Rolf Turner wrote:
On 21/11/2008, at 10:13 AM, Steffy, Elizabeth A. wrote:
I got this error for this equation and i'm not sure what it means or
how to fix it:
Error in S[index] = S[index - 1] + (dSi - dSo - SC) * dt :
nothing to replace with
Does anyone know how to fix this?
No.
PL
Hi
I am trying to figure out, if the matlab style of linear indexing of
an array is the same as in R. i.e.
when
x <- array( 1:24, dim=c(2,3,4) )
x[3]
> 3
and if the same is true in matlab, assuming that
x[n1,n2,n3] in R returns the same as y(n1,n2,n3) when y is a matrix in matlab
I found the
Dear list,
I have following list
[[1]]
Pnr timeCA CACen
1 62083014541 0.008 TRUE
2 62083014542 0.008 TRUE
3 62083014543 0.008 FALSE
4 62083014544 0.013 TRUE
5 62083014545 0.007 FALSE
[[2]]
Pnr timeCA CACen
1 6403147116
>> Error in S[index] = S[index - 1] + (dSi - dSo - SC) * dt :
>> nothing to replace with
>>
Peter Dalgaard wrote:
> ...that her R version is from before September 2008.
Just curious which item in NEWS this comment refers to. Something changed
in [] ?
Dieter
--
View this message in context
Hi Dimitris,
Appreciate for your reply with detailed information, many thanks!
I realize that generating random number won't be so simple more than I
expected, but got some hints from the advice. I am actually hoping to do a
parametric bootstrap likelihood test, because this is the way of testing
Hello,
Sorry to ask again something with ggplot2...
I detect something, perhaps a bug (but more probably a syntax error, I'm
learning...) :
# This is working : x = wt, y = mpg, abline with intercept = 20 and slope =
1
qplot(wt, mpg, data = mtcars) + geom_abline(intercept = 20, slope = 1)
# This
Dear list,
thanks to your help I managed to find means of analysing my data.
However, the whole data set contains 264 variables. Of which some are
factors, others are not. The factors tend to be grouped, e.g.
data$f1304 to data$f1484 and data$f3204 to data$5408.
But there are other types of va
Dieter Menne wrote:
>
>
>>> Error in S[index] = S[index - 1] + (dSi - dSo - SC) * dt :
>>> nothing to replace with
>>>
>
> Peter Dalgaard wrote:
>> ...that her R version is from before September 2008.
>
> Just curious which item in NEWS this comment refers to. Something changed
> in [] ?
No
Hello Peter,
If you want to use R for bioinformatics, you probably want a course
using Bioconductor (www.bioconductor.org). To combine with a
introduction to R, the following should be good:
http://www3.imperial.ac.uk/stathelp/courses/statisticalmicroarrayanalysisusingr
but some time to wait til
There may be some chance that a proper wavelet transform may help me with
signal features extraction.
I know R contains a number of packages implementing wavelett filters and/or
transforms.
Is there an implementation of a trous wavelet transform ?
Thank you so much,
Maura
Alice Messenger ;-) ch
after obtaining the result, the result is displayed below:
individual RS_number Phy_Posi LOH_intensity
1 1718890 rs1496555 2.2241110.8121
2 1668776 rs2376495 3.0849860.786 <---
3 1723597 rs4648462 3.1551270.784 <---
4 1728870 rs10492940 3.1876070.7831
Hi,
I was just *highly* surprised to find out that R 2.8.0 (Windows XP)
changes the timezone-interpretation during operations on time data, as
apparently the timezone attribute is lost and then, for the next
interpretation of the timezone, the system settings are used.
Here is sample code (e
Dear All,
I have a correlation matrix of size 100 x 100 and would like to extract the
diagonal matrix from it. I have used the for loop to store tha correlation
values of the diagonal matrix. Is there a 'R way' of doing this?
Thanks in advance.
Kind regards,
Ezhil
Try
?diag
Rory Winston
RBS Global Banking & Markets
Office: +44 20 7085 4476
-Original Message-
From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] On Behalf Of A Ezhil
Sent: 21 November 2008 12:28
To: r-help@r-project.org
Subject: [R] Extracting diagonal matrix
Dear All,
I have a corre
[EMAIL PROTECTED] wrote:
> Try
>
> ?diag
Or, if he really means the diagonal of a 100x100 correlation matrix,
rep(1,100)
:-)
>
> Rory Winston
> RBS Global Banking & Markets
> Office: +44 20 7085 4476
>
> -Original Message-
> From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] On Beha
ggplot2
ggplot2 is a plotting system for R, based on the grammar of graphics,
which tries to take the good parts of base and lattice graphics and
avoid bad parts. It takes care of many of the fiddly details
that make plotting a hassle (l
Alan Lue wrote:
Is there anyway to label axes in 3D plots with mathematical expressions?
In the code below, I want to replace "delta_yrsed" with what "\Delta
\widehat{yrsed}" represents in TeX, but the [xyz]lab parameters of title3d
appear to only accept character strings.
Unfortunately, th
Hi
[EMAIL PROTECTED] napsal dne 21.11.2008 11:50:52:
> Dear list,
>
> thanks to your help I managed to find means of analysing my data.
>
> However, the whole data set contains 264 variables. Of which some are
> factors, others are not. The factors tend to be grouped, e.g.
> data$f1304 to data
On Fri, 21 Nov 2008, _ wrote:
Hi all,
I have a graphic device divided in 2 areas for a plot.
Is it possible to add lines or points in the first plot after the last one
have been set, without plotting all the data again ?
See ?par, look at mfg. However, you would probably find screen() easier
Thanks to the many people on the list who provided helpful responses,
including those who Emailed me directly.
Several people have suggested that I just pick up R and give it a try.
My reluctance to do this is that I am already very familiar with my
current working method (Python + Numpy) and
Hello,
what I want to do, is, to write every element of a variable into a
separate text-file automatically:
My Variable:
> wull
[1] "Hallo Leute, wie gehts denn euch seid ihr noch alle..."
[2] "Is their anyone how can help me with..."
[3] "mann, mann, mann... das nervt aber.."
Thanks. I would like to extract all the matrix entries below or above the
diagnol. diag(x) simply gives diagonal elements.
Thanks.
Kind regards,
Ezhil
--- On Fri, 11/21/08, Peter Dalgaard <[EMAIL PROTECTED]> wrote:
> From: Peter Dalgaard <[EMAIL PROTECTED]>
> Subject: Re: [R] Extracting diagon
Hi all,
I have a graphic device divided in 2 areas for a plot.
Is it possible to add lines or points in the first plot after the last
one have been set, without plotting all the data again ?
Example
par(mfrow(2,1))
plot(1)
plot(2)
lines(c(1,2)) # should be visible in the first plot
Thanks for
See ?upper.tri and ?lower.tri.
On Fri, Nov 21, 2008 at 1:05 PM, A Ezhil <[EMAIL PROTECTED]> wrote:
> Thanks. I would like to extract all the matrix entries below or above the
> diagnol. diag(x) simply gives diagonal elements.
>
> Thanks.
>
> Kind regards,
> Ezhil
>
> --- On Fri, 11/21/08, Peter D
Well, you could read the R Data Import/Export manual.
Or you could reread the help for write and figure out what
the append argument does.
Or you could reread the help for write and notice that it
should write out your entire "variable" without requiring a loop.
Sarah
On Fri, Nov 21, 2008 at 8:39
You can also do it from first principles:
outer(1:100,1:100,`<`) * m
which generalizes nicely to >, <=, >=, !=, etc.
-s
On 11/21/08, Henrique Dallazuanna <[EMAIL PROTECTED]> wrote:
> See ?upper.tri and ?lower.tri.
__
R-help@r-project.o
pzs wrote:
>
> Several people have suggested that I just pick up R and give it a try.
> My reluctance to do this is that I am already very familiar with my
> current working method (Python + Numpy) and I worry that without a
> course I will work in a Python-centric way, which won't be optimal
Your code isn't changing the filename
Try this
for(i in seq_along(wull)) write(wull[i],
paste("C://Users//zuber//Documents//wull(",i,")",".txt",sep=""))
-Original Message-
From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED]
On Behalf Of [EMAIL PROTECTED]
Sent: Friday, November 21, 2008 7:
Hi Ezhil,
Maybe this will help. There might be an easier way to do this but here is
one solution.
tril <- function(A)
{
A <- as.matrix(A)
cmats <- matrix(rep(1,length(A)),dim(A)[1],dim(A)[2])
upper.tri(cmats,diag=T)
cmats[upper.tri(cmats)] <- 0
out <- A*cmats
return(out)
}
tril(correlation
Dear all,
I have a dataframe with multiple observations and the levels as the last
column, as in:
d <-
data.frame(A=sample(1:100,12),B=sample(1:100,12),levels=c(rep('A',4),rep('B',4),rep('C',4)))
> d
A B levels
1 77 40 A
2 14 18 A
3 56 7 A
4 46 27 A
5 63 35 B
Hi,
I have a vector of Size 7420. I wanna break down in such a way that every 20
elements of it should be as elements of an list.
Ex
EXAM1
ABC, SDF, LMN,ERF,EGC,EFG,WER,FRE,QWE,ERT,DGW,QWE,YUR,ERT,GHJ,FHH,7420
what i want is
Breakdown.list
[[1]]
ABC,SDF,.20
[[2]]
21.40
[[3]]
41
On Thu, Nov 20, 2008 at 5:38 PM, Lizz Metcalfe
<[EMAIL PROTECTED]> wrote:
> I am trying to fit a generalized linear mixed effects model with a binomial
> link function, my response data is binary, using the lme4 R package, for the
> glmer model but with the cauchit link function (CDF of Cauchy dis
> I need to run anova analyis on the group in levels against the merge data
> in the first two columns. I can manually split and join the different
> columns as in
>
> > d.t <-
> rbind(data.frame(value=d[,1],ind=d[,3]),data.frame(value=d[,2],ind=d[,3]))
>
> but I was wondering if there would be a
R uses column-major ordering of multidimensional arrays, as in
Fortran, unlike the row-major ordering of C multidimensional arrays.
This is because the numerical linear algebra code used in R is from
the Eispack, Linpack, BLAS, Lapack family of Fortran subroutine
packages. Even when implementation
On Thu, Nov 20, 2008 at 2:54 PM, Jeff Evans <[EMAIL PROTECTED]> wrote:
> I am making the switch to R and uncertain which of the several packages for
> mixed models is appropriate for my analysis. I am waiting for Pinheiro and
> Bates' book to arrive via inter-library loan, but it will be a week or
No need to use a for loop. Try something like this
dim(EXAM1) <- c(20,7420/20)
test.breakdown.list <- as.list(data.frame(EXAM1))
-Original Message-
From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED]
On Behalf Of Rajasekaramya
Sent: Friday, November 21, 2008 9:48 AM
To: r-help@r-project.o
Also something along the following lines:
> x <- 1:100
> y <- split(x, (seq(along = x) - 1) %/% 5)
HTH,
Giovanni Petris
> Date: Fri, 21 Nov 2008 07:48:09 -0800 (PST)
> From: Rajasekaramya <[EMAIL PROTECTED]>
> Sender: [EMAIL PROTECTED]
> Precedence: list
>
>
> Hi,
>
> I have a vector of Size
another way:
exam.list <- split(EXAM, floor((seq_along(EXAM) - 1) / 20))
On Fri, Nov 21, 2008 at 10:48 AM, Rajasekaramya <[EMAIL PROTECTED]> wrote:
>
> Hi,
>
> I have a vector of Size 7420. I wanna break down in such a way that every 20
> elements of it should be as elements of an list.
>
> Ex
>
Dear list -
Does anyone have any ideas / comments about why I am receiving the following
warning when I run lsoda:
1: lsoda-- at t (=r1), too much accuracy requested in: lsoda(start, times,
model, parms)
2: for precision of machine.. see tolsf (=r2) in: lsoda(start, times,
model,
Hello.
I have one Model (M3) fitted using the lme package, and I have
checked the correlation structure of within-group errors using
plot(ACF (M3,maxLag=10),alpha=0.05)
But now I am not sure how to interpret this plot for the empirical
autocorrelation function.
The problem is that I am used
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
Hi Colleen,
this error was not uncommon and is usually a sign of a numerically
problematic or wrongly implemented model. Please use package deSolve,
the successor of odesolve, that is more robust and has also a bunch
alternative solvers for difficult cases.
I tested your code with deSolve (o
Use something like sprintf if you want trailing zeros.
> x
[1] 0.78
> sprintf("%.4f", x)
[1] "0.7800"
>
On Fri, Nov 21, 2008 at 5:31 AM, Abelian <[EMAIL PROTECTED]> wrote:
> after obtaining the result, the result is displayed below:
>
> individual RS_number Phy_Posi LOH_intensity
> 1 1718
Hi,
Thank you very much for the help. c[lower.tri(c,diag=FALSE)] works fine for me.
Thanks again.
Kind regards,
Ezhil
--- On Fri, 11/21/08, Anup Menon <[EMAIL PROTECTED]> wrote:
> From: Anup Menon <[EMAIL PROTECTED]>
> Subject: Re: [R] Extracting diagonal matrix
> To: [EMAIL PROTECTED]
> Cc:
On Fri, Nov 21, 2008 at 6:12 PM, Douglas Bates <[EMAIL PROTECTED]> wrote:
> R uses column-major ordering of multidimensional arrays, as in
> Fortran, unlike the row-major ordering of C multidimensional arrays.
> This is because the numerical linear algebra code used in R is from
> the Eispack, Linp
Hi, Prof. Ripley,
Thanks for the reply. Mostly I want to capture output as it is written
to the stream. For example, I quite often do the following to view the
progress of a log file from a computationally intensive script.
1. Open a console and type:
Rscript script.R >& script.log
which di
Hi all,
I hope it's not too trivial for the list - I'm trying to concatenate
two factor arrays, and obtain the following:
> f1<-factor(c("a","a","b"))
> f1
[1] a a b
Levels: a b
> f2<-factor(c("b","b","a"))
> f2
[1] b b a
Levels: a b
> c(f1,f2)
[1] 1 1 2 2 2 1
Instead of getting:
[1] a a b b b
Hi,
When I run the following Cox proportional hazards model on the Mayo clinic's
PBC data set (given in the "survival" package), the regression coefficients
do not agree with the results presented in Table 4.6.3 (p. 195) of Fleming &
Harrington's book.
library(survival)
data(pbc)
ans.cox <-
There are a variety of reasons that a question might go unanswered for
over a day on R-help. The question may be so technical or narrow that
only one or two people may be equipped to answer it. Or it may look
like homework. Or it may lack sufficient detail one which to base an
answer. Or it
Hi,
I have a solution to concatenate two factors in one but I don't believe
it is the best one: factor(c(as.character(f1),as.character(f2)))
[1] a a b b b a
Levels: a b
You can always add a level by assigning a new vector at the level vector:
levels(f1) <- c("a","b","c")
f1
[1] a a b
Levels: a
There is a discussion in Appendix D.3 of "Modeling Survival Data" by
Thereau and Grambsch regarding the differences in the datasets
including the fact that "there was significantly more follow-up for
many patients at the time this dataset was assembled". I do not see a
material difference i
Sundar Dorai-Raj wrote:
Hi, Prof. Ripley,
Thanks for the reply. Mostly I want to capture output as it is written
to the stream. For example, I quite often do the following to view the
progress of a log file from a computationally intensive script.
1. Open a console and type:
Rscript script.
Hi David,
I did look at Appendix D.3 of T&G, but am not sure if the data set analyzed
in F&H and that attached with "survival" are different. They both have
n=418 (312 from RCT and 106 observational).
There is a major difference in the coefficient for "edema" 0.66 vs 0.86. In
any case, the po
Hi all!
I tried to perform Shapiro-Wilk test for my sample of 243 values.
> Us
[1] -10.4 -13.1 -12.2 38.1 -18.8 -13.3 -11.7 29.3 49.7 6.8 12.7 16.3
[13] 5.8 -0.7 -29.4 4.1 38.8 -1.4 8.8 15.6 32.9 -5.3 19.1 35.8
[25] 4.0 -1.5 0.6 -4.2 -10.0 -4.0 1.1 48.9 -21.0 -5.
What's the most correct way of doing the equivalent to
apply(x[,2:5],1,sum)
if x is dataframe in which the only numeric fields are
in columns 2:5 ?
(using apply returns a character vector)
Thanks
Agus
--
Dr. Agustin Lobo
Institut de Ciencies de la Terra "Jaume Almera" (CSIC)
LLuis Sole Sabaris
Given
> str(x)
'data.frame': 5284 obs. of 5 variables:
$ COD : chr "0800101001" "0800101002" "0800101003" "0800101004" ...
$ 0-4 : num 79 215 84 58 127 134 15 122 101 99 ...
$ 5-9 : num 76 180 32 56 81 106 10 112 128 96 ...
$ 10-14: num 68 145 39 46 78 81 8 92 142 107 ...
$ 15-19: n
Hello all,
I apologize if this is simple or has already been answered somewhere, but
I'm not sure what to search for although I have tried and didn't come up
with anything so.. Here's my question.
How can I interpolate list names or do I have to do it post list creation.
Since that's not v
I realize this is a little late, but I recently ended up rolling my
own Ryan's Q in R. If anyone is interested, or wishes to make
improvements, you can find it here:
http://homes.msi.ucsb.edu/~byrnes/r_files/ryans_q.r
On Oct 25, 2005, at 10:15 AM, Jarrett Byrnes wrote:
I'm using lm to run
Ben Zuckerberg cornell.edu> writes:
> I have several sets of oscillation data and would like to estimate the
> parameters of a sine function to each set (and hopefully automate
> this).
There is an example using lme (yes, LINEAR) fit on page 239 of
Pinheiro/Bates Mixed Effects Book (ovarian
Agustin Lobo-4 wrote:
>
> Given
> > str(x)
> 'data.frame': 5284 obs. of 5 variables:
> $ COD : chr "0800101001" "0800101002" "0800101003" "0800101004" ...
> $ 0-4 : num 79 215 84 58 127 134 15 122 101 99 ...
> $ 5-9 : num 76 180 32 56 81 106 10 112 128 96 ...
> $ 10-14: num 68
Agustin Lobo-4 wrote:
>
> What's the most correct way of doing the equivalent to
> apply(x[,2:5],1,sum)
>
> if x is dataframe in which the only numeric fields are
> in columns 2:5 ?
> (using apply returns a character vector)
>
Could it be that you meant apply(x[,2:5],2,sum)? Not very easy to
Dear Jorge,
Thanks a lot.
Regards
Anup
On Thu, Nov 20, 2008 at 10:19 PM, Jorge Ivan Velez <[EMAIL PROTECTED]
> wrote:
>
> Dear Anup,
> Try this:
>
> # Data
> A <- c(1,2,3,4,5)
> B <- rnorm(100)
>
> # Results
> t(apply(sapply(A,function(x) pnorm(x*B)),2,function(x)
> c(Mean=mean(x),Var=var(x)))
Thank you for the suggestion, baptiste. segments() does do exactly what I was
wanting and matplot()/matlines() is probably a better solution to what I was
trying to do.
However, I am still concerned about the discrepancy between the documentation
in ?par and the behavior of lines(). Should li
I am having trouble removing entries from a vector based on the value of the
vector and another object value. It works in my pseudo test run:
DEV=400
#Y
CANDS=c(100:105)
#Z
DEVS=c(120,220,320,420,520)
if(DEV>DEVS)
(CANDS=CANDS[which(DEVTAZDetermine_FEET)
Hi,
I am trying to figure out how to save plots produced with
color2D.matplot to a file, any format. I have tried:
jpeg() / png()
color2D.matplot(x,c(1,0),c(1,0),c(1,0))
dev.off()
as well as
dev.new()
dev.print(device=png,file="myfreakinplot2.png") #(and jpeg, ps, pdf,
eps, together with dev.c
Dear all,
I am running anova(lm()) on a series of different data frame and I am getting
the following message
Using dataFrame$levels as id variables
1. Why am I getting that message
2. How do I suppress it (or correct it).
Thanks
Marco
--
Marco Blanchette, Ph.D.
Assistant Investigator
S
Dear all,
I am running anova(lm()) on a series of different data frame and I am getting
the following message
Using dataFrame$levels as id variables
1. Why am I getting that message
2. How do I suppress it (or correct it).
Thanks
Marco
--
Marco Blanchette, Ph.D.
Assistant Investigator
S
Hi all,
I ran a Weibull model, and I am wondering if there is any way to extract
the log likelihood. I tried loglik(model) but it does not seem to work
any help would be greatly appreciated
joe
[[alternative HTML version deleted]]
__
R-help@r
Hi all,
I ran a Weibull model, and I am wondering if there is any way to extract
the log likelihood. I tried loglik(model) but it does not seem to work
any help would be greatly appreciated
joe
[[alternative HTML version deleted]]
__
R-help@r
Ravi Varadhan wrote:
> Hi David,
>
> I did look at Appendix D.3 of T&G, but am not sure if the data set analyzed
> in F&H and that attached with "survival" are different. They both have
> n=418 (312 from RCT and 106 observational).
Well, as David implies, if the observation times are longer an
Brian Diggs wrote:
>Thank you for the suggestion, baptiste. segments() does do exactly what
I was wanting and matplot()/matlines() is probably a better solution to
what I was trying to do.
>
> However, I am still concerned about the discrepancy between the
documentation in ?par and the behavior of
Is there any way to change the orientation of the labels on the end of the
dendrograms to horizontal rather than vertical? If so, how can I do that.
_
Patrick Richardson
Biostatistician - Program of Translational Medicine
Van Andel Research Institute
Is there any way to change the orientation of the labels on the end of the
dendrograms to horizontal rather than vertical? If so, how can I do that.
Below is my code, but I'm not sure which argument(s) I can use to change the
label(s) (if it is possible to do).
a <- hclust(z, method = "comple
> I am having trouble removing entries from a vector based on the value of the
> vector and another object value. It works in my pseudo test run:
>
> DEV=400
> #Y
> CANDS=c(100:105)
> #Z
> DEVS=c(120,220,320,420,520)
> if(DEV>DEVS)
>(CANDS=CANDS[which(DEV CANDS
Hi, I'm using rgl to generate a 3D surface plot and I'm struggling to
get the lighting correct. Currently the surface gets plotted, but is
very 'shiny'. On rotating the view, I get to see parts of the surface
- but overall I don't see much detail because of the spotlight like
lighting.
I'
Peter,
I did check the data in the Appendix of F&H with the data in "survival"
package. I couldn't find any differences in the "time" and "status"
variables.
May be Terry Therneau knows the answer?!
Ravi.
---
Ravi
This does work, but also orients the entire dendrogram as horizontal. If I
could, I'd like the dendrogram to stay vertical and just orient the labels as
horizontal.
Many thanks,
_
Patrick Richardson
Biostatistician - Program of Translational Medicine
Why are you doing the test for normality?
The 2 most common reasons are:
1. want to test if data comes from an exact normal distribution.
2. want to know if tests based on normal assumptions are reasonable to use with
this data.
Technically, the Shapiro-Wilk test does not do either of the above
The $ syntax for working with elements of a list is a magical shortcut for
[[]]. It is a great shortcut when used as intended, but trying to force
magical shortcuts to do things that they were not intended for usually results
in the programming equivalent of turning yourself into a toad.
The c
> library(Kendall)
> Kendall(1:3,1:3)
WARNING: Error exit, tauk2. IFAULT = 12 <<
tau = 1, 2-sided pvalue =1
I believe Kendall tau is well-defined for this case and the reported
value is correct; isn't it a bug to give a warning? (And if, e.g.,
the pvalue is not well-defined in this
hah, I see what you mean. try:
plot(a, leaflab=c("textlike")).
fan
On Nov 21, 2008, at 2:30 PM, Richardson, Patrick wrote:
This does work, but also orients the entire dendrogram as
horizontal. If I could, I'd like the dendrogram to stay vertical
and just orient the labels as horizontal.
2008/11/21 Rajarshi Guha <[EMAIL PROTECTED]>:
> Hi, I'm using rgl to generate a 3D surface plot and I'm struggling to get
> the lighting correct. Currently the surface gets plotted, but is very
> 'shiny'. On rotating the view, I get to see parts of the surface - but
> overall I don't see much detai
hi there,
I have a list of list objects i need to remove the top layer
[[1]]
[1].0
"ABC" "DEF""LMN"
[1].1
"WER" "ERT" "TRY"
[[2]]
[2].0
"ASD","wer""qwe"
[2].1
"wdv""ghj""ggj"
I wanna avoid the top layer...that is [[1]] [[2]] shouldnt be there
just a simple list is wat i need.
[1].0
"ABC" "DEF"
Hi R users
I used the function line(x,y) and line(lowess(x,y)) to see the correlation
between 2 variables (x,y).
Here is my question:
is there a way to ask R to tell me the equation of
-this line : line(x,y)
-this curve: line(lowess(x,y))
Best regards
--
View this message in context:
http://www
Will one of these two solutions work for you:
> x <- list(list(1,2), list(3,4), list(5,6))
> x
[[1]]
[[1]][[1]]
[1] 1
[[1]][[2]]
[1] 2
[[2]]
[[2]][[1]]
[1] 3
[[2]][[2]]
[1] 4
[[3]]
[[3]][[1]]
[1] 5
[[3]][[2]]
[1] 6
>
> lapply(x, "[[", 1)
[[1]]
[1] 1
[[2]]
[1] 3
[[3]]
[1] 5
> lapply(x,
?unlist
-Original Message-
From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED]
On Behalf Of Rajasekaramya
Sent: Friday, November 21, 2008 2:14 PM
To: r-help@r-project.org
Subject: [R] list of list objects
hi there,
I have a list of list objects i need to remove the top layer
[[1]]
[1].0
?unlist
-Original Message-
From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED]
On Behalf Of Rajasekaramya
Sent: Friday, November 21, 2008 2:14 PM
To: r-help@r-project.org
Subject: [R] list of list objects
hi there,
I have a list of list objects i need to remove the top layer
[[1]]
[1].0
Hi,
I created a class (S4) with some slots like value, date, description
(it's actually a financial transaction class). Now I need a method
to convert this class forth and back into a single row data.frame,
where every slots represents a column. This method looks at the
moment like th
I'm a programmer in a biology lab who is starting to use R to automate
some of our statistical analysis of growth rate determination. But I'm
running into some problems as I re-code.
1) Hypotheses concerning Slope similarity/difference:
I'm using R's anova(lm()) methods to analyse a model which
I greatly apoligize, clear in my own mind, just didnt explain well enough.
DEVS should equal 6 values the last being 620, like the data frame:
CANDS DEVS
> [1,] 100 120
> [2,] 101 220
> [3,] 102 320
> [4,] 103 420
> [5,] 104 520
> [6,] 105 620
And yes, my first quest
Greetings,
I have been playing around with the R/Parallel package, which can farm out the
computation of a for-loop among multiple worker processes. Each worker gets a
chunk of the for-loop iterations; for example, if you have two workers and
for(x in 1:1000){...}, one worker would typically g
Outer's arguments are restricted to atomic vectors or arrays built on
atomic vectors (though the documentation is not explicit on this
point). What is the equivalent for lists or arrays built on lists? My
particular application was testing the Kendall tau function. I tried
this
> outer( permn(3)
> > Hi,
> >
> > I created a class (S4) with some slots like value, date, description
> > (it's actually a financial transaction class). Now I need a method
> > to convert this class forth and back into a single row data.frame,
> > where every slots represents a column. This method looks at t
Hi,
I have a dataset that is stratified by 3 levels, with an unequal number of
observations per strata:
AB C
182 18291
The data are approximately normally distrubuted.
Is there any reason why I can NOT use TukeyHSD to do pair-wise comparison of
means?
A
Hi,
I am using the following syntax to enter data and perform a cluster analysis:
x <- read.table ("clstrdbt.csv", header=TRUE, sep = ",",fill = TRUE)
cl<-cclust(x,4,20,verbose=TRUE,method="kmeans")
This is the result I receive:
Error in cclust(x, 4, 20, verbose = TRUE, method = "kmeans") :
(
> DEVS should equal 6 values the last being 620, like the data frame:
>CANDS DEVS
> > [1,] 100 120
> > [2,] 101 220
> > [3,] 102 320
> > [4,] 103 420
> > [5,] 104 520
> > [6,] 105 620
>
> And yes, my first question is do i need a "if" statement at all as the which
>
Cannot reproduce.
Van den Berge Joke wrote:
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-pro
X2R is a bundle of three software libraries for passing complicated data
structures from Fortran, C/C++, or AD Model Builder to R. An update has
been sent to CRAN and should be available from mirrors shortly. From the
menu at the left of the CRAN home page, look under Software / Other.
We als
Take a look at the las() function o ?par. It permits you to choice
labels directions.
__
Rodrigo Aluizio
Em 21/11/2008, às 17:14, "Richardson, Patrick" <[EMAIL PROTECTED]
g> escreveu:
!#x000a
Is there any way to change the orientation of the labels on the end
of the dendrograms
R 2.6.0
Windows XP
I have crated a file containing data with x and y values and have plotted the
data using the output of gam. I would like to overlay an x,y plot of the data
on top of the line returned by gam. I have succeeded in doing this using
par(new=TRUE), unfortunately the y axis of the
1 - 100 of 101 matches
Mail list logo
