Hi all,
I am trying to list all the 4 by 2 tables with some fixed margins.
For example, consider 4 by 2 tables with row margins 1,2,2,1 and
column margins 3,3. I was able to do it using the code below. However,
as seen below, I had to first count the total number of tables with
the specific
t; Cm <- as.matrix( C )
> D <- which( !duplicated( Cm, MARGIN=1 ) )
> nCm <- nrow( Cm )
> F <- lapply( D, function(d) {
>idxrep <- rep( d, nCm )
>which( 0 == unname( rowSums( Cm[idxrep,] != Cm ) ) )
> } )
>
>
> On November 8, 2018 1:42:40 PM PST, l
14" "7-15" "8-16" "1-9"
>> >"2-10"
>> >"3-11" "4-12" "5-13" "6-14"
>> >[15] "7-15" "8-16"
>> >> ## Then maybe:
>> >> tapply(seq_along(vec),vec, I)
&
Hi all,
I use the following example to illustrate my question. As you can see,
in matrix C some rows are repeated and I would like to find the indices of
the rows corresponding to each of the distinct rows.
For example, for the row c(1,9), I have used the "which" function to
identify the row
nction is approximately constant (in
> particular, zero) over nearly all its range it is possible that the result
> and error estimate may be seriously wrong."
>
> You could try to integrate over a finite interval, say (7, 12).
>
> Göran Broström
>
>
> On 2018-02-06
Sorry. I meant in the previous email that the function h() is a monotone
decreasing function. Thanks very much.
2018-02-06 13:32 GMT-05:00 li li <hannah@gmail.com>:
> Hi all,
> The function h below is a function of c and it should be a monotone
> increasing function sinc
Hi all,
The function h below is a function of c and it should be a monotone
increasing function since the integrand is nonnegative and integral is
taken from c to infinity. However, as we can see from the plot, it is not
shown to be monotone. Something wrong with the usage of integrate function?
Dear all,
The function f() below is a function of m1 and m2, both of which are
matrices with 3 rows. The function works sequentially one row after
another.
So altogether there are three stages. I am trying to update the coding to
write a generic function that will work for arbitrary k stages.
applying unique. In this particular case, it seems like all numbers
> are multiples of 1/30, so another idea could be to multiply by 30, round,
> and divide by 30.
>
> -pd
>
> > On 28 Jul 2017, at 17:17 , li li <hannah@gmail.com> wrote:
> >
> > I have the joint
I have the joint distribution of three discrete random variables z1, z2 and
z3 which is captured by "z"
and "prob" as described below.
For example, the probability for z1=0.46667, z2=-1 and z3=-1 is 2.752e-13.
Also, the probability adds up to 1.
> head(z) z1 z2 z3
[1,]
ience
> though implementing that last expression without hardcoding the sequences
> may not be obvious:
>
> expandArray <- function( a ) {
> b <- array( 0, dim = dim( a ) + 1 )
> do.call( `[<-`, c( list( b ), lapply( dim( a ), seq.int ), list( a ) ) )
> }
> expandA
For a data frame, we can add an additional row or column easily. For
example, we can add an additional row of zero and an additional row of
column as follows.
Is there an easy and similar way to add zeros in each dimension? For
example, for
array(1:12, dim=c(2,2,3))?
Thanks for your help!!
> +49 (0) 2631 9772-243
> https://www.researchgate.net/profile/Ivan_Calandra
>
>
> On 07/06/2017 16:32, li li wrote:
>
>> Hi all,
>>In checking my R codes, I encountered the following problem. Is there a
>> way to fix this?
>> I tried to specify options(digi
Hi all,
In checking my R codes, I encountered the following problem. Is there a
way to fix this?
I tried to specify options(digits=). I did not fix the problem.
Thanks so much for your help!
Hanna
> cdf(pmass)[2,2]==pcum[2,2][1] FALSE> cdf(pmass)[2,2][1] 0.758>
> pcum[2,2][1]
That works. Thank you!
2017-06-02 8:00 GMT-04:00 Adams, Jean <jvad...@usgs.gov>:
> Have you tried P2["20", "10", "0"] ?
>
> Jean
>
> On Thu, Jun 1, 2017 at 3:10 PM, li li <hannah@gmail.com> wrote:
>
>> Hi all,
>> I hav
Hi all,
I have a three dimensional array with the corresponding dimension names.
I would like to subset the array according to the dimension names. For
example,
suppose I want to extract the values corresponding to A=20, B=10, C=0. I
know I
can do:
P2[dimnames(P2)$A==20, dimnames(P2)$B==10,
Thanks so much for the help!!
Hanna
2017-06-01 11:02 GMT-04:00 Uwe Ligges :
>
>
> On 01.06.2017 10:03, Rolf Turner wrote:
>
>> On 01/06/17 19:54, Uwe Ligges wrote:
>>
>>> A design flaw, whether the labels are cut depends somewhat on the sizce
>>> of the
Hi all,
I have a question with regard to making plots using function
"scatterplot3d".
Please see the example below. It looks like, for y axis, the tickmark text
was cutoff.
The number "10" does not show up completely. I tried to work with par(mpg).
It does not
seem to work. Hope to get some
I meant that the post in your link above was NOT asked by me.
2017-05-27 16:53 GMT-04:00 li li <hannah@gmail.com>:
> Hi Dave,
> Thanks for your reply but the post in your link above was certainly
> asked by me.
>Hanna
>
> 2017-05-27 16:04 GMT-04:00
Hi Dave,
Thanks for your reply but the post in your link above was certainly asked
by me.
Hanna
2017-05-27 16:04 GMT-04:00 David Winsemius <dwinsem...@comcast.net>:
>
> > On May 27, 2017, at 12:49 PM, li li <hannah@gmail.com> wrote:
> >
> > Hi all,
>
Hi all,
Is there an R function that can be used to enumerate all the contingency
tables with fixed row and column margins. For example, can we list all 3 by
3 tables with row margins 3,6,6 and column margins 5,5,5.
Thanks very much!
Hanna
[[alternative HTML version deleted]]
Yes. If the function f takes the value zero at some discontinuity point,
then the code gives the inf of the set I described.
Otherwise, it is an approximation since we need to worry about numerical
accuracy.
2017-04-10 14:08 GMT-04:00 Bert Gunter :
> Well, I haven't
eley Breathed in his "Bloom County" comic strip )
>
>
> On Sun, Apr 9, 2017 at 1:28 PM, li li <hannah@gmail.com> wrote:
> > Dear all,
> > For a piecewise function F similar to the attached graph, I would like
> to
> > find
> >
Dear all,
For a piecewise function F similar to the attached graph, I would like to
find
inf{x| F(x) >=0}.
I tried to uniroot. It does not seem to work. Any suggestions?
Thank you very much!!
Hanna
F.pdf
Description: Adobe PDF document
Hi all,
I have the following data called "data1". After fitting the ancova model
with different slopes and intercepts for each region, I calculated the
regression coefficients and the corresponding standard error. The standard
error (for intercept or for slope) are all the same for different
,
>
> John
>
> -
> John Fox, Professor
> McMaster University
> Hamilton, Ontario, Canada
> Web: http://socserv.mcmaster.ca/jfox/
>
>
>
>
> On 2017-03-15, 9:43 PM, "R-help on behalf of li li"
> <r-help-boun..
Hi all,
Consider the data set where there are a continuous response variable, a
continuous predictor "weeks" and a categorical variable "region" with five
levels "a", "b", "c",
"d", "e".
I fit the ANCOVA model as follows. Here the reference level is region "a"
and there are 4 dummy variables.
With either of those, area:month and area:group and Residuals add up.
>
>
> On Sun, Mar 12, 2017 at 10:39 li li <hannah@gmail.com> wrote:
>
>> Hi All,
>> I have a dataset which contains 4 variables: area, group, time, y,
>> Area is a factor that has two levels A a
Hi All,
I have a dataset which contains 4 variables: area, group, time, y,
Area is a factor that has two levels A and B, group is a factor that is
nested within area. There are four groups within each area.
y is the response variable, and time refers to different days.
Below is the how data
Hi all,
Is there a function in R that can calculate the quantiles or percentiles
for the weighted sum of independent chi square random variables. I found a
few functions for calculating probability distribution function for such
random variables (e.g. pchisqsum..), but can not find any function
Hi all,
Is there an R function which can handles dependent response in Analysis
of covariance model. The dependence structure is known and is there to
account for it in ANCOVA analysis in R?
Thanks.
Hanna
[[alternative HTML version deleted]]
Hi all,
When using drm function in drc package, we can fit several does response
curves simultaneously or fitting each curve separately. If we fit
simultaneous curves without any constraining condition, for example,
parallelism condition, will we get the same results as the results obtained
by
GMT-04:00 PIKAL Petr <petr.pi...@precheza.cz>:
> Hi
>
> Thanks for code.
> see in line
>
> > -Original Message-
> > From: R-help [mailto:r-help-boun...@r-project.org] On Behalf Of li li
> > Sent: Tuesday, August 30, 2016 5:07 PM
> > To: r-help <r
ty Modeler INTERNET: cl...@math.utah.edu
> Department of Ecology VOICE: (360) 407-6815
> PO Box 47600FAX:(360) 407-7534
> Olympia, WA 98504-7600
>
> USPS: PO Box 47600, Olympia, WA 98504-7600
>
Hi all,
I have the following data. I want to plot the data (y ~ conc)
with different symbols and colors corresponding to different levels of the
factor sample.
I could create a column with color and pch and then do the plot, but I am
sure there are much better ways.
Can anyone make suggestions?
Hi all,
I am trying to use the drc package to fit 4L curve.
I am so confused about the plot.drm function in the drc package.
Particularly, I am confused about the scale of the xaxis in the plots
generated using the plot.drm function. See the example below:
## generate data and fit the model
couple of minor points:
> dim(tmp_a)[1] can be replaced by nrow(tmp_a)
> in the original code, apply() can be replaced by rowSums()
> It might be faster if rowSums() was replaced by .rowSums(); I haven't
> tried that.
>
> -Don
>
>
> --
> Don MacQueen
>
> Lawren
Hi all,
I have four matrix tmp_a, tmp_b, tmp_c and tmp_d whose dimensions are
shown as below.
I want to take one row from each of these matrices and then put the four
selected rows into a matrix.
I want to count the number of such matrices for which the vector of row sum
is less than or equal to
Dear all,
For the data below, I would like to fit a model with common random
slope and common random intercept as shown below. I am interested in
obtaining
separate fixed effect estimates (intercept and slope and corresponding
hypothesis test)
for each method. Instead of performing the
Min Q1 Med Q3 Max
-2.67575293 -0.51633192 0.06742723 0.59706762 2.81061874
Number of Observations: 69
Number of Groups: 7
2016-06-06 11:21 GMT-04:00 Gabriel Baud-Bovy <baud-bovy.gabr...@hsr.it>:
> On 06/06/2016 4:57 PM, li li wrote:
>
Hi all,
After fitting a random slope and random intercept model using lme
function, I want
to test whether each of the fixed slopes is equal to zero (The output of
model is below).
Can this be done (testing each individual slope) using multcomp package?
Thanks much for the help.
Hanna
>
kelinx <thierry.onkel...@inbo.be>
Date: 2016-05-30 4:40 GMT-04:00
Subject: Re: [R] model specification using lme
To: li li <hannah@gmail.com>
Cc: r-help <r-help@r-project.org>
Dear Hanna,
None of the models are correct is you want the same random intercept for
the differe
Hi all,
For the following data, I consider the following random intercept and
random slope model. Denote as y_ijk the response value from *j*th
individual within *i*th method at time point *k*. Assume the following
model for y_ijk:
y_ijk= (alpha_0+ tau_i +a_j(i))+(beta_i+b_j(i)) T_k +
Hi all,
I tried to use the package "XLConnect" to read excel data into R. I got
the following error message:
Error : .onLoad failed in loadNamespace() for 'rJava', details:
call: fun(libname, pkgname)
error: No CurrentVersion entry in Software/JavaSoft registry! Try
re-installing Java and
8e+16 bytes of RAM, i.e. 29327331 GiB =
>> 28640 TiB = 28 PiB. Storing such a large matrix even on file is not
>> possible.
>>
>> In other words, you need to figure out how to approach your original
>> problem in a different way.
>>
>> /Henrik
>>
>>
Hi all,
I am doing some calculation with very large dimension. I need to create a
matrix
with three columns and a very large number of rows
(3195*1290*495*35*35*35*15=1.312083e+15) i
n order to allocate calculation result from a for loop.
R does not allow me to create such a matrix because of
I see. Thanks!
2016-01-22 10:57 GMT-05:00 Rmh <r...@temple.edu>:
> FAQ 7.31
>
> in this case subtract the two numbers and see that
> they differ by about 1e-16
>
> Sent from my iPhone
>
> > On Jan 22, 2016, at 10:46, li li <hannah@gmail.com> wr
Hi all,
I encountered the following strange phenomenon.
For some reason, the obs_p[1] and res1$st_p[89] have
the same value but when I run "==", it returns FALSE.
Can anyone help give some explanation on this?
Thanks very much!
Hanna
> obs_p[1]
[1] 0.002201438
> res1$st_p[89]
[1]
As a follow up, I found out the proc freq in SAS can perform the exact
permutation trend test.
Hanna
2016-01-08 9:30 GMT-05:00 li li <hannah@gmail.com>:
> Thanks Bert.
>
>
> 2016-01-07 13:39 GMT-05:00 Bert Gunter <bgunter.4...@gmail.com>:
>
>> Sorry
Hi all,
I am trying to figure out how to use the function permTREND correctly in
package "perm". There does not seem to be any examples given for this
function. In the help file, it says the following:
## Default S3 method:
permTREND(x, y, alternative = c("two.sided", "less", "greater"), exact
t; Subject: Re: [R] exact trend test (enumerate all possible contingency
> tables with fixed row and column margins)
> To: li li <hannah@gmail.com>
>
>
> I do not know whether there is any package to do what you want.
>
> I **do** know that the algorithms required to
Dewey <li...@dewey.myzen.co.uk>:
> You received a number of suggestions about where to look and packages that
> might be suitable. Did you do that? If you did which ones did you look at
> and why did you reject them?
>
>
> On 07/01/2016 16:29, li li wrote:
>
>> Thanks for a
Hi all,
Is there an R function that does exact randomization trend test?
For example, consider the 2 by 5 contingency table below:
dose0dose 0.15dose 0.5dose 1.5dose 5 row
margin
Yes 43 4 5
8
trouble with having an open mind is that people keep coming along
> and sticking things into it."
> -- Opus (aka Berkeley Breathed in his "Bloom County" comic strip )
>
>
> On Wed, Dec 9, 2015 at 12:27 PM, li li <hannah@gmail.com> wrote:
> > Hi all,
> >
Thanks for your help Giorgio! Both options worked.
2015-12-09 11:46 GMT-05:00 Giorgio Garziano :
> Looking at the source code of the package drc, there is something that may
> somehow explain what
> you are experiencing:
>
> file: plot.drc.R, function addAxes(),
Hi all,
In drc package, is there a function which can be used to fit restricted
4PL curves? For example, we restrict two 4PL curves have the same lower and
upper asymptotes?
Thanks for the help in advance!
Hanna
[[alternative HTML version deleted]]
Hi all,
When trying to load the drc package. I got the following error. Any
suggestions?
Thanks.
Hanna
> install.packages("drc", dependencies=TRUE)
--- Please select a CRAN mirror for use in this session ---
trying URL 'https://cran.fhcrc.org/bin/windows/contrib/3.1/drc_2.5-12.zip'
Thanks.
2015-12-08 15:20 GMT-05:00 Rolf Turner <r.tur...@auckland.ac.nz>:
>
> Please keep communications on-list. Others may have relevant comments and
> suggestions to make.
>
>
> On 09/12/15 09:00, li li wrote:
>
>> Thanks for the reply. So a newer version of
Hi all,
When plotting the dose response curve using plot function as in the
example codes below, the scale of the axis should really be on the log
scale of the dose given the shape of the graph. But as you can see, the
tickmarks of the returned graph represent the original scale. How can I
lot(mod, type="all",log="xy")
>
> Jim
>
>
> On Wed, Dec 9, 2015 at 1:57 PM, li li <hannah@gmail.com> wrote:
>
>> Hi all,
>> When plotting the dose response curve using plot function as in the
>> example codes below, the scale of the ax
cates so the HTML would not corrupt your code.
>
>
> http://stackoverflow.com/questions/5963269/how-to-make-a-great-r-reproducible-example
> --
> Sent from my phone. Please excuse my brevity.
>
> On December 7, 2015 1:05:32 PM PST, li li <hannah@gmail.com> wr
Hi all,
I am trying to use the drm function in drc package to fit a 4 PL or 3PL
curve for an assay response. Please see the listed data below. When I do
the curve fitting, it returns the following error message. Anyone who
familiar with this have any input on what went wrong?
Thanks so much
Hi all,
I have the following data and I fit a log logistic model using the drm
function in DRC package.
I saved the fitted model in an object called "mod". (See below)
I understand that we can use the plot function to plot the regression
curve with the original data points using the code
, and X_2 are both Binomial(n, p). X_0,
X_1, and X_2 are all independent. Then X, Y are correlated and P(X <=
t, Y <= t) can be exactly calculated.
Thanks!
Hanna
2015-10-05 18:00 GMT-04:00, Rolf Turner <r.tur...@auckland.ac.nz>:
> On 06/10/15 04:43, li li wrote:
>>
. Say if X is binomial (n, p1) and Y is
binomial (n, p2) and the correlation between X and Y is rho and we
want to calculate
P(X <= c, Y <= c).
Thanks so much!
Hanna
2015-10-05 11:27 GMT-04:00, li li <hannah@gmail.com>:
>
>
___
__
R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see
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
Hi all,
In R, is there a function for the cumulative distribution function
for multinomial distribution? I only see pmultinom and rmultinom which
are the prabability mass function and the function for generating
multinomial random variables respectively.
Thanks!
Hanna
Thank you! That worked.
2015-09-15 2:50 GMT-04:00 peter dalgaard <pda...@gmail.com>:
>
> > On 15 Sep 2015, at 04:31 , li li <hannah@gmail.com> wrote:
> >
> > Hi Jeff,
> > Thanks for replying. I actually tried "ordered(tmp$type, levels=c("c
Live: OO#.. Dead: OO#.. Playing
> Research Engineer (Solar/BatteriesO.O#. #.O#. with
> /Software/Embedded Controllers) .OO#. .OO#. rocks...1k
> -------
Hi all,
I have the following data "tmp" and I want to plot boxplots for
each level of the factor "type" and the order the factor should be c,
b ,a. In other words, the boxplot corresponding to the level "c"
should be the first and so on.
Any suggestions?
Li
> tmp
result type
1 101
Hi all,
I plotted a dotplot based on the data below and code below. I would
like to add another yaxis on the right with a different col, different
tickmarks and a different label. Can anyone give some help?Thanks very
much!!
Hanna
> tmp1
result lot trt trtsymb trtcol
1 98 lot1
for an answer does not
ensure that a reasonable answer can be extracted from a given body of
data.
~ John Tukey
2015-06-09 21:57 GMT+02:00 li li hannah@gmail.com:
Hi all,
I'd like to fit a random intercept and random slope model. In my
data, there are three groups. I want to have
I got the following warning message when using the lmer function.
Does anyone know what is the implication? Thanks!
Warning message:
In anova(model, ddf = lme4) : bytecode version mismatch; using eval
__
R-help@r-project.org mailing list -- To
Hi all,
I'd like to fit a random intercept and random slope model. In my
data, there are three groups. I want to have different random
intercept for each group but the same random slope effect for all
three groups. I used the following R command.
However, there seems to be some problem. Any
Hi all,
I am fitting a random slope and random intercept model usign lme
fucntion as shown below. Type is factor with two levels. I would like
to to find a confidence interval for mean of this model. Note that the
variance we use in finding the confidence interval should include the
variariance
-project.org] On Behalf Of li li
Sent: Wednesday, 3 June 2015 13:07
To: r-help
Subject: [R] How to add legend to a dotplot
Hi all,
I wanted to add the legend to a dotplot using legend funciton. If
does not seem to be working? Anyone have any suggestions?
Thanks!
Hanna
Hi all,
I wanted to add the legend to a dotplot using legend funciton. If
does not seem to be working? Anyone have any suggestions?
Thanks!
Hanna
__
R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see
Hi all,
I am using the lme function to run a random coefficient model. Please see
output (mod1) as below.
I need to obtain the pvalue for the fixed effect. As you can see,
the pvalues given using the summary function is different from the
resutls given in anova function.
Why should they be
desde mi iPhone
El 26/5/2015, a las 13:18, li li hannah@gmail.com escribió:
Hi all,
I am using the lme function to run a random coefficient model. Please
see
output (mod1) as below.
I need to obtain the pvalue for the fixed effect. As you can see,
the pvalues given using the summary
Hi all,
I am fitting a random slope and random intercept model using R. I
used both lme and lmer funciton for the same model. However I got
different results as shown below (different variance component
estimates and so on). I think that is really confusing. They should
produce close results.
Hi all,
I wrote the following code and have two questions:
(1) As you can see, I would like different colors for different
types. It does not come out that way in the graph from this code.
Anyone know how to fix this?
(2) How do I made the lots number on x axis eligible to read?
(3) How
Hi all,
I have the following data in which there is one factor lot with six
levels and one continuous convariate time.
I want to fit an Ancova model with common slope and different intercept. So
the six lots will have seperate paralell
regression lines.I wanted to find the upper 95% confidence
Hi all,
I have the following data and would like to consider the following model:
value = type + batch (type) + beta1* month +beta2*type*month+ error
Here type is fixed effect and batch is a random effect neste within type,
and month is continuous.
Since there is a random effect in this model.
from my phone. Please excuse my brevity.
On April 15, 2015 7:20:04 PM PDT, li li hannah@gmail.com wrote:
Hi all,
In the following code, I am trying to use uniroot function to solve for
the root (a and b in code below) for function f1.
I am not sure why uniroot function does not give
Thank you.
2015-04-16 12:33 GMT-04:00 William Dunlap wdun...@tibco.com:
Use optimize() to find the minimum and feed that value into uniroot().
Bill Dunlap
TIBCO Software
wdunlap tibco.com
On Thu, Apr 16, 2015 at 7:47 AM, li li hannah@gmail.com wrote:
Hi Jeff,
Thanks
Hi all,
In the following code, I am trying to use uniroot function to solve for
the root (a and b in code below) for function f1.
I am not sure why uniroot function does not give the answer since when we
look the graph, the function does cross 0 twice.
Any suggestion?
Thanks.
Hanna
Hi all,
I have following data. When I perform an anova, the residual sum of
square returns to be zero.
But this is wrong, since we can hand calculate the RSS to be around 0.0308.
Did anyone come across the same problem before? Any suggestions?
Thanks.
Hanna
ydata
Y sample.Y
1
0.003725088 is the mean square error.
2015-04-14 13:44 GMT-04:00 Michael Dewey li...@dewey.myzen.co.uk:
See in-line
On 14/04/2015 18:23, li li wrote:
Hi all,
I have following data. When I perform an anova, the residual sum of
square returns to be zero.
But this is wrong, since we
Hi all,
It looks like the limits for x axis for box function in R is from -1 to
1. How can I change this limit?
Thanks for your help.
Hanna
[[alternative HTML version deleted]]
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Hi all,
I would like to add -70°C ± 10°C/Ambient as the title of my plot.
Could anyone give some help on this?
Thanks.
Hanna
[[alternative HTML version deleted]]
__
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=expression(-70*degree*C%+-%10*degree*C/Ambient))
Bill Dunlap
TIBCO Software
wdunlap tibco.com
On Fri, Feb 27, 2015 at 7:27 PM, li li hannah@gmail.com wrote:
Hi all,
I would like to add -70°C ± 10°C/Ambient as the title of my plot.
Could anyone give some help on this?
Thanks
Hi all,
I am trying to use the mle function in R to find the maximum likelihood
estimator. The ll function below is the negative of the log likelihood.
Suppose x0 is the observed values, I want to find the maximum likelihood
for a and b. After running the code below, I get the error message
, i.e. the
start argument for mle.
Cheers,
Bert
Bert Gunter
Genentech Nonclinical Biostatistics
(650) 467-7374
Data is not information. Information is not knowledge. And knowledge
is certainly not wisdom.
Clifford Stoll
On Sun, Nov 2, 2014 at 10:36 AM, li li hannah@gmail.com wrote
:00 li li hannah@gmail.com:
Thanks Bert for the reply. I still get a message when adding the start
argument.
n - 8
x0 - c(2,3)
ll- function(a,b,x=x0,size=n){
+
-sum(log(gamma((n-1)/2+a-1)/(gamma((n-1)/2)*gamma(a))*1/(2*b^a)*(x/2)^((n-1)/2-1)*(1/b+x/2)^(-((n-1)/2+a-1}
fit
Hi all,
When making a bunch of plots using a loop function, how to add title to
reflect different plots.
Specifically, for the code below, I generated 9 plots. I would like to add
a title to each plot.
For example, the titles will be respectively, plot1, plot 2, ⦠plot 9.
Thank you very
Hi all,
I fit a complete second order model for my response with three
predictors. However, the lack of fit of the model is
still significant. I wish to add the third order terms also. Is there a way
to do that using rsm function?
Thanks.
Hanna
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Hi all,
I am trying to perform an ordinal regression using proportional odds
model.
First I would like to test whether the assumption of proportional odds is
supported by the data.
I got the following error message when I try to fit one of the relevant
models. Can anyone give
some suggestions on
Hi all,
Is there a way to change the color of the boxplot plots outliers?
Thanks.
Hanna
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Hi all,
Does anyone know whether there is an R function available to find the
Bayesian variance components estimators
for random effects or mixed models?
Thank you very much.
Hanna
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Hi, all,
Recently, I met one issue when using socket between R C++ to transmit R
object. Would you pls help give me some suggestions? Many thanks!
[Background]:
I create a socket connection between R C++ binary first, and then, want
to use saveRDS() or save() in R to save the object into
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