Hi Dennis, sorry for the delayed reply and thanks for the article. I digged
into it and found that if you have a GPU, the CUBLAS library beats the
BLAS/ATLAS implementation in the Matrix package for 'large' problems. Here's
what I mean,
its = 2500
dim = 1750
X = matrix(rnorm(its*dim),its, dim)
Hello R,
I have data in txt file of n columns like this:
x y1 .yn
1 100
2 50
3 10
4 200
5 20
I need to find coordinates for each maxima e.g.: [x,y1],[x,y2]...[x,yn]
a=read.table()
attach(a)
b=y1<-x
max(b) returns only value of y but no corresponding x value
--
View this message in
Hello all,
I am working on converting a set of S+ functions to R.
Can anyone tell me how to extract data from an hclust or dendrogram
object in R that is similar to that generated by
P1 = plclust(tree, plot=F)
Any suggestions would be appreciated.
Alan Mitchell, MSc
Biostati
Hi,
I am a student new to R. I would like to estimate the standard error for the
difference in interquantile regression coefficients after but do not know how
to do so. For each quantile I estimated the regression coefficent,
bootstrapped for the SE and saved the coefficient and then take th
Just trying to create returns from prices, and do something like:
returns.z = tail(prices.z,-1)/head(prices.z,-1) - 1 # should be equivalent
to returns = exp(diff(log(prices.z))) - 1
Curiously, I get a zoo object back with zeros everywhere and also with the
index having one fewer element than it
This is where which.max() comes in handy
n<-length(x)
x.c<-rep(0,n)
for(i in 1:n){
x.c[i]<-which.max[y1]
}
x.c is then a vector of x coordinates for the maximum for columns
y1,y2,...,yn
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Hi,
I have created a dataframe (lets call is df) that contains the following
variables
"Dependent1" "Dependent2" Dependent3" "Independent1"
I want to do the following regressions:
z<- glm( df$Dependent1 ~ df$Independent1)
z<- glm( df$Dependent2 ~ df$Independent1)
z<- glm( df$Dependent3 ~ df$I
I am trying to use some code from opencv in an r package, using Rcpp to build
the package. When I compile the c code on my machine, it works fine. However,
when I try to include it in my package, it gives me a bunch of error messages
like:
"error: opencv2/core/operations.hpp: No such file or di
Mr. Giles Crane,
I am new to R (only a month in). My response is as best as I understand the
workings of R (so if I'm wrong more experienced people plese help me out).
AOV is not really appropriate for an unbalanced model. This is where you can
rely on the lm() function using these steps:
I'm sure I used to be able to do
my.dependencies<- c("Depends", "Imports", "LinkingTo", "Suggests", "Enhances")
install.packages("animation", lib = Sys.getenv("R_LIBS_USER"), dependecies =
my.dependencies)
but now I get
Error in download.file(url, destfile, method, mode = "wb", ...) :
unused
Hello, Indrajit,
a "quick and dirty" solution could be to loop over the names of the
relevant variables of the data frame (in your case over the first three
variables) and use "[[ ]]" for indexing:
for (i in names( df)[ 1:3]) {
z <- glm( df[[ i]] ~ df$Independent1)
print( z)
}
An output
On 03/15/2011 06:57 PM, Allan Engelhardt wrote:
I'm sure I used to be able to do
...
unused argument(s) (dependecies = c("Depends", "Imports", "LinkingTo",
check the spelling.
Jim
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/lis
Hello all
I'm trying to run some path analysis in either sem or lavaan (preferably lavaan
because I find its interface easier to use). Most of my variables are
continuously distributed and fairly well-behaved but I have a single exogenous
variable (sex) which is not continuously distributed. Pr
Groan! Thanks. New baby mean no sleep. I'll make some more coffee.
Allan
On 15/03/11 08:18, Jim Lemon wrote:
On 03/15/2011 06:57 PM, Allan Engelhardt wrote:
I'm sure I used to be able to do
...
unused argument(s) (dependecies = c("Depends", "Imports", "LinkingTo",
check the spelling.
Jim
On 14/03/11 16:00, David Winsemius wrote:
On Mar 14, 2011, at 6:36 AM, Allan Engelhardt wrote:
On 14/03/11 02:00, Raoni Rosa Rodrigues wrote:
[...]
I'm working in a project with a software that register date and time
data in serial time format. This format is used by excel, for
exemple. In
Hi,
> returns.z = tail(prices.z,-1)/head(prices.z,-1) - 1 # should be equivalent
> to returns = exp(diff(log(prices.z))) - 1
>
> Curiously, I get a zoo object back with zeros everywhere and also with the
> index having one fewer element than it should.
>
> Does anyone know how to pointwise divid
On 15/03/11 05:17, Joshua Wiley wrote:
Hi Laura,
?is.vector
Consider methods::is if you might ever use attributes:
x<- 1:10
is.vector(x)
# [1] TRUE
comment(x)<- "Mine!"
is.vector(x)
# [1] FALSE
is(x, "vector")
# [1] TRUE
Allan
__
R-help@r-projec
Hi
I am trying to plot two simple graphs with a grid in background. The axis
and grid appears in correct position, but the actual data are not there
Can somebody provide me a hint what is missing?
The code is:
pln <- read.table(file="PLN.txt", header=TRUE, dec=",")
par(mfrow=c(1,2))
plot(
Hi Ajay,
thanks for this comparison, which prodded me to give CUDA another try on my now
somewhat aging MacBook Pro.
> Hi Dennis, sorry for the delayed reply and thanks for the article. I digged
> into it and found that if you have a GPU, the CUBLAS library beats the
> BLAS/ATLAS implementation
On Tue, Mar 15, 2011 at 12:01:45PM +0100, Sara Szeremeta wrote:
> Hi
>
> I am trying to plot two simple graphs with a grid in background. The axis
> and grid appears in correct position, but the actual data are not there
> Can somebody provide me a hint what is missing?
>
> The code is:
>
Hi all,
Does anyone have any advice or experience storing package settings
between R runs? Can I rely on the user's home directory (e.g.
tools::file_path_as_absolute("~")) to be available and writeable
across platforms?
Hadley
--
Assistant Professor / Dobelman Family Junior Chair
Department of
Hi there,
I try to model some dose response curves (drc-package). In most cases
it is fine but now I got some data which produces me the following
error:
load("drmData.RData")
library(drc)
drmObj <- drm(value ~ concentration, cmpd_respvar, data = drmData, fct = LL.4())
predict(drmObj)
>> Error
--- included text --
I have done model selection between candidate Cox models, using AICc
calculated with penalized log likelihoods. Then model averaging was done
to
obtain model averaged parameter estimates. Is there a way to plot
survival
curve from the averaged model, by estimating baseline haza
Marlin Keith Cox wrote:
>
> Thank you, but I need only the area under the llm colored. What if there
> were two lm lines?
>
For the well-behaved case you could find the range to be colored with
range = which(diff(resid(llm)>0)!=0)
Dieter
x=seq(0,5,len=100)
y=-(x-5)^2
llm<-lm(y~x)
range = w
I tried to use lsode with precompiled C code instead of an R function
defining the derivatives. No problem so far.
However, now I need to implement an ODE that contains spline functions,
i.e. the derivatives at given time points depend on the value of a
spline function at this time point. What is
Hi all,
I have a question regarding the estimation of a two-sided truncation
regression model using maximum likelihood estimation. I want to
estimate a linear model where the dependent variable is greater 0 and
upper truncated at 1. Does anyone know a R-package that contains a routine
that could
On Mon, Mar 14, 2011 at 10:16 PM, Adam Gustafson wrote:
> Just trying to create returns from prices, and do something like:
>
> returns.z = tail(prices.z,-1)/head(prices.z,-1) - 1 # should be equivalent
> to returns = exp(diff(log(prices.z))) - 1
>
> Curiously, I get a zoo object back with zeros
It occured to me:
x[max(y)] simple isn't it?
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Sent from the R help mailing list archive at Nabble.com.
__
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That actually won't work. max(y) will give a value, not a coordinate, so
x[max(y)] is definitely not what you want.
--
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Sent from the R help mailing list archive at Nab
Hi,
I am trying to recursively apply a function to a selection of columns
in a dataframe. I've had a look around and from what I have read, I
should be using some version of the apply function, but I'm really
having some headaches with it.
Let me be more specific with an example.
Say I have a dat
Hi,
This seems like there should be a simple answer, but having spent most of
the day trying to find it, I'm becoming less convinced and as such am asking
it here.
Here's a sub-set of my data (a data.frame in R):
myDF
time value trial
1 2011-03-01
How could one get started with a self-written function?
I have a function written in .txt format, but can not find the way to import
it to the R space.
Would be very appreciated for help.
--
View this message in context:
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I was wondering if there is a way to get read in a single keystroke at a
time in R as a string, akin to ncurses-style interfaces. I looked into
readLines, readChar, etc. using stdin, but these all require the use of an
end of line. Has anyone ever had need to do this or have any ideas on how
to
On Mar 15, 2011, at 2:17 AM, Raoni Rosa Rodrigues wrote:
Hello Mr. Allan and Mr. David!
Thanks very much for your time and lessons.
Works pretty well, Mr. Allan, specially after Mr. David explanation
(and the 2 days-bug correction).
Just a curious behavior of data tranformation: the R-tra
?strsplit
x <- "ThisIsaString"
y<- strsplit(x,"")
This gives a list, which you can convert to a vector by unlist(y)
Incidentally, you could have found out about strsplit via R's
help.search("character string") (or similar) or even googling "R
string function" . Please use R's native Help capabil
Right.
?which.max
is what's needed.
-- Bert
On Tue, Mar 15, 2011 at 6:00 AM, nblarson wrote:
> That actually won't work. max(y) will give a value, not a coordinate, so
> x[max(y)] is definitely not what you want.
>
> --
> View this message in context:
> http://r.789695.n4.nabble.com/Finding-
Hello,
I have written a 'for' loop which on the first run makes nearest neighbour
calculations for my dataset 'A' in relation to dataset 'B', then based on
these results, some of the rows from A are moved into dataset B, and the
calculation is repeated on the remaining rows in A. Therefore a s
I have a dataset like this:
q25_1 q25_2 q25_3 q25_4 q25_5 q25_6 q25_7 q25_8 q25_9 q25_10 q25_11 q25_12
1 2 2 1 1 2 1 2 1 1 2 1 3
2 2 2 2 1 2 1 2 1 1 2 1 2
3 2 1 1 1 2
Try this:
data.frame(count = colSums(x == 1), percentage = colSums(x == 1) / nrow(x))
On Tue, Mar 15, 2011 at 11:13 AM, Orvalho Augusto wrote:
> I have a dataset like this:
>
> q25_1 q25_2 q25_3 q25_4 q25_5 q25_6 q25_7 q25_8 q25_9 q25_10 q25_11
> q25_12
> 1 2 2 1 1 2
try the following:
DF <- as.data.frame(matrix(sample(2, 120, TRUE), 10, 12))
Results <- data.frame(
var = names(DF),
count = colSums(DF == 1),
percentage = colMeans(DF == 1)
)
I hope it helps.
Best,
Dimitris
On 3/15/2011 3:13 PM, Orvalho Augusto wrote:
I have a dataset like thi
On Sun, Mar 13, 2011 at 10:00 PM, Raoni Rosa Rodrigues
wrote:
> Hello R Help,
>
> I'm working in a project with a software that register date and time data in
> serial time format. This format is used by excel, for exemple. In this
> format, 40597.3911423958 is 2011/2/23 09:23:15. First part is
I apologize for being unclear in my original post. What I am trying to
achieve is to capture the stdin() connection and read in a single
keystroke - including arrow keys and the like - without having to have to
use the return key each time. My goal is to create a small UI for rapidly
looking th
On Tue, Mar 15, 2011 at 10:24 AM, Gabor Grothendieck
wrote:
> On Sun, Mar 13, 2011 at 10:00 PM, Raoni Rosa Rodrigues
> wrote:
>> Hello R Help,
>>
>> I'm working in a project with a software that register date and time data in
>> serial time format. This format is used by excel, for exemple. In t
On Tue, Mar 15, 2011 at 10:29 AM, Jonathan P Daily wrote:
> I apologize for being unclear in my original post. What I am trying to
> achieve is to capture the stdin() connection and read in a single
> keystroke - including arrow keys and the like - without having to have to
> use the return key ea
?"while"
You don't want a for loop. You need a while loop.
--
Jonathan P. Daily
Technician - USGS Leetown Science Center
11649 Leetown Road
Kearneysville WV, 25430
(304) 724-4480
"Is the room still a room when its empty? Does the room,
the thing itself have purp
Solved. Thanks.
--
Jonathan P. Daily
Technician - USGS Leetown Science Center
11649 Leetown Road
Kearneysville WV, 25430
(304) 724-4480
"Is the room still a room when its empty? Does the room,
the thing itself have purpose? Or do we, what's the word... imbue it.
This is!
Thank you everyone.
Caveman
On Tue, Mar 15, 2011 at 4:24 PM, Dimitris Rizopoulos <
d.rizopou...@erasmusmc.nl> wrote:
> try the following:
>
> DF <- as.data.frame(matrix(sample(2, 120, TRUE), 10, 12))
>
> Results <- data.frame(
>var = names(DF),
>count = colSums(DF == 1),
>
source("myfunction.R") is the usual approach. You could also
just copy your code and paste it into the R console.
Sarah
On Tue, Mar 15, 2011 at 9:21 AM, bra86 wrote:
> How could one get started with a self-written function?
> I have a function written in .txt format, but can not find the way to
On Mar 15, 2011, at 9:21 AM, bra86 wrote:
How could one get started with a self-written function?
I have a function written in .txt format, but can not find the way
to import
it to the R space.
Would be very appreciated for help.
It would be more typical to use a file extension of .r but I
Readers,
For a data set:
text1,23,text2,45
text1,23,text3,78
text1,23,text3,56
text1,23,text2,45
The following command was entered:
datasubset<-data.frame(dataset[,3]=="text3")
The result of
datasubset
is
TRUE
TRUE
The required result is
text1,23,text3,78
text1,23,text3,56
What is the co
You need to specify the group aesthetic - that defines how
observations are grouped into instances of a geom.
Hadley
On Tue, Mar 15, 2011 at 8:37 AM, joeP wrote:
> Hi,
>
> This seems like there should be a simple answer, but having spent most of
> the day trying to find it, I'm becoming less con
Thanks! I've never come across 'while' before and it's perfect.
On Mar 15 2011, Jonathan P Daily wrote:
?"while"
You don't want a for loop. You need a while loop.
--
Jonathan P. Daily
Technician - USGS Leetown Science Center
11649 Leetown Road
Kearneysville
did you try to fit your data with a skew-normal/skew-t distribution?
If that works, you can use simulation.
Kjetil
On Tue, Mar 15, 2011 at 2:31 AM, Lao Meng wrote:
> Hi all:
> I have a question on sample size calculation of 2 groups of data. If 2
> groups of data are all normal distribution, the
I am trying to calculate monthly means by year of phosphates and nitrates from
a multi year data set. Can anybody help me out with the most effective way to
do this?
My data looks like this:
Collection_Date Test.Name Value
2000-01-24 17:00:00 Phosphate
64 Bit R w/JAGS seems to be stalling out as well, I ran a test run of 100
iterations and it's been hanging for 8 hours so that doesn't seem to be the
solution. I'll take a look at PYMC. That CppBUGS package looks pretty
interesting, I'll keep my eye on it.
My C programming book arrives today fro
Does it mean that regsubsets doesn't work with categorical variables?
It's because I'm trying the following and I don't know what happens. Any
help would be appreciated.
varin <- data.frame(v1,v2,...,v7, factor1,..., factor4)
Dependent variable: height
Then:
eu.subsets <- regsubsets(varinc ,
you could run source(directory/textfilename.txt) [and then you can use it in R]
or paste the function into the console,
or highlight and pass the function from the built in R text editor,
or highlight and pass the function from your external text editor (notepad++,
emacs, textwrangler, etc.)
I have just started using R so forgive me if this question is very
simple. I have a data set (in a data frame called dm) that looks like
this
x (Cells)y(males)
1 0
2 2
3 7
4 12
5 12
6 19
7 22
8 23
9 25
10 23
11 23
12 11
13
Hi, all,
Does R have a "const object" concept like which is in C++ language? I want to
set some data frames as constant to avoid being modified unintentionally.
Thanks!
xiagao1982
2011-03-15
[[alternative HTML version deleted]]
__
R-help@
Dear R useRs,
I am a newbie. Currently, my script is quick long, therefor I am trying to
improve the efficiency of my script.
I use often a version of the following code, the script below is just a
simple example, but I especially try to improve the last three lines:
matrix1<-diag(100)
a=2
try<
?source
. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .
Wolfgang Raffelsberger, PhD
IGBMC,
1 rue Laurent Fries, 67404 Illkirch Strasbourg, France
wolfgang.raffelsberger (at) igbmc.fr
De : r-help-boun...@r-project.org [r-help-boun
Try using which()
Something like data.frame(dataset[which(dataset[,3]=="text3"),])
e-letter wrote:
>
> Readers,
>
> For a data set:
>
> text1,23,text2,45
> text1,23,text3,78
> text1,23,text3,56
> text1,23,text2,45
>
> The following command was entered:
>
> datasubset<-data.frame(dataset[,3]
Dear R useRs,
I have a problem with nls.lm function of minpackl.lm package.
I need to fit the Van Genuchten Model to a set of data of Theta and hydraulic
conductivity with nls.lm function of minpack.lm package.
For the first fit, the parameter estimates keep changing even after 1000
iteration
Hi,
Le 15/03/11 16:09, e-letter a écrit :
> Readers,
>
> For a data set:
>
> text1,23,text2,45
> text1,23,text3,78
> text1,23,text3,56
> text1,23,text2,45
>
> The following command was entered:
>
> datasubset<-data.frame(dataset[,3]=="text3")
datasubset<-subset(dataset,dataset[,3]=="text3")
> The
Hi,
What you are after is:
datasubset <- dataset[ dataset[,3] == "text3", ]
Equivalently, you can use:
datasubset <- subset(dataset, subset = dataset[,3] == "text3")
HTH,
Francisco
On Tue, Mar 15, 2011 at 3:09 PM, e-letter wrote:
> Readers,
>
> For a data set:
>
> text1,23,text2,45
> text1
Hi Claus,
On Tue, Mar 15, 2011 at 9:33 AM, Claus O'Rourke wrote:
> Hi,
> I am trying to recursively apply a function to a selection of columns
> in a dataframe. I've had a look around and from what I have read, I
> should be using some version of the apply function, but I'm really
> having some h
R-help,
I'm trying to read a data file with plenty of columns.
I just need the first 5 but it doe not work by doing something like:
> mycols <- rep(NULL, 430) ; mycols[c(1:4)] <- NA
> read.table(myfile, skip=2, colClasses=mycols)
Any suggestions?
Thanks in advance
_
On Tue, 15 Mar 2011, Hadley Wickham wrote:
Hi all,
Does anyone have any advice or experience storing package settings
between R runs? Can I rely on the user's home directory (e.g.
tools::file_path_as_absolute("~")) to be available and writeable
across platforms?
No. First, please use path.e
Hi Keith,
how about this:
fit<-function(x,coefs) x*coefs[2]+coefs[1]
plot(1:10,1:10,type="n")
x<-par("usr")[1:2]
cf1<-c(a=3,b=.5) #eg coefs from a lm-object
cf2<-c(a=1,b=.9) #another line, coloring only intersection
polygon(c(x,rev(x)),c(fit(x,cf1),fit(rev(x),cf2)),col="red")
setting one of the c
Dear all,
I need to plot an cumulative distribution plot of a variable and then to fit a
distribution to that, probably a weibull or lognormal.
I have plotted the ecdf as
> plot(ecdf(x))
but I haven't managed to fit the distribution. I have as well attached the data.
I would appreciate if you
On Mar 15, 2011, at 12:00 PM, Lathouri, Maria wrote:
Dear all,
I need to plot an cumulative distribution plot of a variable and
then to fit a distribution to that, probably a weibull or lognormal.
I have plotted the ecdf as
plot(ecdf(x))
but I haven't managed to fit the distribution. I
hi: you also may want to look at the admit package. it does metropolis
hastings using a weighted mixture of t-distributions so you just need to
write a function for the likelihood you're trying to get the parameters for.
I don't know about it's speed or efficiency for large data sets but
you could
On Tue, 2011-03-15 at 07:24 -0700, Carl Nim wrote:
> I am trying to calculate monthly means by year of phosphates and nitrates
> from a multi year data set. Can anybody help me out with the most effective
> way to do this?
>
> My data looks like this:
>
> Collection_Date Test.Name Value
> 2000
One things you should do, as Kate suggested, is to check whether the Jacobian
functiions are correctly code. You can do this with "numDeriv" package:
require(numDeriv)
?jacobian
Compare the jacobian from numDeriv with your jacobian for a few reasonable
parameter vectors.
Ravi.
_
Actually I have already done a search on that but it was not much of a help.
That is why I posted it in the r-help in case it was much more helpful.
Thank you.
Kind regards
Maria
From: David Winsemius [dwinsem...@comcast.net]
Sent: 15 March 2011 16:19
T
On Mar 15, 2011, at 12:23 PM, Lathouri, Maria wrote:
Actually I have already done a search on that but it was not much of
a help. That is why I posted it in the r-help in case it was much
more helpful.
There is a tutorial by Ricci with the exact title I suggested looking
for. At the Baro
Hi,
One "easy" way to do that is by using the "nls" (nonlinear least square)
library.
Basically you will test with that if your distribution can be adjusted to
the distribution you say it matches (Weibull?) through the adjustment of
some parameters.
The base package includes the "nls" library, b
Hi Jon, I read your question differently. Is the answer? - Rex
> ch=scan(stdin(),what=character(0),n=1)
1: f
Read 1 item
> ch
[1] "f"
>
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
Behalf Of Bert Gunter
Sent: Tuesday, March 15, 2011 10
I think you need to read an introduction to R.
For starters, read.table returns its results as a value, which you are not
saving.
The probable answer to your question:
Read the whole file with read.table, and select columns you need, e.g.:
tab <- read.table(myfile, skip=2)[,1:5]
-Original Mes
On Mar 15, 2011, at 1:11 PM, wrote:
I think you need to read an introduction to R.
For starters, read.table returns its results as a value, which you
are not saving.
The probable answer to your question:
Read the whole file with read.table, and select columns you need,
e.g.:
tab <- read.
You could use the by() function after a little data manipulation.
The first line will create a field just of the date portion of your datetime
field. Then you can use the by() function to use the indices you desire to
calculate the mean.
mSamp$cDT <- chron(unlist(strsplit(as.character(mSamp$Col
On 15.03.2011 15:53, xiagao1982 wrote:
Hi, all,
Does R have a "const object" concept like which is in C++ language? I want to
set some data frames as constant to avoid being modified unintentionally. Thanks!
Although there is almost never a "No" in R, the best short answer is: "No".
Best,
On 15/03/2011 2:23 PM, Uwe Ligges wrote:
On 15.03.2011 15:53, xiagao1982 wrote:
> Hi, all,
>
> Does R have a "const object" concept like which is in C++ language? I want
to set some data frames as constant to avoid being modified unintentionally. Thanks!
Although there is almost never a "No
It would be nice to have a standard directory where R can write things
this way. A semi-standard directory is given by
Sys.getenv("R_LIBS_USER"), which defaults to ~/R/.../. Maybe ~/R/
could serve as that convention? That way we (various developers etc)
would also not clutter up users home direc
On 03/15/2011 11:34 AM, Duncan Murdoch wrote:
On 15/03/2011 2:23 PM, Uwe Ligges wrote:
On 15.03.2011 15:53, xiagao1982 wrote:
> Hi, all,
>
> Does R have a "const object" concept like which is in C++ language?
I want to set some data frames as constant to avoid being modified
unintentionally. Th
On Tue, Mar 15, 2011 at 10:53 AM, xiagao1982 wrote:
> Hi, all,
>
> Does R have a "const object" concept like which is in C++ language? I want to
> set some data frames as constant to avoid being modified unintentionally.
> Thanks!
>
>
LockBinding will prevent a variable from being modified (but
Dear All:
I am a newbie to R. I am trying to use the lmsqreg.fit pack, although I
have installed(I think) it I am not able to invoke it when write the code
for it. Here's what I got when I installed it.
package 'lmsqreg' successfully unpacked and MD5 sums checked
And here is what I got wit
I am a new R user (am using it through the Rcmdr package) and have struggled
to find out how to report OR and RR directly when running GLM models (not
only reporting coefficients.)
Example of the syntax that I have used:
GLM.2 <- glm(diarsev ~ treatmentarm +childage +breastfed,
family=binomial(lo
On 15.03.2011 19:35, Martin Morgan wrote:
On 03/15/2011 11:34 AM, Duncan Murdoch wrote:
On 15/03/2011 2:23 PM, Uwe Ligges wrote:
On 15.03.2011 15:53, xiagao1982 wrote:
> Hi, all,
>
> Does R have a "const object" concept like which is in C++ language?
I want to set some data frames as constan
OR <- exp(coef(GLM.2)[-1])
OR.ci <- exp(confint(GLM.2)[-1,])
-Aaron
On Tue, Mar 15, 2011 at 1:25 PM, lafadnes wrote:
> I am a new R user (am using it through the Rcmdr package) and have
> struggled
> to find out how to report OR and RR directly when running GLM models (not
> only reporting coef
On Mar 15, 2011, at 1:25 PM, lafadnes wrote:
I am a new R user (am using it through the Rcmdr package) and have
struggled
to find out how to report OR and RR directly when running GLM models
(not
only reporting coefficients.)
Example of the syntax that I have used:
GLM.2 <- glm(diarsev ~
On Mar 15, 2011, at 1:03 PM, YAddo wrote:
Dear All:
I am a newbie to R. I am trying to use the lmsqreg.fit pack,
although I
have installed(I think) it I am not able to invoke it when write
the code
for it. Here's what I got when I installed it.
package 'lmsqreg' successfully unpacked
The Lattice auto.key argument has a bug in R.12.2.
R version 2.12.2 (2011-02-25)
Platform: i386-pc-mingw32/i386 (32-bit)
other attached packages:
[1] lattice_0.19-17
loaded via a namespace (and not attached):
[1] grid_2.12.2
If I set up my plot parameters as
require(lattice)
superpose.line
Dear R helpers
Suppose I have a vector as
vect_1 = c("AAA", "AA", "A", "BBB", "BB", "B", "CCC")
vect_1_id = c(1:length(vect_1))
Through some process I obtain
vect_2_id = c(2, 3, 7), then I need a new vector say vect_2 which will give me
vect2 = ("AA", "A", "CCC") i.e. I need the subset of ve
vect_1[vect_2_id]
and perhaps a readthrough of some of the very nice
intro to R materials available online.
Sarah
On Tue, Mar 15, 2011 at 3:41 PM, Vincy Pyne wrote:
> Dear R helpers
>
> Suppose I have a vector as
>
> vect_1 = c("AAA", "AA", "A", "BBB", "BB", "B", "CCC")
>
> vect_1_id = c(1:leng
Hi All,
is there any effiective and dense/compact method to calculate the mean of a
list of - of course coincident - matrices on an element by element basis? The
resulting matrix' [i, j]-th element is the mean of the list's matrices' [i,
j]-th elements respectively...
Iterating by for statement
Peter,
as the matrices in the list have the same shape, you can unlist them
into an array and then use rowMeans.
HTH
Claudia
Am 15.03.2011 21:17, schrieb hihi:
Hi All,
is there any effiective and dense/compact method to calculate the mean of a
list of - of course coincident - matrices on a
Try this:
l <- list(matrix(rnorm(9), 3), matrix(rnorm(9), 3), matrix(rnorm(9), 3))
Reduce('+', l) / length(l)
On Tue, Mar 15, 2011 at 5:17 PM, hihi wrote:
> Hi All,
> is there any effiective and dense/compact method to calculate the mean of a
> list of - of course coincident - matrices on an e
On 15/03/2011 2:56 PM, Gene Leynes wrote:
The "getSrcFilename" function is exactly what I was trying to describe, and
I'm excited to know that it's on it way!
I have tried to create that type of function, but I didn't think it was
possible with currently available functions. I would be interest
On Tue, Mar 15, 2011 at 1:17 PM, hihi wrote:
> Hi All,
> is there any effiective and dense/compact method to calculate the mean of a
> list of - of course coincident - matrices on an element by element basis? The
> resulting matrix' [i, j]-th element is the mean of the list's matrices' [i,
> j]
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