similar to: significance testing for the difference in the ratio of means

Saludos, tengo un problema que no puedo resolver dentro del R Estoy creando una funcion que a partir de un objeto compuesto por diferentes matrices como el que esta a continuacion: [[1]] M1 M2 M3 M4 sp1 2 0 1 8 sp2 4 5 2 4 sp3 0 0 4 0 sp4 5 7 5 0 sp5 0 4 0 0 [[2]] M3 M2 M4 M1 sp1 1 0 8 2 sp2 2 5 4 4 sp3 4 0 0 0 sp4 5 7 0 5 sp5 0 4 0 0 [[3]]

Dear Bert, thank you for your response. here it is the piece of R code : given 3 data frames below --- N <- data.frame(N=c("n1","n2","n3","n4")) M <- data.frame(M=c("m1","m2","m3","m4","m5")) C <- data.frame(n=c("n1","n2","n3"),

Hi, May be this helps: m1<- as.matrix(read.table(text=" y1 g24 y1 0 1 g24 1 0 ",sep="",header=TRUE)) m2<-as.matrix(read.table(text="y1 c1 c2 l17 ?y1 0 1 1 1 ?c1 1 0 1 1 ?c2 1 1 0 1 ?l17 1 1 1 0",sep="",header=TRUE)) m3<- as.matrix(read.table(text="y1 h4??? s2???? s30 ?y1 0 1 1 1 ?h4 1 0 1 1 ?s2 1 1 0 1 ?s30 1 1 1

Hi Bogdan, Kinda messy, but: N <- data.frame(N=c("n1","n2","n3","n4")) M <- data.frame(M=c("m1","m2","m3","m4","m5")) C <- data.frame(n=c("n1","n2","n3"), m=c("m1","m1","m3"), I=c(100,300,400))

Hi, I have a data-matrix: > PID sato hrs fim health 214 3 4.376430 6.582958 5 193 6 4.361825 3.138525 6 8441 6 4.205771 3.835886 7 7525 6 4.284489 3.245139 6 6806 7 4.168926 2.821833 7 5682 7 1.788707 1.212653 7 5225 6 1.651463 1.436980 7 4845 6 1.692710 1.267359 4 4552 5 1.686448 1.220539 6

> On Jun 6, 2017, at 4:01 AM, Jim Lemon <drjimlemon at gmail.com> wrote: > > Hi Bogdan, > Kinda messy, but: > > N <- data.frame(N=c("n1","n2","n3","n4")) > M <- data.frame(M=c("m1","m2","m3","m4","m5")) > C <-

Thank you David. Using xtabs operation simplifies the code very much, many thanks ;) On Tue, Jun 6, 2017 at 7:44 AM, David Winsemius <dwinsemius at comcast.net> wrote: > > > On Jun 6, 2017, at 4:01 AM, Jim Lemon <drjimlemon at gmail.com> wrote: > > > > Hi Bogdan, > > Kinda messy, but: > > > > N <-

Hello all, Probably my concepts about the data.frame and matrix and array in R are not clear, I need some clarification to help me understand them better. >M <- read.table("test1.csv",sep=",",row.names=NULL,header=T) gives me: M as M1 M2 M3 M4 M5 M6 M7 M8 M9 M10 1 9 11 14 15 18 20 20 20 20 20 2 3 4 8 9 11 12 14 15 15 15 3 4 5 8 8 9 9 9 9 9 9 4 4

Simple matrix indexing suffices without any fancier functionality. ## First convert M and N to character vectors -- which they should have been in the first place! M <- sort(as.character(M[,1])) N <- sort(as.character(N[,1])) ## This could be a one-liner, but I'll split it up for clarity. res <-matrix(NA, length(M),length(N),dimnames = list(M,N)) res[as.matrix(C[,2:1])] <-

Dear all, please could you advise on the R code I could use in order to do the following operation : a. -- I have 2 lists of "genome coordinates" : a list is composed by numbers that represent genome coordinates; let's say list N : n1 n2 n3 n4 and a list M: m1 m2 m3 m4 m5 2 -- and a data frame C, where for some pairs of coordinates (n,m) from the lists above, we have a

Here's another approach: N <- data.frame(N=c("n1","n2","n3","n4")) M <- data.frame(M=c("m1","m2","m3","m4","m5")) C <- data.frame(n=c("n1","n2","n3"), m=c("m1","m1","m3"), I=c(100,300,400)) # Rebuild the factors using M and N C$m <-

For a pedagogical purpose, I was trying to show how the formula for a simple regression line (~1+x) could be crossed with a factor (~1:group + x:group) to fit separate regressions by group. For example: set.seed(201108) dat <- data.frame(x=1:15, y=1:15+rnorm(15), group = sample(c('A','B'), size=15, replace=TRUE)) m1 <- lm(y~ 1 + x, data=dat)

Hello all! I have a problem that calls for a better understanding, than mine, of how lme() uses the random part of the call. The dataset consists of eleven field trials (Trial) with three replicates (Block) and four fertiliser treatments (Treat). Analysing for example yield with lme() is easy: m1 <- lme(Yield ~ Treat, data=data, random =~1| Trial/Block) giving estimates of

Two questions related to creating animated movies with rgl: 1. I've created an rgl scene with 5 different views I want to display in a movie, but I'd like to insert pauses (say, 5 seconds) at each view. How can I do this? I first created 5 userMatrix's, then play3d( par3dinterp( userMatrix=list(M1, M2, M3, M4, M5)), ,duration=2*60/5) ) then tried simply repeating each twice,

I have a big database which more classes in class variable. I want to make each class to one dataset, for example: x1 x2 x3 x4 class a b a c c M1 c b b c c M4 c c a c c M2 c a c a a M2 c c a a a M1 c a b c a M3 c c a b c M3 c a c a b M2 c c a b a M1 How can I make, like: x1 x2 x3 x4 class a b a c c M1 c c a a a M1 c

Reproducible example, please. -- In particular, what exactly does C look ilike? (You should know this by now). -- 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 Mon, Jun 5, 2017 at 6:45 PM, Bogdan Tanasa <tanasa at gmail.com>

Hi All I can specify whatever inits, it has no effect on the estimation. I am replicating a textbook example. The result is completely trash, having estimates of -58.7 (sd=59.3), where it should be closer to an ml estimate of 0.585 (SE=0.063). The two chains within one run are different, but with different inits for different runs, I get exactly the same chains, and I mean exactly. If I set

Hello I have 4 1000*1000 matrix A,B,C,D. I want to use the corresponding element of the 4 matrices. Using the "for loop" as follow: E<-o for (i in 1:1000) {for (j in 1:1000) { E<-fisher.test(matrix(c(A[i][j],B[i][j],C[i][j],D[i][j]),2))#call fisher.test for every element } } It is so time-consuming Need vectorization Yours sincerely ZhaoXing Department of

On Tuesday 05 August 2008 12:39, Dan Gohman wrote: > On Tue, August 5, 2008 8:32 am, David Greene wrote: > > On Monday 04 August 2008 17:56, Dan Gohman wrote: > >> The applymask approach leverages use-def information rather than > >> what can be thought of as duplicating a subset of it, making the IR > > > > I don't understand what you mean by

Kevin Wright wrote: > library(nlme) > m2 <- gnls(conc ~ t1*(1-t2*exp(-k*time)), > data = df.Chloride, > start = list( > t1 = 35, > t2 = 0.91, > k = 0.22)) So my error was to use nls instead that gnls. Thanks a lot, Kevin. > summary(m2) > plot(m2) > lag.plot(resid(m2), do.lines=FALSE) >