Hi, I'd like to normalize a dataset by dividing each row by the first row. Very simple, right? I tried this: > expt.fluor X1 X2 X3 1 124 120 134 2 165 163 174 3 52 51 43 4 179 171 166 5 239 238 235 >first.row <- expt.fluor[1,] > normed <- apply(expt.fluor, 1, function(r) {r / first.row}) >normed [[1]] X1 X2 X3 1 1 1 1 [[2]] X1 X2 X3 1 1.330645 1.358333 1.298507 [[3]] X1...

...the name of the dependent variable first in the in "dataClasses". It caused an unexpected behavior in my code, so (as usual) the bug may be mine, but in my heart, I believe it belongs to delete.response. To illustrate, here's a terms object from a regression. > tt y ~ x1 * x2 + x3 + x4 attr(,"variables") list(y, x1, x2, x3, x4) attr(,"factors") x1 x2 x3 x4 x1:x2 y 0 0 0 0 0 x1 1 0 0 0 1 x2 0 1 0 0 1 x3 0 0 1 0 0 x4 0 0 0 1 0 attr(,"term.labels") [1] "x1" "x2" "x3"...

...ariables when the full hierarchy of effects is not present. Depending on which lower level interactions are specified, the factor encoding changes for a higher level interaction. Consider the following minimal reproducible example: -------------- > runmatrix = expand.grid(X1=c(1,-1),X2=c(1,-1),X3=c("A","B","C"))> model.matrix(~(X1+X2+X3)^3,data=runmatrix) (Intercept) X1 X2 X3B X3C X1:X2 X1:X3B X1:X3C X2:X3B X2:X3C X1:X2:X3B X1:X2:X3C 1 1 1 1 0 0 1 0 0 0 0 0 0 2 1 -1 1 0 0 -1 0 0...

Hi R-helpers! I have the following dataframe: firm<-c(rep(1:3,4)) year<-c(rep(2001:2003,4)) X1<-rep(c(10,NA),6) X2<-rep(c(5,NA,2),4) data<-data.frame(firm, year,X1,X2) data So I want to obtain the same dataframe with a variable X3 that is: X1, if X2=NA X2, if X1=NA X1+X2 if X1 and X2 are not NA So my final data is X3<-c(15,NA,12,5,10,2,15,NA,12,5,10,2) finaldata<-data.frame(firm, year,X1,X2,X3) I've tried this finaldata<-ifelse(data$X1==NA,ifelse(data$X2==NA,NA,X2),ifelse(data$varvendas==NA,X1,X1+X2)) But I g...

...m I totally missing something (I'm quite new to this, as I'm sure is obvious). Why such a difference in GCV change between poisson and negbin? Thanks in advance! Jordan library(mgcv) set.seed(0) n<-400 sig<-2 x0 <- runif(n, 0, 1) x1 <- runif(n, 0, 1) x2 <- runif(n, 0, 1) x3 <- runif(n, 0, 1) f0 <- function(x) 2 * sin(pi * x) f1 <- function(x) exp(2 * x) f2 <- function(x) 0.2*x^11*(10*(1-x))^6+10*(10*x)^3*(1-x)^10 f <- f0(x0) + f1(x1) + f2(x2) g<-exp(f/4) y<-rpois(rep(1,n),g) summary(gam(y~s(x0)+s(x1)+s(x2)+s(x3),family=poisson,scale=-1))$gcv summa...

Dette er en melding med flere deler i MIME-format. --=_alternative 004613C000257091_= Content-Type: text/plain; charset="US-ASCII" And some more informastion I forgot. R does not crash if I write out the formula: set.seed(123) x1 <- runif(1000) x2 <- runif(1000) x3 <- runif(1000) x4 <- runif(1000) x5 <- runif(1000) x6 <- runif(1000) x7 <- runif(1000) x8 <- runif(1000) y <- rnorm(1000) fit <- lm(y~(x1*x2*x3*x4*x5*x6*x7*x8)^2) -> R crashes fit <- lm(y~x1+x2+x3+x4+x5+x6+x7+x8 +x1:x2+x1:x3+x1:x4+x1:x5+x1:x6+x1:x7+x1:...

Forgive me if this is a trivial question, but I couldn't find it an answer in former forums. I'm trying to reproduce some SAS results where they set two parameters equal. For example: y = b1X1 + b2X2 + b1X3 Notice that the variables X1 and X3 both have the same slope and the intercept has been removed. How do I get an estimate of this regression model? I know how to remove the intercept ("-1" somewhere after the tilde). But how about setting parameters equal? I have used the car package to...