Daniel Malter wrote:
Second, given the figure, a linear specification is obviously a
misspecification of your model, unless you account for autocorrelation.
I've decided to use this as a learning opportunity. I looked up
autocorrelation: It does not apply in any way. By your standards,
Thank you Brian.
Sorry for being such a noob. I am not a programmer and just learning R by
myself. This is was I typed, but ended up with a couple error messages.
df -structure(list(year = c(1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L,
+ 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L, 1L,
Greetings,
I've been struggling for some time with a problem concerning a big database
that i have to deal with.
I'll try to exemplify my problem since the database is really big.
Suppose I have the following data:
AA = c(4,4,4,2,2,6,8,9)
A1 = c(3,3,5,5,5,7,11,12)
A2 = c(3,3,5,5,5,7,11,12)
A =
Naomi,
You can reply to messages in a Digest provided you have set
your list options so that you choose MIME rather than
Plain Text for the option Get MIME or Plain Text Digests?
provded your email client supports MIME Digests.
In that case, when you open the Digest email, you will be
able to
On 2011-07-25 10:30, Manojit Roy wrote:
Dear all,
I am trying to create a 6-plot layout - 3 rows and 2 columns - so that
only the top two plots have variable widths, all else with their default
setting. Using
layout(matrix(c(1,2,3,4,5,6),3,2,byrow=T),widths=c(5,2)) rescales
column #2 of all
Hi all:
There's a question about glht function.
My data:data_ori,which inclue CD4, GROUP,time.
f_GROUP-factor(data_ori$GROUP)
f_GROUP is a factor of 3 levels(0,1,2,3)
result - lme(sqrt(CD4) ~ f_GROUP*time ,random = ~time|ID,data=data_ori)
glht(result, linfct = mcp(f_GROUP=Tukey) )
Error in
On 2011-07-26 00:16, Lao Meng wrote:
Hi all:
There's a question about glht function.
My data:data_ori,which inclue CD4, GROUP,time.
f_GROUP-factor(data_ori$GROUP)
f_GROUP is a factor of 3 levels(0,1,2,3)
result- lme(sqrt(CD4) ~ f_GROUP*time ,random = ~time|ID,data=data_ori)
glht(result,
For question (a), do:
which(AA%in%BB)
Question (b) is very ambiguous to me. It makes little sense for your example
because all values of BB are in AA. Therefore I am wondering whether you
meant in question (a) that you want to find all values in BB that are in AA.
That's not the same thing. I am
Steven's solution is great, but it will only work if the rows are really
duplicates. If the data frame contains another variable whose values vary,
it will not work because then the rows are obviously unique.
df-data.frame(df,value=rnorm(11))
unique(df)
You would then have to make a decision,
Dear Ben,
Maybe the model converges more slowy than other models. Running more iterations
might solve the problem. Have a look at ?lme and ?lmeControl.
Best regards,
Thierry
PS R-sig-mixedmodels is a better list for questions on lme().
Hi,
Well, i need some help, practical and theoretical. I am wondering why the
fitdistplus (mle function) is returning an error for this code:
[code]
x1 - c(100,200,140,98,97,56,42,10,2,2,1,4,3,2,12,3,1,1,1,1,0,0);
plotdist(x1);
descdist(x1, boot =1000);
y- sum(x1);
d= as.vector(length(x1));
On Jul 26, 2011 Lao Meng wrote:
glht(result, linfct = mcp(f_GROUP=Tukey) )
Error in `[.data.frame`(mf, nhypo[checknm]) : undefined columns selected
It is almost certainly the underscore in the name (_) that is causing the
problem. Try putting the term in quotes (f_GROUP).
Regards, Mark.
On Tue, 26 Jul 2011, Lao Meng wrote:
Hi all:
There's a question about glht function.
My data:data_ori,which inclue CD4, GROUP,time.
f_GROUP-factor(data_ori$GROUP)
f_GROUP is a factor of 3 levels(0,1,2,3)
result - lme(sqrt(CD4) ~ f_GROUP*time ,random = ~time|ID,data=data_ori)
This is a theoretical issue. It is impossible for beta-distributed values to
take the value of 0 or 1. Hence, an attempt to fit a beta distribution to a
vector containing these values fails.
HTH,
Daniel
baxy77 wrote:
Hi,
Well, i need some help, practical and theoretical. I am wondering
Hi,
I have an intraday timeseries of financial data (see below) which has gaps
due to market opening and closing hours. I am trying to plot it, but the
time gap is always visible in the plot. I tried converting data to xts, zoo,
timeSeries and plotting it with different functions i.e. plot.xts,
On 2011-07-26 01:13, Achim Zeileis wrote:
On Tue, 26 Jul 2011, Lao Meng wrote:
Hi all:
There's a question about glht function.
My data:data_ori,which inclue CD4, GROUP,time.
f_GROUP-factor(data_ori$GROUP)
f_GROUP is a factor of 3 levels(0,1,2,3)
result- lme(sqrt(CD4) ~ f_GROUP*time
On Tue, Jul 26, 2011 at 12:23 AM, Amelia McNamara
amelia.mcnam...@stat.ucla.edu wrote:
I am trying to create a plot that has multiple plot characters for
each point (e.g. a point within a triangle, a triangle within a
square, etc). The workaround I have found to do this is by plotting
twice,
On Tue, 26 Jul 2011, Peter Ehlers wrote:
On 2011-07-26 01:13, Achim Zeileis wrote:
On Tue, 26 Jul 2011, Lao Meng wrote:
Hi all:
There's a question about glht function.
My data:data_ori,which inclue CD4, GROUP,time.
f_GROUP-factor(data_ori$GROUP)
f_GROUP is a factor of 3 levels(0,1,2,3)
Hi,
I am trying to do a linear regression but I want one of my variables to not
generate a coefficient. E.g. what I want to do is fit for y=a+b+c but
forcing the coefficient of b to be 1. Is this possible?
I have been fitting y-b=a+c but I have found that when I recalculate y it is
not close
Dear Joe,
You need to use offset()
lm(y ~ a + offset(b) + c)
Best regards,
Thierry
ir. Thierry Onkelinx
Instituut voor natuur- en bosonderzoek
team Biometrie Kwaliteitszorg
Gaverstraat 4
9500 Geraardsbergen
Belgium
Date: Mon, 25 Jul 2011 11:39:22 -0700
From: lukescore...@gmail.com
To: r-help@r-project.org
Subject: [R] Is there an R program that produces optimal solution/mix of
multiple samples' varying volumes and values
Sorry about the lengthy subject
At 15:47 25/07/2011, Joshua Wiley wrote:
Hi Martin,
Off hand I do not know of a pre-defined function to do it, but the
z-transformation is just the inverse hyperbolic function, the mean
is just the mean, and the back transformation is the hypoerbolic
function so...
x - c(.5, .4)
Date: Mon, 25 Jul 2011 19:03:08 +0100
From: jbustosm...@yahoo.es
To: r-help@r-project.org
Subject: [R] Life Cycle Assessment with R.
Hello everyone,
There's something really important about climate change and how many
institutions around the globe
Try the logLik function with your model !
Arnaud
Date: Mon, 25 Jul 2011 12:16:37 +0100
From: Partha Pratim PATTNAIK p.pattn...@sms.ed.ac.uk
To: r-help@R-project.org
Subject: [R] How to find the likelihood of a null model in R
Message-ID: 20110725121637.jo1u2ctuxy8kw...@www.sms.ed.ac.uk
On Tue, Jul 26, 2011 at 4:48 AM, djdjoko djdj...@googlemail.com wrote:
Hi,
I have an intraday timeseries of financial data (see below) which has gaps
due to market opening and closing hours. I am trying to plot it, but the
time gap is always visible in the plot. I tried converting data to xts,
On 07/26/2011 02:40 AM, Naomi Robbins wrote:
Hello!
It's a shoot in the dark, but I'll try. If one has a total of 100
(e.g., %), and three components of the total, e.g.,
mytotal=data.frame(x=50,y=30,z=20), - one could build a pie chart with
3 sectors representing x, y, and z according to their
On 07/25/2011 05:49 PM, ATANU wrote:
i am trying to make 3-d barplots,pie-charts in R,just like Excel. i have used
rgl , but that does not produce beautiful graphs like excel(i dont need to
rotate the graph). can anyone help me to produce graphs,just like excel.
thanks in advance
Hi ATANU,
See
On Tue, Jul 26, 2011 at 7:11 AM, Gabor Grothendieck
ggrothendi...@gmail.com wrote:
On Tue, Jul 26, 2011 at 4:48 AM, djdjoko djdj...@googlemail.com wrote:
Hi,
I have an intraday timeseries of financial data (see below) which has gaps
due to market opening and closing hours. I am trying to plot
On 26-Jul-11 11:26:14, Jim Lemon wrote:
On 07/26/2011 02:40 AM, Naomi Robbins wrote:
Hello!
It's a shoot in the dark, but I'll try. If one has a total of 100
(e.g., %), and three components of the total, e.g.,
mytotal=data.frame(x=50,y=30,z=20), - one could build a pie chart with
3 sectors
And at that point, one is essentially doing a meta-analysis. For example:
library(metafor)
ri - c(.5, .4)
ni - c(40, 25)
res - rma(ri=ri, ni=ni, measure=ZCOR, method=FE)
predict(res, transf=transf.ztor, digits=2)
pred se ci.lb ci.ub
0.46 NA 0.24 0.64
You also get the CI (in addition to
Hello,
I'd like to compare the predictive power of two independent simple
regression models, which relate to data sets with different numbers of
points (n1 and n2). I think I could use a F-test to compare both
residual variances, but I don't know whether it's a good idea or not.
To my
Surv(entry-time, last-follow-up-time, status-at-last-follow-up)
This works in survfit, coxph, and coxme. Left truncation is currently
not implemented for survreg.
In your case both times above would be time since event.
Terry T.
--- begin included message --
I have a fairly simple
Hi,
I wanted to know if there existed an good implementation in R of the
following classical subset Integers :
$P_{n,m} = {k_1, \ldots, k_m \in \mathbb{N} : k_1 + \ldots + k_m = n }$
for any integers $m n$. There is an obvious not optimal code which would
be to run through $m$ sums and put a
I am using the latest version of R 2.13.1 and need to load the following
packages;
library(maps)
library(mapdata)
library(mapproj)
library(lattice)
library(tgp)
library(spatstat)
library(akima)
I get the follwing error when loading the packages, maps, mapdata and
mapproj
Error: package 'maps'
You could doa manual uninstall. Read windows FAQ for Windows on the
entries the R installation program makes to the windows registry. It
also describes how to remove the registry entries. Once you have done
this you can simply delete the R directory. You will be missing a
very good piece of
There is something very odd about your data to give se values that are
so very large. Usually, this means that the data is deterministic: some
combination of predictors is able to separate a subset with no events
from all the others. This leads to a problem where the log-likelihood
is maximum
[Follow-up -- see at end]
On 26-Jul-11 12:03:46, Ted Harding wrote:
On 26-Jul-11 11:26:14, Jim Lemon wrote:
On 07/26/2011 02:40 AM, Naomi Robbins wrote:
Hello!
It's a shoot in the dark, but I'll try. If one has a total of 100
(e.g., %), and three components of the total, e.g.,
Hello,
I'm new to R and work with psychometrics (Item response theory). I was using
a package called cirt to model a data that contains only 0s, 1s and 9s. The
package is basically used for Monte Carlo Markov Chain estimation of
parameters and I specified 5000 iterations. The model seems to
I don't know if it could help, but under Mac os, you have to re-install all
your packages after installing a new version of R.
just use the following command :
install.packages(package.name,dependencies = TRUE);
--
View this message in context:
Well, please install them for arch=i386 !
And do note what the rw-FAQ and the @ReadMe said about not reporting
here, and what you needed to do if a package was not available.
(FWIW, my machine correctly finds that arch for those packages. Are
you *sure* you installed them for R 2.13.1? Try
Hi
this might be a little bit off topic, but here it goes: lets assume I have
the following:
set.seed(13)
dat1 - rnorm(2000, mean=10, sd=10)
dat2 - rnorm(100, mean=10, sd=20)
d.all - density(dat, n=1024)
d.co - density(x[[v]], , from=min(d.all$x),
On Jul 26, 2011, at 8:43 AM, Terry Therneau wrote:
There is something very odd about your data to give se values that are
so very large. Usually, this means that the data is deterministic:
some
combination of predictors is able to separate a subset with no events
from all the others. This
This is not very confusing. It is the exact same error in the sense that this
time the values of x1 are not only outside the interval (0-1) but within
[0-1] as in your first example, but this time they are also outside [0-1].
The reason is that you did not divide x1 by sum(x1) this time. In other
1. This thread belongs on the r-sig-mixed-models list ,
2. I would bet that for some of the data sets, the model is
unbalanced/overspecified, causing te convergence issues (imagine
climbing a narrow ridge by zig-zagging back and forth across it).
Cheers,
Bert
On Tue, Jul 26, 2011 at 12:56 AM,
For the list to have any hope of helping, we need at the very least a small
reproducible example, one that comes with data and gives the same
error. We don't even know what commands you issued.
Sarah
On Tue, Jul 26, 2011 at 8:48 AM, Rohini Sen r...@nbme.org wrote:
Hello,
I'm new to R and
On 26/07/2011 8:48 AM, Rohini Sen wrote:
Hello,
I'm new to R and work with psychometrics (Item response theory). I was using
a package called cirt to model a data that contains only 0s, 1s and 9s. The
package is basically used for Monte Carlo Markov Chain estimation of
parameters and I
On Mon, Jul 25, 2011 at 8:03 PM, Jose Bustos Melo jbustosm...@yahoo.es wrote:
Who knows if there's people working in Life Cycle Assesment (carbon emition)
with R? or If there's someone interested in doing a package about it, please
let me know!
Some of the places to check would be:
rseek.org
I am trying to used ordgee from geepack for an ordinal dataset.
When I write the code it returns
Warning message:In binomial(link) : use of binomial(link=link) is deprecated ,
but the program runs.
Even when I run your example for ohio and respdis, it returns the same
error.
Please guide me
Hi, but why we do the difference : ltemp - 2 * diff(tfit$loglik[1:2]) ??
Where I can find information about Integrate Likelihooh and null like
lihood??
Thank you very much,
Roby
--
View this message in context:
Daniel, thanks for the answer.
I will try to make myself i little bit clearer. Doing step by step I would
have (using a loop trough the lines of 'A'):
1. AA[1] is 4. As so, I would have to compare A1[1] = 20 and A2[1] =3 with
B1 B2 B3
B[3,2:4] 7 11 NA
beacause BB[3]=4. Since there is
DODamen represents the dissolved oxygen values and is a valid predictor. I'm
not sure how the with(test1, table(Depart, DODamen)) helps. It does show that
the DODamen column is populated and in use since it lists out a table of
relation between the Depart and DODamen columns.
Thanks for the
Hello,
I have a population of 2000+ zoo time series (but my environment also
contains objects that are not zoo time series). I'm trying to calculate the
latest 90 days Z-Score of all zoo time series, using the following code:
LZS-function(ser) {
temp-window(ser,start=Sys.Date()-90)
Hi Merik,
Please keep the mailing list copied.
On Tue, Jul 26, 2011 at 6:44 AM, Merik Nanish merik.nan...@gmail.com wrote:
You can convert my data into a dataframe simply by dat - data.frame(id,
month, value). That doesn't help though.
Can you be more specific? What is the problem you are
William Dunlap wrote:
$ cut(c(20.8, 21.3, 21.7, 23, 25), 2, dig.lab=1)
[1] (21,23] (21,23] (21,23] (23,25] (23,25]
Levels: (21,23] (23,25]
So the first number, 20.8, get put in the interval (21,23], which seem
strange. I can see why this could happen, though, as perhaps the 20.8 is
On a point of information, the beta distribution is indeed
defined for x = 0 and, respectively, for x = 1 so long as
the parameters a=shape1 and b=shape2 are respectively
not less than 1:
dbeta(x,a,b) = (x^(a-1))*((1-x)^(b-1))/Beta(a,b)
When a=1 and b=1 we have the uniform distribution on
Hi
Re: [R] Big data and column correspondence problem
Daniel, thanks for the answer.
I will try to make myself i little bit clearer. Doing step by step I
would
have (using a loop trough the lines of 'A'):
I am not sure if you are successful in your clarifying.
1. AA[1] is 4. As so, I
On Tue, Jul 26, 2011 at 9:48 AM, thierrydb thierr...@gmail.com wrote:
Hello,
I have a population of 2000+ zoo time series (but my environment also
contains objects that are not zoo time series). I'm trying to calculate the
latest 90 days Z-Score of all zoo time series, using the following
Hi,
I've done a linear fit on my data and I would like to get back the a (time)
coefficient ...
mod-lm(res_sql2$Lx0x~0+time)
result-data.frame()
result-coef(mod)
print(result)
print(result)
[1] result
time
0.02530191
But I would like just the value 0.02530191 ... I tried result$time but
The ls() function return the names of the objects not the objects.
So if you have :
a - c(1,2,3)
b - c('A','B','C')
ls() will return :
a b which only 2 letters.
-
Christophe Poulet
GIGA-Research.
Human Genetics Dept.
Liège, Belgium.
--
View this message in context:
Hello
I have been trying to figure out how to randomly select a number of records. I
have a column ID which contains 16 individuals (numbered 1-16), each
individual has a few hundred rows of GPS locations associated with it. I am
trying to select a random sample of these individuals (with all
Dear R-Gurus
I am a PhD student from South Africa working on chimpanzee behaviour.
I am looking at patterns of shade utilization and am using generalized
linear mixed models to examine the effects of various factors on
whether chimpanzees choose to spend time in the sun or shade. I
realise that
# for, e.g., four random individuals
x.sub-data[data$ID %in% sample(1:16, 4),]
On Tue, Jul 26, 2011 at 10:10 AM, Soanes, Louise
louise.soa...@liverpool.ac.uk wrote:
Hello
I have been trying to figure out how to randomly select a number of records.
I have a column ID which contains 16
On Tue, Jul 26, 2011 at 4:21 PM, ascoquel ascoq...@yahoo.fr wrote:
Hi,
I've done a linear fit on my data and I would like to get back the a (time)
coefficient ...
mod-lm(res_sql2$Lx0x~0+time)
result-data.frame()
result-coef(mod)
print(result)
print(result)
[1] result
time
Will:
result$coef[[2]]
Give you want you want?
Jeremy
On 26 July 2011 08:21, ascoquel ascoq...@yahoo.fr wrote:
Hi,
I've done a linear fit on my data and I would like to get back the a (time)
coefficient ...
mod-lm(res_sql2$Lx0x~0+time)
result-data.frame()
result-coef(mod)
Thank you, this works well - overall, I now have pretty good control over the
appearance of the L-hand side of the plot and relative distance between axis
line, ticks, tick labels and axis title by adjusting pad1, pad2 and the plot
position in the print statement. I am sure it is not the best
It was
result[[1]]
I have no intercept
Thanks !!!
--
View this message in context:
http://r.789695.n4.nabble.com/function-lm-get-back-the-coefficient-tp3696109p3696230.html
Sent from the R help mailing list archive at Nabble.com.
__
I am making final adjustments to a multi-plot figure using basic and lattice.
In the lattice plot, I would like to include a legend that matches the
format of the legends in the other plots, which consist of appropriately
colored squares (pch 22) followed by text. In lattice, I was able to get a
I am trying to used ordgee from geepack for an ordinal dataset.
When I write the code it returns
Warning message:In binomial(link) : use of binomial(link=link) is deprecated ,
but the program runs.
fit - ordgee(ordered(phy) ~ age+BMI++convert, id=id, data=newdata,
int.const=FALSE)
Warning
Please do *NOT* repeat post, nor send HTML (see the posting guide).
This is repeat of
https://stat.ethz.ch/pipermail/r-help/2011-July/284829.html
There is no error here. If there had been, the posting guide asks you
to contact the package maintainer, who will (I hope) be able to
explain the
On 2011-07-26 08:47, marcel wrote:
I am making final adjustments to a multi-plot figure using basic and lattice.
In the lattice plot, I would like to include a legend that matches the
format of the legends in the other plots, which consist of appropriately
colored squares (pch 22) followed by
On Jul 26, 2011, at 12:28 PM, Anera Salucci wrote:
I am trying to used ordgee from geepack for an ordinal dataset.
When I write the code it returns
Warning message:In binomial(link) : use of binomial(link=link) is
deprecated ,
but the program runs.
fit - ordgee(ordered(phy) ~
OK, Ive done more research, and I think that what I am looking for is
repeated cross section or pseudo-panel estimators. Does anyone know if
these have been implimented inany r package?
--
View this message in context:
On Tue, 2011-07-26 at 16:43 +0100, Barry Rowlingson wrote:
On Tue, Jul 26, 2011 at 4:21 PM, ascoquel ascoq...@yahoo.fr wrote:
Hi,
I've done a linear fit on my data and I would like to get back the a (time)
coefficient ...
mod-lm(res_sql2$Lx0x~0+time)
result-data.frame()
Not quite, Gavin. You have to assign the value of unname back:
z - structure(2, names=a)
unname(z)
[1] 2
z
a
2
zz - unname(z)
zz
[1] 2
names(z) - NULL
z
[1] 2
Cheers,
Bert
On Tue, Jul 26, 2011 at 1:18 PM, Gavin Simpson gavin.simp...@ucl.ac.uk wrote:
On Tue, 2011-07-26 at 16:43 +0100,
On Tue, 2011-07-26 at 13:42 -0700, Bert Gunter wrote:
Not quite, Gavin. You have to assign the value of unname back:
Yes; thought that would be a given. The point was to give a simple
example to show that `unname()` removes names/dimnames.
G
z - structure(2, names=a)
unname(z)
[1] 2
z
Hi,
I'm trying to get a plot that looks somewhat like the attached image
(sketched in word).
I think I need somthing called a rose diagram? but I can't get it to do what
I want. I'm happy to use any library.
Essentially, I want a circle with degree slices every 10 degrees with 0 at
the top
Hi
I'm trying to replicate Smith et al.'s
(http://www.sciencemag.org/content/330/6008/1216.abstract) findings by
fitting their Gompertz and logistic models to their data (given in
their supplement). I'm doing this as I want to then apply the
equations to my own data.
Try as a might, I can't
This is much clearer. So here is what I think you want to do. In theory and
practice:
Theory:
Check if AA[i] is in BB
If AA[i] is in BB, then take the row where BB[j] == AA[i] and check whether
A1 and A2 are in B1 to B3. Is that right? Only if both are, you want the
indicator to take 1.
Here
If I understand you correctly, you are looking for partitions of an integer, so
look at Robin Hankin's package 'partitions'
David L. Reiner
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
Behalf Of David Pham
Sent: Tuesday, July 26, 2011
Hello
I have a huge file (not an R-file) in which the first column is a string
with date, hour, minutes and seconds (For instance, 31-Jul-2010 23:59:00).
I tried as.Date but the error msg was Error in charToDate(x) :
character string is not in a standard unambiguous format.
I have checked
Making the second layer of legend have full transparency seems to have fixed
the issue. My first layer has to have a white background because I have a
colored grid behind (sorry that didn't make it into my toy example).
I did try to use type=Xlib, but I get the error libpng warning:
Application
I am trying to tweak how my categorical x-axis labels are formatted in
my bar graph. Specifically, I would like to a) decrease the spacing
between lines (e.g. spacing between Dialium and guianensis) b) right
justify the text and c) have each species name align with the center
of the
I am a graduate student who's just starting to use more advanced
statistics and is completely new to R. I'm looking for whether
parasite prevalence varies with region. I'd like to try using just a
subset of the data defined by one these covariates, sex, looking at
data only from adults, without
Hi guys,
I need your help with the boxplot.
I've to create a boxplot starting from a table (.csv) in which there are the
result of a test.
Each column is a question and the rows are the answer of the respondents
(from 0 to 5).
Some answer is missing and has been filled with NA.
the table has been
Dear r-helpers,
I would be very grateful if you could post the message below on the r-help
discussion board. Thank you very much!
Best Wishes,
Pawel
Hello R community,
I am generating lots of results using the fisher.test function, testing many
2x2 tables of SNPs for association with a
?strptime
--
Clint BowmanINTERNET: cl...@ecy.wa.gov
Air Quality Modeler INTERNET: cl...@math.utah.edu
Department of Ecology VOICE: (360) 407-6815
PO Box 47600FAX:(360) 407-7534
Olympia, WA 98504-7600
Tia:
Please first read the Help file -- ?bwplot
and especially note the examples at bottom.
-- Bert
On Tue, Jul 26, 2011 at 12:39 PM, Tia Molte moltenigiuse...@alice.it wrote:
Hi guys,
I need your help with the boxplot.
I've to create a boxplot starting from a table (.csv) in which there are
Hello again
I do apologize for the previous email without any useful information that
can lead to an answer. To those who felt offended by to no avail, I do
apologize again. I guess I will be always a newbie as far as R is
concerned.
The date format was wrong and although I have tried to use
On Tue, Jul 26, 2011 at 7:18 PM, Eduardo Mendes emammen...@gmail.com wrote:
Hello again
I do apologize for the previous email without any useful information that
can lead to an answer. To those who felt offended by to no avail, I do
apologize again. I guess I will be always a newbie as far
Roland Sookias r.sookias at gmail.com writes:
Hi
I'm trying to replicate Smith et al.'s
(http://www.sciencemag.org/content/330/6008/1216.abstract) findings by
fitting their Gompertz and logistic models to their data (given in
their supplement). I'm doing this as I want to then apply the
This is clearly a message for the R-help mailing list, since it was
sent to the R help mailing list.
fisher.test(x)[1]
Jeremy
On 26 July 2011 14:51, Zmarz, Pawel pawel.zmar...@imperial.ac.uk wrote:
Dear r-helpers,
I would be very grateful if you could post the message below on the r-help
Hi:
Here are a few options. The most important thing I'd recommend is to
rotate the bar plot so that the species can be read easily. In the
process, you also get the right justification you want in a readable
text size. Here's an example with the barplot() function in base R:
set.seed(103)
x -
Yes.
According to your suggestion,I modified my code.It works well.
Thanks you very much.
My best.
2011/7/26 Peter Ehlers ehl...@ucalgary.ca
On 2011-07-26 00:16, Lao Meng wrote:
Hi all:
There's a question about glht function.
My data:data_ori,which inclue CD4, GROUP,time.
Dear R User,
I am wondering if there is a way to generate correlated multivariate
non-normal distribution?
For example, I want to generate four correlated negative binomial
series with parameters r=10, p=0.2, based on the correlation
coefficient matrix
| 1 0.9 0.8 0.8 |
| 0.9 1 0.8
Afternoon R help,
I want to run Rasch/IRT analyses using the ltm package, however, I am
using large scale survey data which requires weighting for accurate
results. I attempted to create a weighted object to insert into the
formulae of the ltm packages, however, the survey data only includes
30
On Wed, Jul 27, 2011 at 2:58 PM, Erica Crome erica.cr...@mq.edu.au wrote:
Afternoon R help,
I want to run Rasch/IRT analyses using the ltm package, however, I am
using large scale survey data which requires weighting for accurate
results. I attempted to create a weighted object to insert into
Hello,
I have indicators for the present of absent of a snps in columns and the
categorey (case control column). I would like to extract ONLY the tables and
the indices (SNPS) that give me 2 x 3 tables. Some gives 2x 2 tables when
one of the allelle is missing. The data look like the matrix snpmat
Hi Jim,
Here is one way:
# data
x - structure(list(category = structure(c(1L, 1L, 1L, 2L, 2L, 2L,
1L, 1L, 2L), .Label = c(case, control), class = factor),
SNP1 = c(1L, 0L, 1L, 0L, 1L, 0L, 1L, 0L, 1L), SNP2 = c(0L,
1L, 2L, 1L, 2L, 0L, 0L, 1L, 0L), SNP3 = c(2L, 1L, 2L, 0L,
1L, 0L, 2L,
Dear all,
Does any one know if any R package or function can do Ordinary Least Products
regression? Many thanks!
Bill
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide
99 matches
Mail list logo
