Hi Josh,
Thank you for your reply.
The reason I thank "Y" (0 and 1) here as p is because I think each
observation is just a bernulli trial, so in this case the binomial n=1. And
yet R still fits it (with the logit link) . I know the expression for the
logit link, so I assumed I can take y here as
Thanks! Exactly what I wanted, as the same as Jorge also suggested.
At 12:49 AM 6/5/2011, Dennis Murphy wrote:
Hi:
Try this:
> dd <- data.frame(a = factor(rep(1:5, each = 4)),
+ b = factor(rep(rep(1:2, each = 2), 5)),
+ y = rnorm(20))
> str(dd)
'data.frame':
Thanks for your help.
As far as your question below is concerned, the data frame arose as a
result of some data cleaning on an original data frame, which was
changed into a table, modified, and changed back to a data frame:
ttcrmean<- as.table(by(ngbe[,'Log10'],
list(Time=ngbe$Time,Temp=ngbe
Exactly! Thanks.
At 12:49 AM 6/5/2011, Jorge Ivan Velez wrote:
Dr. LaBudde,
Perhaps
as.numeric(as.character(x))
is what you are looking for.
HTH,
Jorge
On Sun, Jun 5, 2011 at 12:31 AM, Robert A. LaBudde <> wrote:
I have a data frame:
> head(df)
Time Temp Conc ReplLog10
10 -20
Hi Robert,
Try this:
## Example data converting mtcars to factors
testdf <- as.data.frame(lapply(mtcars, factor))
str(testdf)
## taking advantage of assignment methods to avoid an explicit call to
as.data.frame
## convert factor to numeric using the technique recommended in ?factor
testdf[] <- l
Dr. LaBudde,
Perhaps
as.numeric(as.character(x))
is what you are looking for.
HTH,
Jorge
On Sun, Jun 5, 2011 at 12:31 AM, Robert A. LaBudde <> wrote:
> I have a data frame:
>
> > head(df)
> Time Temp Conc ReplLog10
> 10 -20H1 6.406547
> 22 -20H1 5.738683
> 3
Hi:
Try this:
> dd <- data.frame(a = factor(rep(1:5, each = 4)),
+ b = factor(rep(rep(1:2, each = 2), 5)),
+ y = rnorm(20))
> str(dd)
'data.frame': 20 obs. of 3 variables:
$ a: Factor w/ 5 levels "1","2","3","4",..: 1 1 1 1 2 2 2 2 3 3 ...
$ b: Factor w/ 2 l
I have a data frame:
> head(df)
Time Temp Conc ReplLog10
10 -20H1 6.406547
22 -20H1 5.738683
37 -20H1 5.796394
4 14 -20H1 4.413691
504H1 6.406547
774H1 5.705433
> str(df)
'data.frame': 177 obs. of 5 variabl
Hi,
Y is not the same as P. P is the conditional probability given the
data matrix. So theoretically, P can take on any value in [0, 1],
which means the odds can be anywhere from [0, +infty], not just 0 or
undefined. In logistic regression, the logit link is pretty standard,
so I do not think y
On Sat, Jun 4, 2011 at 6:44 PM, Abraham Mathew wrote:
> Let's say that I have a string and I want to know if a single word
> is present in the string. I've written the following function to see if
> the word "Geico" is mentioned in the string "Cheap Geico car insurance".
> However, it doesn't work
Hi all,
I have a problem with binary response data in GLM fitting.
The problem is that the "y" take only 1 or 0, and if I use logit link, it is
the log of the odds ratio, which is p/(1-p). In my situation, think "y" is
"p", so sometimes the odds is 0, sometimes it is "1/0", which is (should be)
un
Let's say that I have a string and I want to know if a single word
is present in the string. I've written the following function to see if
the word "Geico" is mentioned in the string "Cheap Geico car insurance".
However, it doesn't work, and I assume it has something to do with the any()
function.
Hey there!
In normal Regression if p>=alpha, there is no significance.
If i get this in quantile regression (for every tau), can I conclude, that
there is no relationship between x and y?
--
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This is not really an R question, and it indicates that you have a good deal
of studying to do about quantile regression before you rely on it.
Frank
-
Frank Harrell
Department of Biostatistics, Vanderbilt University
--
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Jim -
In what sense does cbind(A,B,C) not work?
A = matrix(rnorm(10),5,2)
B = matrix(rnorm(15),5,3)
C = matrix(rnorm(20),5,4)
cbind(A,B,C)
[,1] [,2] [,3] [,4] [,5][,6]
[1,] -0.54194873 -1.1105170 -0.479010 0.619911163 0.1610162 0.49028633
[2,]
A, B, C should have the same number of rows.
mlist = replicate(3, matrix(rnorm(6), 2), simplify=FALSE)
names(mlist) = LETTERS[seq_along(mlist)]
with(mlist, cbind(A,B,C))
or,
do.call(cbind, mlist)
HTH,
baptiste
On 5 June 2011 11:14, Jim Silverton wrote:
> How can I cbind three or more matrice
do.call(cbind, list(A, B, C))
On Sat, Jun 4, 2011 at 7:14 PM, Jim Silverton wrote:
> How can I cbind three or more matrices like A,B and C. This does not work:
>
> cbind(A,B,C)
>
>
> --
> Thanks,
> Jim.
>
--
Sarah Goslee
http://www.functionaldiversity.org
_
How can I cbind three or more matrices like A,B and C. This does not work:
cbind(A,B,C)
--
Thanks,
Jim.
[[alternative HTML version deleted]]
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the
Actually they say they used SAS, and Googling for "SAS local linear
logistic" suggests they used PROC GAM with the LOESS() smoother.
Probably quite similar to gam::gam().
-thomas
On Sun, Jun 5, 2011 at 9:12 AM, Thomas Lumley wrote:
> The Stanform gam() [gam package] has choices of spline or
The Stanform gam() [gam package] has choices of spline or
local-polynomial (defaulting to local-linear) smoothers. That's
probably the best match for the description. It shouldn't be
necessary to guess -- the paper should have cited the package -- but
we know that is often missed.
-thomas
There's code at http://tuvalu.santafe.edu/~aaronc/powerlaws/
-thomas
On Sun, Jun 5, 2011 at 9:01 AM, fernando del bon
wrote:
>
>
>
>
>
>
> p { margin-bottom: 0.08in; }
>
> Dear All,
>
> I will appreciate some
> suggestions of R packages for "ESTIMATION OF THE EXPONENT OF
> POWER-LAW FR
p { margin-bottom: 0.08in; }
Dear All,
I will appreciate some
suggestions of R packages for "ESTIMATION OF THE EXPONENT OF
POWER-LAW FREQUENCY DISTRIBUTIONS". I have been searching at
the R-help list several keywords for this subject and I did not find
a ver
Hi,
I'm using the Tps function of the fields package to plot 2D surfaces. My
problem is that some arrays have a lot of zeroes and this function fits the
data in a way that contour lines for zero are all over the place. Sometimes
there are lines where there shouldn't be any that extend from the pa
Dear R-Users,
I have fitted a cross-classified generalized linear mixed model using the
lmer package with the following code.
Mod<-lmer(y~x+(1|a)+(1|b)+ (1|c), family=binomial)
In this case, only including a covariate (x) as a fixed effect.
The fitted values, using fitted(mod), correspond to t
I too have this problem. Everything worked fine last year, but after
updating R and packages I can no longer do word stemming.
Unfortunately, I didn't save the old binaries, otherwise I would just
revert back.
Hoping someone finds a solution for R on Windows. Thanks!
There is a potential solution
Take a look at packages mgcv or gam (and probably others). Different
smoothers are used, but it's nonlinear, nonparametric logistic
regression. which is usually the important part. It also penalizes,
which can be even more important than which smoother is used.
-- Bert
On Sat, Jun 4, 2011 at 9:02
Dear R List-Members,
I wish to find the nonlinear least squares of function defined by an
integral which must be
evaluated numerically.
Is that possible to implement in R?
If it is possible, which problems I need to consider first?
Many Thanks,
Dejan
--
View this message in context:
http://r
What have you actually tried? Please read the posting guide. It also
tells you to specify version numbers and OS. New vesions are on its way
to CRAN.
So we can neitehr reproduce (without your code) nor know which version
you are trying.
Uwe Ligges
On 04.06.2011 19:46, hawkhandler wrote:
i
On 2011-06-04 11:11, Peter Ehlers wrote:
On 2011-06-03 13:34, Jason024 wrote:
I have a data frame like this:
col1 col2
r1 21
r2 43
r3 65
r4 87
r5109
r612 11
r714 13
r816 15
r918 17
r10 20 19
I want to modify this data fr
On 2011-06-03 13:34, Jason024 wrote:
I have a data frame like this:
col1 col2
r1 21
r2 43
r3 65
r4 87
r5109
r612 11
r714 13
r816 15
r918 17
r10 20 19
I want to modify this data frame, for example, assign every row in colum
i'm trying to create an new OAuthFactory variable for twitter but when i run
the handshake i get the following error:
Error in FUN(c("key", "secret", :
unused argument(s) (post.amp = TRUE)
has anyone seen this before or have any suggestions.
thanks ahead
--
View this message in context:
h
On Fri, Jun 3, 2011 at 7:03 PM, Matias Salibian-Barrera
wrote:
> Hello,
>
> This simple SVD calculation (commands are copied immediately below) crashes
> on my Ubuntu machine (R 2.13.0). However it works fine on my Windows 7
> machine, so I suspect there's a problem with (my?) Ubuntu and / or R.
On Jun 4, 2011, at 10:33 AM, kfl wrote:
I will be pleased to know, how to get the classic parametrisation in
a two
way analysis of varians:
Classic parametrisation:
Observed = intercept + row-effect + col-effect+ error, where sum af
row-effect=0 and sum of col_effect=0
?contrasts # which h
I will be pleased to know, how to get the classic parametrisation in a two
way analysis of varians:
Classic parametrisation:
Observed = intercept + row-effect + col-effect+ error, where sum af
row-effect=0 and sum of col_effect=0
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View this message in context:
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On Jun 4, 2011, at 11:41 AM, zhu yao wrote:
Dear UseRs:
Recently, I have read an article regarding the association between
age and
lymph node metastases.
http://jco.ascopubs.org/content/27/18/2931.long
In statistical analysis, the authors stated "Because a nonlinear
relationship between age
Dear UseRs:
Recently, I have read an article regarding the association between age and
lymph node metastases.
http://jco.ascopubs.org/content/27/18/2931.long
In statistical analysis, the authors stated "Because a nonlinear
relationship between age and lymph node involvement was expected based on
e
On Jun 3, 2011, at 7:20 PM, lana wrote:
Hi,
I have tried numerous methods and packages, but thus far cannot
seem to
find a solution. I am looking to essentially draw a filled colored
shape
around subsets on my data points on a scatter plot where none of the
shapes
overlap but instead
Thank you so much all for your invaluable inputs.
On Sat, Jun 4, 2011 at 3:36 AM, Patrick Burns wrote:
> A common thing to do is the Ljung-Box
> test on the residuals. For garch it
> would be the residuals squared.
>
> Actually for garch it should be the
> rank of the squared residuals -- see
>
On 04.06.2011 09:37, Petar Milin wrote:
Dear Uwe,
Please, can you help me with this?
I simplified the situation:
I am using Debian testing i386 (also, I have VirtualBox with Win XP on it)
myPackage is the name of the package
and foo.c is in the src/
in NAMESPACE I put:
useDynLib("foo")
export("
Dear R Group
I was trying to use the triangle.test function in SensoMineR and strangely i
encounter a error in the output of preference matrix from the analysis.
To illustrate, pl see the following dataframe of a design with the response
and preference collected as shown below:
design<-structure(l
On Ubuntu 10.04 it ran fine, albeit in a machine with lots of memory, it seems
to work
fine. Here's the output:
> rm(list=ls())
> gc()
used (Mb) gc trigger (Mb) max used (Mb)
Ncells 131881 7.1 35 18.7 35 18.7
Vcells 128838 1.0 786432 6.0 559631 4.3
> p <- 500
> n <
Peter et, al:
As a minor note ,,,
On Fri, Jun 3, 2011 at 8:00 PM, Peter Ehlers wrote:
> David has given you the answer. I'll just add that you might
> want to widen the strips a bit if you use superscripted factor levels:
>
> xyplot(Sepal.Length ~ Sepal.Width | Species, data = iris,
> st
Dear R-users,
The following question is more a statistic question than a R issue.
I would like to know the difference between the two following non
parametric technics : the bootstrap and the Wilks formula.
I understand well the theory about the two technics but cannot find
anything about their
I asked a radiographer friend of mine to examine this
suggestion, but he said it wouldn't scan.
Ted.
On 04-Jun-11 04:04:43, John wrote:
> Last line, try
>
> "but one can't, for Pi is transcendental."
>
>
> On Friday, June 03, 2011 04:12:07 AM Jim Lemon wrote:
>> On 06/01/2011 10:14 AM, baptis
On Fri, Jun 03, 2011 at 01:54:33PM -0700, Ned Dochtermann wrote:
> Petr,
> This is the code I used for your suggestion:
>
> k<-6;kk<-(k*(k-1))/2
> x<-matrix(0,5000,kk)
> for(i in 1:5000){
> A.1<-matrix(0,k,k)
> rs<-runif(kk,min=-1,max=1)
> A.1[lower.tri(A.1)]<-r
On Fri, 3 Jun 2011, someone ashamed of his/her real name wrote:
Hello,
I am trying to save a .gif movie using movie3d from the package {rgl}. I am
using the following code combined with the globe example on the ?movie3d
page. I've installed ImageMagick and the directory seems to be working
pr
I want to Fit a logistic growth model for
y = k *eb0+b1(age)/1 + eb0+b1(age), can some one help on how to get the
initial coefficients b0 and b1? I need to estimate in order to do the
regression analysis. When I run using b0=0.5 and b1=3.4818, I get the
following error
397443.8 : 0.5 3.4818
HI I want to Fit a logistic growth model for this data set where
y = k*exp(b0+b1(age))/1 + exp(bo+b1(age)),
start = list(b0 = 3, b1 = 3.5 ),
trace = TRUE)
I need to find the initial valued for b0 and b1. K =3. When I run using b0=3
abd b1 = 3.5, or any number I get the follo
Hi,
I have tried numerous methods and packages, but thus far cannot seem to
find a solution. I am looking to essentially draw a filled colored shape
around subsets on my data points on a scatter plot where none of the shapes
overlap but instead bend around each other if necessary. I finally ca
Hello,
I am trying to save a .gif movie using movie3d from the package {rgl}. I am
using the following code combined with the globe example on the ?movie3d
page. I've installed ImageMagick and the directory seems to be working
properly, i.e. when I do Sys.getenv("PATH"), C:\\Program Files
(x86)\
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