search for: fac3

...s same output as SAS anova(mlmfit, mlmfit0, M = ~ fac1 + fac2, X = ~ fac2, idata = idata, test = "Wilks") % test fac1*fac2 interaction, also works, also the same output as SAS anova(mlmfit, mlmfit0, X = ~ fac1 + fac2, idata = idata, test = "Wilks") Three Factors: fac1, fac2, fac3 mlmfit <- lm(mydata~1) mlmfit0 <- update(mlmfit, ~0) % test fac1, works, same as SAS anova(mlmfit, mlmfit0, M = ~ fac1 + fac2 + fac3, X = ~ fac2 + fac3, idata = idata, test = "Wilks") Now, I try to compute the interactions the same way, but this doesn't work: % fac1*fac2...

Dear List, I am using function recode, from package car, within sapply, as follows: L3 <- LETTERS[1:3] (d <- data.frame(cbind(x = 1, y = 1:10), fac1 = sample(L3, 10, replace=TRUE), fac2 = sample(L3, 10, replace=TRUE), fac3 = sample(L3, 10, replace=TRUE))) str(d) d[, c("fac1", "fac2")] <- sapply(d[, c("fac1", "fac2")], recode, "c('A', 'B') = 'XX'", as.factor.result = TRUE) d[, "fac3"] <- recode(d[, "fac3"], "c(...

...quot;i","j")]<-"C" levels(fac1) print(fac1) ##second method fac2<-fac? levels(fac2)[c(1,2,3)]<-"A" levels(fac2)[c(2,3,4)]<-"B" # not c(4,5,6) levels(fac2)[c(3,4,5,6)]<-"C" # not c(7,8,9,10) levels(fac2) print(fac2) #third method fac3<-fac levels(fac3)<-list("A"=c("a","b","c"),"B"=c("d","e","f"),"C"=c("g","h","i","j")) levels(fac3) print(fac3) I first tried method 1 and had no luck with it at a...

...m writing to the nested functions. There must be a way, but I have not been able to figure it out. An example is below. Any advice would be greatly appreciated. Thanks, Dan # some example data df=expand.grid(rep=1:4, fac1=c("a","b"), fac2=c("c","d"), fac3=c ("e","f")) df$resp1=rnorm(length(df$fac1)) df # define a funciton to compute mean, std error and n. meanstderr <- function(x) c(Mean=mean(x,na.rm=TRUE),se=sqrt(var (x,na.rm=TRUE)/length(na.omit(x))), n=length(na.omit(x))) # what I want to wrap into a single funtction:...

Hey, hopefully there is an easy way to solve my problem. All that i think off is lengthy and clumsy. Given a data.frame d with columns VALUE, FAC1, FAC2, FAC3. Let FAC1 be something like experiment number, so that there are exactly the same number of rows for each level of FAC1 in the data.frame. Now i would like to scale all values according to the center of its experiment. So i can apply s <- by(d[1], FAC1, scale). But i don't want to lose the...

..."contr.poly" > scores <- rep(seq(-2, 2), 3); scores [1] -2 -1 0 1 2 -2 -1 0 1 2 -2 -1 0 1 2 > fac <- gl(3, 5); fac [1] 1 1 1 1 1 2 2 2 2 2 3 3 3 3 3 Levels: 1 2 3 > > # I get this: > model.matrix(~ scores:fac) (Intercept) scores:fac1 scores:fac2 scores:fac3 1 1 -2 0 0 2 1 -1 0 0 3 1 0 0 0 4 1 1 0 0 5 1 2 0 0 6 1 0 -2...

...function in the rms package, which at least generates confidence intervals similar to what is observed in SPSS. However, the p-values associated with each predictor in the model are not really close in many instances. Here is the code I am using: formula=Surv(months, recidivate) ~ fac1 + fac2 + fac3 + fac4 + fac5 + fac6 + fac7 + fac8 fit=cph(formula, data=temp, x=T, y=T) validate(fit, method="boot", B=9999, bw=F, type="residual", sls=0.05, aics=0,force=NULL, estimates=TRUE, pr=FALSE) out=bootcov(fit, B=9999, pr=F, coef.reps=T, loglik=F) for (i in 1:8) { print(quantile(out$b...

...function in the rms package, which at least generates confidence intervals similar to what is observed in SPSS. However, the p-values associated with each predictor in the model are not really close in many instances. Here is the code I am using: formula=Surv(months, recidivate) ~ fac1 + fac2 + fac3 + fac4 + fac5 + fac6 + fac7 + fac8 fit=cph(formula, data=temp, x=T, y=T) validate(fit, method="boot", B=9999, bw=F, type="residual", sls=0.05, aics=0,force=NULL, estimates=TRUE, pr=FALSE) out=bootcov(fit, B=9999, pr=F, coef.reps=T, loglik=F) for (i in 1:8) { print(quantile(out$b...

Stefan Roepcke <stefan.roepcke at metagen.de> writes: > Hey, > > hopefully there is an easy way to solve my problem. > All that i think off is lengthy and clumsy. > > Given a data.frame d with columns VALUE, FAC1, FAC2, FAC3. > Let FAC1 be something like experiment number, > so that there are exactly the same number of rows for each level of FAC1 > in the data.frame. > > Now i would like to scale all values according to the center of its > experiment. > So i can apply s <- by(d[1], FAC1, scale)...

...t) plot(b,pages=1) summary(b) Family: gaussian Link function: identity Formula: y ~ fac + s(x2, by = fac) + s(x0) Parametric coefficients: Estimate Std. Error t value Pr(>|t|) (Intercept) 1.1784 0.1985 5.937 6.59e-09 *** fac2 -1.2148 0.2807 -4.329 1.92e-05 *** fac3 2.2012 0.2436 9.034 < 2e-16 *** --- Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1 Approximate significance of smooth terms: edf Ref.df F p-value s(x2):fac1 5.364 6.472 2.285 0.0312 * s(x2):fac2 4.523...

...hat there is rounding of the p-values to zero, if the p-value is less than about 1.0e-16. Is there a way we can obtain the exact p-values from lme without rounding? used commands: library(nlme) g<-lme(value~factor(fac1)+factor(fac2)+factor(fac1):factor(fac2),data=mydataframe,random=~1|factor(fac3)) ag<-anova(g) kind regards, R. Alberts

...;, ??????? ylab = "individual estimates") boxplot(coefs ~ factor, data = temp, naxt="n",add = TRUE, ??????? boxwex = 0.25, at = 1:7 + 0.15, ??????? subset = experiment == "second", col = "green") axis(at=1:7,side=1,c("fac1","fac2","fac3","fac4","fac5","fac6","fac7")) legend(6,-0.5, c("experiment1", "experiment2"), ?????? fill = c("red", "green")) Does anyone know how I can surpress these labels for the second boxplot? Thanks in advance, Karin

...c) ## create an example data.frame test.df <- data.frame(sex = gl(2, 110, labels = c("Male", "Female")), fac1 = sample(gl(11, 100, labels = paste("V1 Level", 1:11))), fac2 = sample(gl(11, 100, labels = paste("V2 Level", 1:11))), fac3 = sample(gl(11, 100, labels = paste("V3 Level", 1:11))), fac4 = sample(gl(11, 100, labels = paste("V4 Level", 1:11))), fac5 = sample(gl(11, 100, labels = paste("V5 Level", 1:11))), fac6 = sample(gl(11, 100, labels = paste("V6 Level"...

...t) plot(b,pages=1) summary(b) Family: gaussian Link function: identity Formula: y ~ fac + s(x2, by = fac) + s(x0) Parametric coefficients: Estimate Std. Error t value Pr(>|t|) (Intercept) 1.1784 0.1985 5.937 6.59e-09 *** fac2 -1.2148 0.2807 -4.329 1.92e-05 *** fac3 2.2012 0.2436 9.034 < 2e-16 *** --- Signif. codes: 0 '***' 0.001 '**' 0.01 '*' 0.05 '.' 0.1 ' ' 1 Approximate significance of smooth terms: edf Ref.df F p-value s(x2):fac1 5.364 6.472 2.285 0.0312 * s(x2):fac2 4.523...

Hello, I am trying to understand why the FUNCTION used in several codes, won't create the object after it finishes running the code. For instance, look at the following: Number<- function(x) {MyNumberIs<-x} When I run Number(5) Everything goes well, except that if I try to call the object MyNumberIs, I won't find it. I understand that this function can assume many parameters,

Sorry for the stupid question, but is there the factorial function in R? I tried to find it using help.search('factorial') but got nothing appropriate. Many thanks, -Serge

Hello again, I was too quick before. What I was looking for was a function that constructs the design (or incidence) matrix (X in a linear model) from a factor. Uwe Ligges suggested using model.matrix and this does almost what I want, but it is first necessary to construct a data variable. It also asigns ones to all rows of the first column (because this is set to be the contrast, not really what

I''m running Xen 4.1.4 on Fedora 17. I have some CentOS 6 DomUs - an haproxy machine and some tomcat VMs. When clients send requests with: POST /ProposalInterface HTTP/1.1 Content-Type: text/xml Host: www.myhost.co.uk Content-Length: 2099 Expect: 100-continue Connection: Keep-Alive The "continuation" doesn''t happen. The POST is truncated at 1449 bytes, and the

...4 of the variables and can't test than 17 variables (see below). I am wondering if there is a work around this problem. Please help, Mahesh gam.object <- gam(YVAR~1, data=gam.data) step.object <-step.gam(gam.object, scope=list( "FAC1"=~1+FAC1, "FAC2"=~1+FAC2, "FAC3"=~1+FAC3, "NUM1"=~1+NUM1+ns(NUM1,df=2), "NUM2"=~1+NUM2+ns(NUM2,df=2), "NUM3"=~1+NUM3+ns(NUM3,df=2), "NUM4"=~1+NUM4+ns(NUM4,df=2), "NUM5"=~1+NUM5+ns(NUM5,df=2), "NUM6"=~1+NUM6+ns(NUM6,df=2), "NUM7"=~1+NUM7+ns(NUM7,df=2), &quo...