Hi,
I am a beginner in R. I have a query as below:
Is it possible to develop a Windows based statistical software
(user-friendly) like SPSS using R as a programming language?
Otherwise, is it possible to use R code directly (no command-line
execution) in Windows based programming language such a
Hi,
In the decomposition of the dwt When I generate the out their levels goes
of the 0 to 15 in the decompositions And i like to known how i do to
visualise In the out the most concern levels for me
for exemple levels 7 to 14. I like to can say what levels I want visualise.
Is it possible in t
My question is how to use R to program google search.
I found this information:
"The SOAP Search API was created for developers and researchers
interested in using Google Search as a resource in their
applications." Unfortunately google no longer supports that. They
are supporting the AJAX Search
On Sep 4, 2010, at 1:20 AM, Dennis Murphy wrote:
Hi:
Do you mean something like
RSiteSearch('loess predict')
[opens up a web page with 53 matches to the request] or
package(sos)
findFn('multiple imputation')
? If not, could you be more specific about what you're after?
One further op
Hi:
Do you mean something like
> RSiteSearch('loess predict')
[opens up a web page with 53 matches to the request] or
package(sos)
findFn('multiple imputation')
? If not, could you be more specific about what you're after?
HTH,
Dennis
On Fri, Sep 3, 2010 at 2:23 PM, Waverley @ Palo Alto <
Hi, all
I have a huge object that use almost all of available memory.
R> rm(a_huge_object)
R> gc()
doesn't free memory and ?gc doesn't show anything.
Are there any suggestion?
Thanks in advance,
Regards,
Hyunchul
[[alternative HTML version deleted]]
Hi:
Here's a slightly different approach, using both mapply() in base (a la
David) and mlply() in plyr.
My thought was to put the arguments together into a three column data frame;
the names were chosen to correspond to the first three arguments of runif():
params <- data.frame(n = as.vector(rat
Thanks for the input
Adding "print" took care of the first problem. The output looks like what I
would expect, so I think the code is doing what I would like it to for the
first 44 observations.
> print(results.df)
DR D.1 R.1 V1V2dif V1V4dif
1 68.92500 75.00
Well done! I was totally misled and trapped by the error message.
Acturally the absence of "1.00" in the upper panel is an implicit
indication that error occurred there.
On 2010-9-3 13:02, Shi, Tao wrote:
Hi Dejian,
Thanks for the reply!
I finally found the problem. It is actually in the "
first of all if you are 'sourcing' the file, put explicit print calls
on your data frame; e.g., print(dataframe). Just because your input
is 2149 lines long does not mean you results will necessarily be that
long. You are using 'tapply' which will be aggregating your data.
You need to provide som
On 03/09/2010 4:28 PM, Horace Tso wrote:
Folks, the rgl 0.91 binary for windows seems corrputed. WinZip complains the
downloaded zip file is not a valid archive. I had no luck with R-forge either.
Could someone point me to the latest production ready binary?
Why would you use WinZip? Just i
Hi all,
I have a short R code file that I am using to perform calculations on a
dataset. I am having a few issues with output:
1. Although my input data file is 2149 lines long, when I type "results.df"
from the command line, I get the appropriate calculation results for only
the first 46 rows.
It occurred to me after my initial post that you will need a little more
"fixing" of your axis labels if you have data that is not trivial integers
as in your example. Consider the following solution for some irrational
random numbers:
dat=rnorm(20)
dat1=dat-min(dat)
barplot(dat1,axes
First, thanks very much for making rmate available. I'm having trouble porting
it to e-texteditor for windows, though, (specifically the send-line command)
because osascript is mac-only. Is there a way to send selections in
e-texteditor to the windows version of R.app? Thanks!
Dan
_
Hi,
Can someone help as how to use R to program google search in the R
code? I know that other languages can allow or have the google search
API
If someone can give me some links or sample code I would greatly appreciate.
Thanks.
--
Waverley @ Palo Alto
__
Hi,
I think you just need to add the format = argument. Does this help?
> x <- factor("01-11-2007")
> as.character(x)
[1] "01-11-2007"
> as.Date(as.character(x), format = "%d-%m-%Y")
[1] "2007-11-01"
Cheers,
Josh
On Fri, Sep 3, 2010 at 2:11 PM, André de Boer wrote:
>
> Hello,
>
> I have a da
Hello,
I have a dataframe with data such as:
> dat$BEGINDATUM[3]
[1] 13-09-2007
> dat$BEGINDATUM[4]
[1] 01-11-2007
> class(dat$BEGINDATUM[3])
[1] "factor"
Now I need to make calculation with these dates.
But I get these result:
> as.date(as.character(dat$BEGINDATUM[3]))
[1]
> as.date(as.charact
Thank you very much for your help, Greg!
Here's my ludicrous vision/attempt/whatever :o)
### Idea for making a density() with confidence interval ###
xx <- faithful$eruptions
xx.hist <- hist(xx, breaks="FD", freq=F)
# plot(xx.hist$mids, xx.hist$density)
# gives a rough "plot(density(xx))" out
Folks, the rgl 0.91 binary for windows seems corrputed. WinZip complains the
downloaded zip file is not a valid archive. I had no luck with R-forge either.
Could someone point me to the latest production ready binary?
Thks.
H
[[alternative HTML version deleted]]
__
I believe that you will have to draw them in ggplot2 as someone mentioned or in
base graphics. Here is a rough first attempt that may give you some ideas.
xx <- -3:4
yy <- rep(-3,length(xx))
plot(xx,xx, type="n", xlim=c(-4, 5))
rect(xx, yy, xx+.5, xx )
==
Here is a simple approach that uses bootstrapping (this could probably be
improved by using better bootstrap estimates and not ignoring the dependence
between points):
xx <- faithful$eruptions
fit1 <- density(xx)
fit2 <- replicate(1, { x <- sample(xx, replace=TRUE);
density(x, fro
The package or wavelets in general? What do you want to do?
2010/9/3 Marize Simões :
>
> Hi user's
>
> Does anybody work with wavelets on R?
> Please I need some help.
>
> Atte
> Marize Simões
> --
> View this message in context:
> http://r.789695.n4.nabble.com/Package-wavelets-tp2526023
Hi user's
Does anybody work with wavelets on R?
Please I need some help.
Atte
Marize Simões
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_
As is often the case in R, just because you shouldn't do something, doesn't
mean you can't do it. Still, I'd urge you to consider the "visual honesty"
of what you propose.
If you're still insistent:
dat <- (-3:4)
dat1=dat-min(dat)
barplot(dat1,axes=FALSE)
axis(2,dat1,labels=dat)
Typically, ze
You installed the package but probably forgot to load the library.
This works fine for me.
=
library(reldist)
x<-c(541, 1463, 2445, 3438, 4437, 5401, 6392, 8304, 11904, 22261)
G<-gini(x)
===
If A is a squared matrix, solve(A) gives the inverse of A; if you have a
system of linear equation AX=B, solve(A,B) gives the solution to this system
of equations. For example:
x-2y =1
-2x+3y=-3
> A=matrix(c(1,-2,-2,3), ncol=2, byrow=T)
> B=c(1,-3)
>
> # to get the inverse of A
> solve(A)
On 03-Sep-10 17:12:55, Zoppoli, Gabriele (NIH/NCI) [G] wrote:
> Dear r-help mailing list,
>
> this seems stupid, but I actually don't find the solution:
>
> if I have a vector of numbers x of length n, ranging, say,
> from -3 to 4, if I do
>
> barplot (x)
>
> all the values below 0 go downwards
On Sep 3, 2010, at 1:12 PM, Zoppoli, Gabriele (NIH/NCI) [G] wrote:
Dear r-help mailing list,
this seems stupid, but I actually don't find the solution:
if I have a vector of numbers x of length n, ranging, say, from -3
to 4, if I do
barplot (x)
all the values below 0 go downwards, and al
Dear Gabriele,
I suspect the reason you are having difficulty finding the solution is
because barplots were meant to be anchored at 0.
What information are you really trying to convey? There is probably a
very clear, aesthetically pleasing way to achieve your goal without a
barplot.
For instanc
In the bar plot, the vertical axis is a numerical axis representing the
frequency (the height of the vertival bar -= frequency). If you really want
to have vertical bar corresponding to the negative values go downward, you
need to make your own function to achieve the goal.
--
View this message i
One can write an R function to produce a kernel density curve with a
confidence band. See, for example, the steps of doing this in a technical
report at
http://fmwww.bc.edu/repec/usug2003/bsciker.pdf
--
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?system #look at the wait parameter
?.First.lib
I only checked the above for windows, you did not specify what platform you are
using but referring to .exe implies windows.
--
Gregory (Greg) L. Snow Ph.D.
Statistical Data Center
Intermountain Healthcare
greg.s...@imail.org
801.408.8111
>
Dear r-help mailing list,
this seems stupid, but I actually don't find the solution:
if I have a vector of numbers x of length n, ranging, say, from -3 to 4, if I do
barplot (x)
all the values below 0 go downwards, and all the positive values go upward. How
can I make them all begin from the m
Bert,
Your null is still simple (it is your alternative that does not fit the
standard), for any test statistic you just need to estimate the distribution of
it under the null. If you cannot do this theoretically it is easy enough to
simulate. Then all the theory around p-values still holds.
Hi Peng,
I did that i installed the package RELDIST, but nothing happened. R does not
recognize this function.
Still looking for the solution.
Thanks,
Marcio
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you need install and load package {reldist} before you call function gini().
HTH.
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R
?persp
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https://stat.ethz.ch/mailman/listinfo/r-help
Luis Felipe Parra wrote:
>
> Hello I have and 11 by 2 by 1200 matrix from which I would like to make a
> surface graph.
Check the example in the docs of surface3d or persp3d from the rgl package.
Note that attachments cannot be sent to the list (pdf and a few others are
the exception)
Dieter
Some thoughts on your question about having pairs increase the size of the
device:
The pairs function is fairly old (and plot even older) and fully adjustable
graphics devices are actually fairly recent. I remember a time when the only
way to get high quality color plots was to send the plot t
Hi Dimitris,
I have already seen your code in another post. But, I would like to weight
my data. So, I wish I could use the following command:
gini(x, weights=rep(1,length=length(x)))
Thanks anyway and I am trying to understand your gini function in order to
apply a weigth.
Marcio
--
View this
for the Gini coefficient you can use this function:
gini <- function(x, unbiased = TRUE, na.rm = FALSE){
if (!is.numeric(x)) {
warning("'x' is not numeric; returning NA")
return(as.numeric(NA))
}
if (any(na.ind <- is.na(x))) {
if (!na.rm)
stop("'x'
On 2010-09-03 2:05, Niels Richard Hansen wrote:
In the Value section of the 'seq' help page it says
'seq_along' and 'seq_length' always return an integer vector.
I believe it should be
'seq_along' and 'seq_len' always return an integer vector.
as there are no seq_length functi
Hello I have and 11 by 2 by 1200 matrix from which I would like to make a
surface graph. the first element of the first column represents the date,
which means I've got 1200 dates. I would like to graph the 11 elements of
the second column for each date. Does anybody know how can I do this?
attache
Hi listers,
Does it necessary to install any package in order to use the GINI or INEQ
functions.
If I use the following command the R tells me that didn't find the GINI
function.
x<-c(541, 1463, 2445, 3438, 4437, 5401, 6392, 8304, 11904, 22261)
G<-gini(x)
Thanks in advance,
Marcio
--
View this
Hello R users & R friends,
I just want to ask you if density() can produce a confidence interval,
indicating how "certain" the density() line follows the true frequency
distribution based on the sample you feed into density().
I've heard of loess.predict(loess(y ~ x), se=TRUE) which gives yo
Hi All,
I am using vglm & the fitted values is NA. As in the R documentation that
for Pareto1, if the estimate of k is less than or equal to unity then the
fitted values will be NA.
what is NA means ?
how to solve it ?
how to get the k estimate ? (in the R document the estimate of alpha is
f.
Sarah,
Here's a version using two nested loops:
rate_number = matrix(c(5, 15, 60, 15, 5, 0, 20, 60, 20,0, 10, 20, 40,
20, 10), nrow = 5, ncol = 3)
range_mat = matrix(c(6.25, 6.75, 7.25, 8.75, 9.25, 9.75, 8.5, 9, 9.5,
10.5, 11, 11.5, 4.25, 4.75, 5.25, 5.75, 6.25, 6.75), nrow = 6, ncol = 3)
set.s
On Sep 3, 2010, at 8:39 AM, raje...@cse.iitm.ac.in wrote:
I have a socket connection where I do
-socketConnection
-writeLines
and then wait for the server to send data through the socket. so I
have to wait on a readLines.But when I say
str<-readLines(con) it executes immediately and str gets
Ah, the plot thickens! The p-value imbroglio again.
I won't comment except to note that all of this so far assumes a
simple null. What if you have a composite null? --e.g. My null is that
the data are drawn from a normal with unknown mean and variance versus
they are drawn from mixture of 2 normal
> My main doubts are:
> 1.Is the use of "by" and "te" right with the negative binomial
> distribution and with the binomial distribution?
-- yes! These things specify the `linear predictor' of the model --- the
correctness of the linear predictor does not usually depend on the
response di
Thanks
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Ted,
I agree that we are measuring discrepancies and that large discrepancies
correspond to p-values near 0 and small discrepancies correspond to large
p-values. But interpreting discrepancies on a p-value scale leads more to
confusion than understanding. If you are interested in the discrepa
On 3 September 2010 12:03, zbynek.jano...@gmail.com
wrote:
>
> I am having problem using output of lm() function for further analysing using
> systemfit package.
>
> Basicaly, the problem s following - I generate several formulas using lm()
>> fo1 <- lm(r98[,2] ~ f98[,1] + f98[,2] + ... + f98[,43]
Hi,
I have a socket connection where I do
-socketConnection
-writeLines
and then wait for the server to send data through the socket. so I have to wait
on a readLines.But when I say
str<-readLines(con) it executes immediately and str gets nothing.what do i do
for this?
[[alternative HT
On Sep 3, 2010, at 6:32 AM, Sarah Sanchez wrote:
Dear R helpers
I have following dataset
rate_number = matrix(c(5, 15, 60, 15, 5, 0, 20, 60, 20,0, 10, 20,
40, 20, 10), nrow = 5, ncol = 3)
range_mat = matrix(c(6.25, 6.75, 7.25, 8.75, 9.25, 9.75, 8.5, 9,
9.5, 10.5, 11, 11.5, 4.25, 4.7
Hello R users,
I am working with the GAM to inspect the effect of some factors (year, area)
and continuous variables (length, depth, latitude and longitude) on the
intensity and prevalence of the common parasite Anisakis. I would like
introduce interaction in my models, both "continuous variabl
After the R CMD check is done and an R CMD INSTALL is done, where does the
external dll have to be?..what can i do to make it move around with the
package? can I zip it with the package? if so in what folder?
- Original Message -
From: Duncan Murdoch
To: raje...@cse.iitm.ac.in
Cc: r-he
Try this:
> result <- list() # holds the numbers for each column
> for (i in seq(ncol(rate_number))){
+ ans <- NULL
+ for (j in seq(nrow(rate_number))){
+ ans <- c(ans, runif(rate_number[j, i], range_mat[j, i],
range_mat[j + 1, i]))
+ }
+ result[[i]] <- ans
+ }
> result
[[
On Sep 3, 2010, at 9:33 AM, Hyunchul Kim wrote:
Hi, all
is there a built-in function to compare two numbers?
something like following function
cmp <- function(x, y){
value <- 0
if (x > y){
value <- 1
}else if (x == y){
value <- 0
}else {
value <- -1
}
r
> x <- 2
> y <- 3
> ifelse(x > y, 1, ifelse(x < y, -1, 0))
[1] -1
On Fri, Sep 3, 2010 at 9:33 AM, Hyunchul Kim wrote:
> Hi, all
>
> is there a built-in function to compare two numbers?
>
> something like following function
>
> cmp <- function(x, y){
> value <- 0
> if (x > y){
> valu
You could potentially use sign()
> sign(3 - 5)
[1] -1
> sign(5 - 3)
[1] 1
> sign(5 - 5)
[1] 0
But... this could fail when you think two numbers are equal and the
computer doesn't, due to floating point precision. (Your version could
fail in exactly the same way.)
> x <- .3 - .2
> x
[1] 0.1
> sig
Try this:
sapply(dat, '[', 1:max(sapply(dat, length)))
On Fri, Sep 3, 2010 at 10:31 AM, Ron Michael wrote:
> Thanks Henrique and Marc for your reply, both worked.
>
> However I need one more suggestion on from list to matrix
>
> dat <- vector("list", 3)
> dat[[1]] <- rnorm(5)
> dat[[2]] <- rnor
On Sep 3, 2010, at 9:08 AM, syrvn wrote:
Hi,
the R code:
a <-
matrix
(c(1,5,4,3,7,10,34,4,3,8,6,5,12,17,45,3,2,45,46,47,3,4,22,12,21),
nrow=5)
rownames(a) <- c("a","b","c","d","e")
a
a[which(a[,3] < 8), ]
a[which(a[,3] < 6), ]
It wasn't exactly clear and possible that you wanted only r
Hi!
Maybe something like this:
x <- 2
y <- 3
#since FALSE will be converted to 0 and TRUE to 1 you can do
as.numeric(x>y)
as.numeric(x Hi, all
>
> is there a built-in function to compare two numbers?
>
> something like following function
>
> cmp<- function(x, y){
> value<- 0
> if (x>
Dear R helpers
I have following dataset
rate_number = matrix(c(5, 15, 60, 15, 5, 0, 20, 60, 20,0, 10, 20, 40, 20, 10),
nrow = 5, ncol = 3)
range_mat = matrix(c(6.25, 6.75, 7.25, 8.75, 9.25, 9.75, 8.5, 9, 9.5, 10.5, 11,
11.5, 4.25, 4.75, 5.25, 5.75, 6.25, 6.75), nrow = 6, ncol = 3)
> rate_
I am having problem using output of lm() function for further analysing using
systemfit package.
Basicaly, the problem s following - I generate several formulas using lm()
> fo1 <- lm(r98[,2] ~ f98[,1] + f98[,2] + ... + f98[,43])
> fo2 <- lm(r98[,1] ~ f98[,1] + f98[,2] + ... + f98[,43])
and than
Thanks Henrique and Marc for your reply, both worked.
However I need one more suggestion on from list to matrix
dat <- vector("list", 3)
dat[[1]] <- rnorm(5)
dat[[2]] <- rnorm(15)
dat[[3]] <- rnorm(3)
> dat
[[1]]
[1] 0.7615421 0.4151374 0.6534402 1.5999338 1.4131837
[[2]]
[1] 0.7364765 2.078
Dear r-help,
>
> I am using CAH. I would cut my dendogram. What is the command in R that
> allows draw a graph of the semi-partial R-squared ?
>
I did this:
d3=read.table("dd.txt", header=TRUE,dec=',',row.names='variable')
don<-scale(d3, center = TRUE, scale = TRUE)
#Calcul du Matrice de simi
Hi, all
is there a built-in function to compare two numbers?
something like following function
cmp <- function(x, y){
value <- 0
if (x > y){
value <- 1
}else if (x == y){
value <- 0
}else {
value <- -1
}
return(value)
}
Thanks in advance,
Hyunchu
Hi!
I don't really understand your goal, but if you want to extract the row
names,
rownames(a[which(a[,3] < 8), ])
does it.
But I think you've already tried it. Maybe it's just me...
HTH
Ivan
Le 9/3/2010 15:08, syrvn a écrit :
> Hi,
>
> the R code:
>
> a<- matrix(c(1,5,4,3,7,10,34,4,3,8,6,5,
R doesn't simply treat a row vector as a matrix.
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R-
Hi,
I do not understand the code right now but it does its job.
Thanks a lot!
Best,
syrvn
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On Sep 3, 2010, at 8:59 AM, Ravi S. Shankar wrote:
Hi David,
Thank you for your response. But that wasn't what I was looking for.
names=c("Ravi", "Raj","Shubha","Nivriti")
select.list(names) gives a pop up with the 4 names. I just wanted in
the pop up a heading (say BOYS) followed by two names
Try this:
a[a[,3] < 6,, drop = FALSE]
On Fri, Sep 3, 2010 at 10:08 AM, syrvn wrote:
>
> Hi,
>
> the R code:
>
> a <-
> matrix(c(1,5,4,3,7,10,34,4,3,8,6,5,12,17,45,3,2,45,46,47,3,4,22,12,21),
> nrow=5)
> rownames(a) <- c("a","b","c","d","e")
> a
> a[which(a[,3] < 8), ]
> a[which(a[,3] < 6), ]
>
Hi,
the R code:
a <- matrix(c(1,5,4,3,7,10,34,4,3,8,6,5,12,17,45,3,2,45,46,47,3,4,22,12,21),
nrow=5)
rownames(a) <- c("a","b","c","d","e")
a
a[which(a[,3] < 8), ]
a[which(a[,3] < 6), ]
produces the following output:
> a
[,1] [,2] [,3] [,4] [,5]
a1 10633
b5 345
Try this:
unclass(as.data.frame(mat))
On Fri, Sep 3, 2010 at 9:48 AM, Ron Michael wrote:
> Dear all, suppose I have following matrix:
>
> > mat <- matrix(rnorm(25), 5)
> >
> > mat
> [,1] [,2][,3] [,4] [,5]
> [1,] 0.97056228 -1.3278509 -0.73511792 2.16506
On Sep 3, 2010, at 7:48 AM, Ron Michael wrote:
> Dear all, suppose I have following matrix:
>
>> mat <- matrix(rnorm(25), 5)
>>
>> mat
> [,1] [,2][,3] [,4] [,5]
> [1,] 0.97056228 -1.3278509 -0.73511792 2.1650629 -0.4411997
> [2,] 0.58613700 -0.2559899 -1
On Sep 3, 2010, at 7:02 AM, Lathouri, Maria wrote:
Dear all,
I want to plot 8 chemical variables in the same plot. I have used
the interaction.plot command:
interaction.plot(speclong$Date,speclong$time2,speclong
$outcome,xaxt="n",type="l",pch=20,xlab="", log="y",
ylab="Concentration (ug
Hi David,
Thank you for your response. But that wasn't what I was looking for.
names=c("Ravi", "Raj","Shubha","Nivriti")
select.list(names) gives a pop up with the 4 names. I just wanted in
the pop up a heading (say BOYS) followed by two names and another
heading (say GIRLS) and the remaining two
Dear all, suppose I have following matrix:
> mat <- matrix(rnorm(25), 5)
>
> mat
[,1] [,2] [,3] [,4] [,5]
[1,] 0.97056228 -1.3278509 -0.73511792 2.1650629 -0.4411997
[2,] 0.58613700 -0.2559899 -1.18334248 -1.4990907 1.8138846
[3,] -1.0313 2.0227887
On Sep 3, 2010, at 3:55 AM, Ravi S. Shankar wrote:
Hi R,
I am using select.list
names=c("Ravi", "Raj","Shubha","Nivriti")
select.list(names) provides a drop down to choose one of the 4 names.
However I would like to know if it is possible to create a
classification something like this
Ben Bolker gmail.com> writes:
>
> Qiu, Weiyu ualberta.ca> writes:
>
> >
> > Hi,
> >
> > I'm doing a generalized linear mixed model, and I currently use an
> > R function called "glmm".
>
> If you could get by with an AR or ARMA structure then you could
> use lme() with the 'correlation'
raje...@cse.iitm.ac.in wrote:
After the R CMD check is done and an R CMD INSTALL is done, where does the
external dll have to be?
See my message below.
..what can i do to make it move around with the package? can I zip it with the
package? if so in what folder?
R isn't designed to hand
Peng, C wrote:
Is there a complete list of these very handy and power functions in the base
R?
ls("package:base") will list all the objects exported from base, but it
doesn't rate them to let you know which ones are handy and which ones
are a waste of space. We like to think they're all use
Dear all,
I want to plot 8 chemical variables in the same plot. I have used the
interaction.plot command:
> interaction.plot(speclong$Date,speclong$time2,speclong$outcome,xaxt="n",type="l",pch=20,xlab="",
> log="y", ylab="Concentration (ug/L)",trace.label="",col=rainbow(8))
However, what it giv
Thanks to Chris and Dennis. My problem is solved although it will take
some additional time before I fully understand the solution.
Cheers
Geert De Meyer
_
Bayer BioScience N.V.
BioInformation Management
Zwijnaarde
Company Identification Number BE042252
On Thu, Sep 2, 2010 at 2:33 PM, Greg Snow wrote:
> Suggestion: use the power of R.
>
> If x and y are independent then sorting y based on x is meaningless.
>
> If sorting y based on x is meaningful, then they are not independent.
>
> Trying to force non-independent things to pretend that they are
Lewis G. Dean wrote:
>
>> post-hoc power analysis on a Wilcoxon test.
>
There is a (somewhat dated) list of "why-not" papers in
http://www.childrens-mercy.org/stats/size/posthoc.asp
Dieter
--
View this message in context:
http://r.789695.n4.nabble.com/Power-analysis-tp2524729p2525333.h
Geert,
thanks for providing a nice example. When use see "groups" in xyplot, you
should switch to the documentation (and use) panel.superpose. Which has a
somewhat different philosophy (your looks more like ggplot2 would do it).
The docs of panel.superpose say (tersely...)
col: graphical param
Hi:
The 'easy' solution is to define the colors corresponding to the genotypes
directly in lattice:
xyplot( result ~ time | location, data=dataset, groups=genotype, pch=19,
type="b", col = c('red', 'green', 'blue'))
The problem with trying to use color as a variable to 'match' to genotype is
tha
Hi Geert
###
dataset <- data.frame ( time = rep(1:5,times=9),
genotype = factor(rep(rep
(c("A","B","C"),each=5),times=3 )),
location= factor(rep (paste("LOC",1:3),each=15)),
color = rep (rep
(c("red","green","blue"),each=5),times=3
),
In the Value section of the 'seq' help page it says
'seq_along' and 'seq_length' always return an integer vector.
I believe it should be
'seq_along' and 'seq_len' always return an integer vector.
as there are no seq_length function?
Best, Niels
--
Niels Richard Hansen
> "o" == omerle
> on Thu, 2 Sep 2010 10:30:49 +0200 (CEST) writes:
o> Hi,
o> I didn't find any post on this subject so I ll ask you some advices.
o> Let's say that I have two library trees.
o> Number 1 is the default R library tree on path1
o> Number 2 is anoth
Hi R,
I am using select.list
names=c("Ravi", "Raj","Shubha","Nivriti")
select.list(names) provides a drop down to choose one of the 4 names.
However I would like to know if it is possible to create a
classification something like this
select.list(names) should give
Boys
Ravi
> "NM" == Norm Matloff
> on Thu, 2 Sep 2010 12:20:44 -0700 writes:
NM> Tal wrote:
>> A friend recently brought to my attention that vector assignment actually
>> recreates the entire vector on which the assignment is performed.
NM> ...
NM> I brought this up in r-
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