This should do the trick :
colnames(y)[1:2]=c("a","a")
y2=rbind(y[,-1], y[,-2]) #duplicating the "y" matrix so the identifiers
are only in 1 column
merged = merge(x,y2)
merged
a b d e1 e2
1 aa 1 10 100 101
2 aa 2 20 200 201
3 ab 1 30 100 101
4 ab 2 40 200 201
5 ba 1 50 300 301
6 ba 2 60
Or, assuming you only have 4 different elements :
mat<- matrix(rep(c(1,2,"A", "B"),each=10),10,10, byrow=F)
mat2 <- as.data.frame(mat)
mat
[,1] [,2] [,3] [,4] [,5] [,6] [,7] [,8] [,9] [,10]
[1,] "1" "2" "A" "B" "1" "2" "A" "B" "1" "2"
[2,] "1" "2" "A" "B" "1" "2" "A" "B"
r(j in which(cols)){
if(j %% 2){
y[,j]=1
}else{y[,j]=2}
}
for(j in which(!cols)){
if(j %% 2){
y[,j]="A"
}else{y[,j]="B"}
}
if you really wants to use this code.
Cheers,
Eloi
> Cheers,
> Claudia
> On Mon, Jul 30, 2012 at 5:38
if(surviv<=S){
> stay=runif(1,0,1)
> if(observable==1){
>if(stay<=PsiOO){
>observable=1
> }else{observable=0}
>}else{
> return=runif(1
Plz provide a reproducible example (using dput).
Anyway, I suspect your data to be 'factor' and not numeric.
y=rbinom(100,1,0.5)
x=runif(100)
plot(x,y) #OK
plot(x,as.factor(y))
See ?factor and ?as.numeric.
Regards,
Eloi
On 12-08-01 11:41 AM, Georgiana May wrote:
> Hello,
> Anyone know why the
Hi Chris,
I personally found the documentation rather...indigestible. At least 3
years ago when I developed my R package. It seems they have nicely
improved this part of the documentation though.
I'm not an expert at all, but at least you will have the point of view
of a beginner.
First, i
for (n in LIST2) print(as.character(n))
[1] "Michael"
[1] "Dana"
[1] "Rob"
Eloi
On 12-04-24 12:22 PM, R. Michael Weylandt wrote:
Don't you just want
LIST<- as.data.frame(LIST, stringsAsFactors = FALSE) # Avoid making factors
for(n in LIST) print(n)
But I can't understand why you want to do i
hc.obj is not a dendrogram.
plot(dend1, xlab='',ylab='',sub='', nodePar=list(pch = c(1,NA), cex=0.8,
lab.cex = 0.8))
See ?plot.hclust and ?plot.dendrogram. They are not using the same
arguments.
Regards,
Eloi
On 12-04-30 06:36 AM, farscape2012 wrote:
Hi
I have a distance matrix which i
Hello,
When i generate data with the code below there appear NA as part of the
generated data, i prefer to have zero (0) instead of NA on my data.
Is there a command i can issue to replace the NA with zero (0) even if it is after generating the data?
Thank you
?Surv
time2: ending time of the
See ?sparseMatrix from package Matrix.
Eloi
On 12-05-07 02:23 PM, Lucas wrote:
> Dear R people.
> I´m facing a big problem.
> I need to create a matrix with 10.000 columns and 750.000 rows.
> matrix<- as.data.frame(matrix(data=0L, nrow=75, ncol=1)
> as you can see, the data frame has huge
You did not specify any object in the function. Thus R is building the
package "test" with all the objects present in your session when you are
calling the package.skeleton function. I suppose that one of these
objects is causing problem. I suggest you list all the
variables/function necessary
On 12-05-17 04:34 PM, Rich Shepard wrote:
hco32 <- qqmath(~ log10(HCO3 | factor(basin), data = surfchem.cast,
main = 'Bicarbonate (Log10)',
prepanel = prepanel.qqmathline,
panel = function(x, ...) {
panel.qqmathline(x, ...)
panel.qqmath(x, ...)
})
Missing a closing parenthesis af
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