the p-value of 0.004332?
Sorry for these questions.
Best wishes,
Amor
--- Peter Dalgaard schrieb am Fr, 13.11.2009:
Von: Peter Dalgaard
Betreff: Re: [R] p-value > 1
An: "amor Gandhi"
CC: r-h...@stat.math.ethz.ch
Datum: Freitag, 13. November 2009, 15:24
amor Gandhi wrot
Dear all,
I am trying to use SAMr-library(samr), it gives me p-value = 1.001, any idea
why?
Many thanks,
Amor
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R-help@r-projec
Dear all,
Is there any bayesian t-test in R?
Many thanks,
Gandhi
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https://s
Hi all,
I would like to write the following latex code in a plot x=\sqrt[4]{x[0]} using
x0 <- rgamma(1000,2,1)
y0 <- rgamma(100,2,1)
x <- x0^0.25
y <- y0^0.25
plot(x, y,xlab=expression(x=sqrt[4]{x[0]}),ylab=expression(y=sqrt[4]{y[0]}))
Many thanks,
Amor
[[alternative HTML vers
Hi,
Can you please tell me what does the the function logged2 in R do im list or..?
As I have
> ?logged2
No documentation for 'logged2' in specified packages and libraries:
you could try '??logged2'
> ??logged2
No help files found with alias or concept or title matching âlogged2â
using fuzzy
Hi everyone,
Is there any counter function in R for the following purpose:
x <- matrix(c(1,1,0,2,1,0,0,2,0,1,2,1,2,1,0,1),nrow=4)
As I would like to know how many zeros, ones, and twos in each row of x?
Many thank in advance,
Amor
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Hello,
I am using
mod1 <- lrm(y~x1+x2,na.action=na.pass,method="lrm.fit")
summary(mod1)
and I've got the following error:
Error in summary.Design(mod1) : adjustment values not defined here or with
datadist for x1 x2
Many thank,
Amor
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Many thanks
--- milton ruser schrieb am Mo, 8.6.2009:
Von: milton ruser
Betreff: Re: [R] mean
An: "amor Gandhi"
CC: r-h...@stat.math.ethz.ch
Datum: Montag, 8. Juni 2009, 14:27
oops.
If x1 is the individual try
x2.mean<-aggregate(data[2:2], list(x1), mean)
x2.mean
you can
Hi,
I have gote the following data
x1 <- c(rep(1,6),rep(4,7),rep(6,10))
x2 <- rnorm(length(x1),6,1)
data <- data.frame(x1,x2)
and I would like to compute the mean of the x2 for each individual of x1, i. e.
x1=1,4 and 6?
Thank you very much in advance,
Amori
[[alternative HTML
Hi,
I have the following data and I would like to delete douple names, it is almost
similar to SAS PROC SORT nodupkey! Is there any function in R does this?
x1 <- rnorm(11,5,1)
x2 <- runif(11,0,1)
nam <-paste("A", c(1:4,4,5:9,9), sep=".")
mydata <- data.frame(x1,x2)
crownames(mydata) <- nam
Thanks, I can't use the name as a variable because I need to create a matrix
which does include only numerical values!
--- Uwe Ligges schrieb am Di, 28.4.2009:
Von: Uwe Ligges
Betreff: Re: [R] duplicate 'row.names' are not allowed
An: "amor Gandhi"
CC: r-h...@stat
: [R] duplicate 'row.names' are not allowed
An: "amor Gandhi"
CC: r-h...@stat.math.ethz.ch, "S Ellison"
Datum: Dienstag, 28. April 2009, 14:52
amor Gandhi wrote:
> Thank you for your reply :)! Is it possible to solve the problem by using
> A.4, A.4a, A.9
.ch, "amor Gandhi"
Datum: Dienstag, 28. April 2009, 14:17
You have used A.4 and A.9 twice.
Look at
> nam
# or
> duplicated(nam)
>>> amor Gandhi 28/04/2009 13:04:03 >>>
Hi everyone,
*á
I have got the following problem:
*á
x1 <- rnorm(10,5,1)
x2 <- runif(10
Hi everyone,
Â
I have got the following problem:
Â
x1 <- rnorm(10,5,1)
x2 <- runif(10,0,1)
nam1 <- paste("A",1:4,sep=".")
nam2 <- paste("A",6:9,sep=".")
nam <- c(nam1,"A.4",nam2,"A.9")
mydata <- data.frame(x1,x2)
rownames(mydata) <- nam
Error in `row.names<-.data.frame`(`*tmp*`, value = c("A.1",
Hello Richie,
I would like to do three (or k) swap steps in each step just 2 ID recursive
swaping
x <- 1:10
swap <- function(x){
 a <- sample(x,2)
 x[x==a[1]] <- swap[2]
 x[x==a[2]] <- swap[1]
 return(x)
 }
 swap(swap(swap(x))) -> mix
Â
Is this possible?
Thanks you in advance!
Amor
Hello everybody,
Â
I wonder if there is any swap function in R that does the following:
x <- seq(1,10,12)
x1 <- swap(x)
x1
1 8 3 4 5 6 7 2 10
Thank you very much!
Â
Amor
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Hello,
I wrote
> setwd("D:/")
> dd <- read.table(file="test.txt",header=TRUE)
> attach(dd)
> boxplot(x)
> outlier <- function(y){
+ out <- boxplot(y, range = 1)$out
+ outliers <- which(y == out)
+ dev.off()
+ return(out,outliers)
+ }
> outlier(x)
$out
[1] 1.950208 2.082025 4.7
Hello,
I wrote
> setwd("D:/")
> dd <- read.table(file="test.txt",header=TRUE)
> attach(dd)
> boxplot(x)
> outlier <- function(y){
+ out <- boxplot(y, range = 1)$out
+ outliers <- which(y == out)
+ dev.off()
+ return(out,outliers)
+ }
> outlier(x)
$out
[1] 1.950208 2.082025 4.7
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