Homework? Ask your prof.
cheers,
Rolf Turner
On 26/07/12 11:48, stuco wrote:
Hey, I'm an R noobie and I have been trying calculate SSEr and SSEc in order
to determine if there is sufficient evidence to include second-order terms
in my model, but I have no idea what command to use
Having had a quick look at the source code for read.table.ffdf, I
suspect that using 'NULL' in the colClasses argument is not allowed.
Could you try to see if you can use read.table.ffdf with specifying
the colClasses for all columns (thereby reading in all columns in the
file)? If that w
On Jul 26, 2012, at 09:07 , Rolf Turner wrote:
>
> Homework? Ask your prof.
Not necessarily, Rolf. Project work perhaps, but could also be completely
extracurricular (knows theory and textbook notation, now has a real data set).
I think a few hints could be in order:
(a) Figure out how to f
On Jul 25, 2012, at 14:56 , arun wrote:
> Hi,
>
> From the ANOVA results, you could get MSE and MS of group. MSE is basically
> sigma^2 error. MS group of MS between group contains sigma^2
> error+replication*sigma^2group (please check the formula. It can be slightly
> different when the m
Hello,
Simple, look carefully at what you have, rho is a function (type
closure), the vector is rhoes.
And you are trying to print rho[i], the i-th function in the "vector"
called rho.
print( rhoes[i] ) # works
Hope this helps,
Rui Barradas
Em 26-07-2012 06:40, Rafael_Leon escreveu:
Hi e
David's ?measurement
measurement(mz$age_variable) <- "interval"
# where age_variable is the unstated item in that "select" list
is what I use in similar circumstances.
Where it seems to come from is the SPSS users habit of setting value
labels on various categories of user-missing values - so a su
Dear all,
I would like to calculate the optimal cut off (threshold) of a test using
the Epi package. Here I am presenting some data based on the output of two
tests. I am interested in identifying the optimal cut off an its 95% CI.
Running the ROC() function with the Epi package I obtain a nice
On Wed, Jul 25, 2012 at 4:34 PM, mdvaan wrote:
> Thanks Gabor. That worked really well. I have been reading about the use of
> POSIX and regular expressions and I tried to use your example to see if I
> could ignore all matches in which the character preceding (rather than
> following) the match
Dear all,
I would like to save few variable-names with their values in a tabular form,
with that I mean
that files can be printed easily in R in a tabular form and also saved in a
ascii file that when one opens it see also the variables in a nice tabular
format.
IS that possible? Below a small
I have updated from ggplot2 0.8.9 on Windows to ggplot 0.9.0 (and then
0.9.1) and I can't find scale_y_logit() anymore ... On Mac, I can't
see scale_y_logit() in ggplot 0.9.0 either.
Am I missing anything?
Thanks
__
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ht
Hi Alex,
sprintf with left alignment might be a start:
cat(sprintf("%-7s| %-12d|
%-5d",paste("City",1:3,sep=""),(12345)%/%10^c(1:3),c(2,5,54433)),sep="\n")
#cat has also a file option, see ?cat.
but acutally your given output doesn't look like a conventional table
but much more like this:
cat(sp
example
but a wild guess: if your class information is numeric the default is to do
eps-regression, not classification. Use factors or specify the type you want to
use.
?svm
might help there.
On 25.07.2012, at 15:31, Meffy wrote:
> Hello users!
> I'm calculating a simple model using svm(
Dear Waheed,
As you correctly inferred, these are just warnings and dont need to bother you
now. The maintainer/author of the taylor.diagram function should be more
worried.
These warnings just say that in upcoming versions of R the functions within
taylor.diagram() will not work anymore. How
HI Namit,
I did posted a reply on nabble. Is that the solution you were looking for??
A.K.
- Original Message -
From: namit
To: r-help@r-project.org
Cc:
Sent: Thursday, July 26, 2012 1:32 AM
Subject: Re: [R] Arranging the data
Hi Arun,
Its the same question,but i need to arrange t
I got a regression model trained using the 'cubist' function of the
'Cubist' package. This works fine. I'm now searching for a graphical
representation of the regression model:
As the Cubist Algorithm is based on a forrest of trees (or a single tree
if 'committees==1'), is it somehow possible to a
Dear R helpers,
I try to use the 'delaunayn' function in the 'geometry' package for
Delaunay triangulation in 2 dimensions.
For the four following points, I get a warning message :
> coord=matrix(ncol=2,byrow=TRUE,c(622633,7073452,
+ 621228,7073517,
+
Hi guys,
does anyone know if there is the possibility to fit a gamma distribution
using ugarch?honestly i don't know if maybe is possible to fix some
parameters that reduce ghyp or ged in a gamma distribution..
thanks a lot
sara
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Dear Jan, thank you for your answer.
I am basically following the code Ive been using with read.table, where
x.class <- c('NULL', 'numeric','NULL','NULL','NULL', 'NULL', 'NULL')
has been working fine.
Reading all columns works with me but take much longer than allowed time
constrains.. (460 such
Hi,
I have some data that is log-normally distributed and I am using a glm, type
Gaussian to fit the logs.
I would like to know the expected values within the context of the lognormal
model.
I am unsure whether I have to use:
Expected_values= exp(fitted(model)+sqrt(summary(model)$dispersion)
Good afternoon,
I am using the function gcFitModel() from grofit package for modelling grass
growth pattern over time.
Besides the usual parameters that come out from the model, i.e. A, mu and
lambda, I am interested in finding the timepoint at which A is reached (what
I call the day_max) .
Look
Dear all,
I am new to R in general and ways to retrieve XML or JSON data in
particular. I have tried to get information through the XML package
and various websites without being able to do exactly what I want. I
hope someone of you can give me some help.
I want to retrieve information about movi
Hello all,
I am a newbie at R, with some experience in PERL.
I have a database table that contains the following data:
Name | Score
=== | =
Sachin T | 25
Sachin T | 53
Sachin T | 57
Sachin T | 34
Rahul D | 38
Rahul D | 31
Rahul D | 53
Ricky P | 7
Ricky P | 45
Ricky P
Dear all,
I was willing to use the library "rgl" to plot some 3D graphics, but
unfortunately, I wan't able to instal the library. The error message is below.
I would be very grateful if you could give me any clues about how I can solve
this.
Below you will find :
- installation from binaries
*..plus I get the following message after reading the whole set (all 7
columns):*
> read.csv.ffdf(file=csvfile, header=FALSE, skip=100, first.rows=1000,
> next.rows=1e7, VERBOSE=TRUE)
read.table.ffdf 1..1000 (1000) csv-read=0.02sec ffdf-write=0.08sec
read.table.ffdf 1001..10001000 (1000) cs
I am attempting to fit a partial credit model using the PCM() function in eRm.
My data set comprises 88 items in 43 individuals with possible scores 0,1,2,3.
Attempting either a PCM or RSM model fit simply hangs - no error messages but
no return to command prompt and a continuously spinning "wo
hey guys
I want two plots in one window with an overall title and with individual
titles for each plots.
my code:
par(mfrow=c(2,1))
bp_dirverq1=boxplot(dirverq1, col="orange",horizontal=TRUE, main="Q1
2012",cex.main=0.7)
bp_dirverq2=boxplot(dirverq2, col="orange",horizontal=TRUE, main="Q2
2012",
Hi,
I have a query on regression output generated by R.
> result=lm( Y~X , data=trail)
> summary(result)
After running this 2 statements the following output is generated.
Call:
lm(formula = Y ~ X, data = trail)
Residuals:
Min 1Q Median 3Q Max
-245.30 -90.77
HI,
Not sure if this is what you are looking for.
Try:
dat1<-read.table(text="
Postal Code | Superb
City1 | 2134 | 2
City2 | 254 | 5
City3 | 12 | 54433
",sep="|",header=TRUE)
Postal.Code1<-paste0("|",dat1$Postal.Code)
Supertb1<-paste0("|",d
I don't think it is provided and gamma is neither a special case of ghyp
nor ged.
Is there a reason for you to use the gamma distribution? The gamma
distribution only has support for positive number and thus impossible
for stock return.
Cheers,
M
On 26/07/12 09:52, saraberta wrote:
Hi guys
Does
> delaunayn(scale(coord),options="Qbb")
help?
Keith J
On 26/07/2012 08:26, Jean-Luc Dupouey wrote:
Dear R helpers,
I try to use the 'delaunayn' function in the 'geometry' package for
Delaunay triangulation in 2 dimensions.
For the four following points, I get a warning message :
> co
Looking at the source code for read.table.ffdf what seems to happen is
that when reading the first block of data by read.table (standard 1000
lines) the specified colClasses are used. In subsequent calls the
types of the columns of the ffdf object are used as colClasses. In
your case the
You probably have a character (which is converted to factor) or factor
column with a large number of distinct values. All the levels of a
factor are stored in memory in ff.
Jan
threshold schreef:
*..plus I get the following message after reading the whole set (all 7
columns):*
read.c
It is not appropriate to seek a cutoff for a test unless every patient is
known to have exactly the same utility function, which is extremely rare.
There is nothing wrong with developing probability models and stopping at
that point, deferring the decision to the possessor of the utility function.
Something like that should work, though you might need to construct
the formula as a string:
paste("y ~", names(x)[i])
instead.
More worrisome is the methodology: doing 10k regressions on a single
response is almost guaranteed to give spurious results. This
methodological mistake has different n
On Thu, Jul 26, 2012 at 3:59 AM, guruappa wrote:
> Hello all,
>
> I am a newbie at R, with some experience in PERL.
>
> I have a database table that contains the following data:
> Name | Score
> === | =
> Sachin T | 25
> Sachin T | 53
> Sachin T | 57
> Sachin T | 34
> Rahul D |
On Wed, Jul 25, 2012 at 11:41 PM, Manish Gupta
wrote:
> Hi,
>
> I am working on reporting and need some fancy image instead of barplot, pie
> etc. Like Clock and Speedometer. How can i draw myself in R?
Well, I don't know, because I have no idea what you look like :-)
But you might want to
fortune()
On Fri, Jul 27, 2012 at 12:26 AM, R. Michael Weylandt
wrote:
> On Wed, Jul 25, 2012 at 11:41 PM, Manish Gupta
> wrote:
>> Hi,
>>
>> I am working on reporting and need some fancy image instead of barplot, pie
>> etc. Like Clock and Speedometer. How can i draw myself in R?
>
> Well, I
On Thu, Jul 26, 2012 at 4:28 AM, phillen wrote:
> hey guys
>
> I want two plots in one window with an overall title and with individual
> titles for each plots.
>
> my code:
>
> par(mfrow=c(2,1))
> bp_dirverq1=boxplot(dirverq1, col="orange",horizontal=TRUE, main="Q1
> 2012",cex.main=0.7)
> bp_dirv
Dear All,
I would like to read the data file via read.csv (the 3rd line of the
following program) and the file name is stored in a dataframe. Since I have
several files to read, I store the file names as well as the sample period
inside a file “B_M2Q.csv” and I read the file name first, and th
Let me count the ways...
R supplies a number of different ways. Here is sample using basic R and some
other packages. Youprobably will need to install the packages (
?install.packages) to run any but aggregate().
mydata <- structure(list(Name = structure(c(4L, 4L, 4L, 4L, 2L, 2L, 2L,
3L, 3
1. m2q already is a data frame, so the m2qldf statement that follows
is completely unnecessary.
2. Please read ?read.table carefully, and especially the bit about the
stringsAsFactors argument. The problem is that by default character
strings are read in as factors, not characters strings. So you
Try changing your plot and outer margins.
par(mfrow=c(2,1), mar=c(4, 4, 2, 1), oma=c(0, 0, 2, 0))
Jean
phillen wrote on 07/26/2012 04:28:58 AM:
>
> hey guys
>
> I want two plots in one window with an overall title and with individual
> titles for each plots.
>
> my code:
>
> par(mfrow=c(2
On Thu, Jul 26, 2012 at 4:18 AM, Richard Ohrvall
wrote:
> Dear all,
>
> I am new to R in general and ways to retrieve XML or JSON data in
> particular. I have tried to get information through the XML package
> and various websites without being able to do exactly what I want. I
> hope someone of y
Sorry, got cut off...
On Thu, Jul 26, 2012 at 8:07 AM, Bert Gunter wrote:
> 1. m2q already is a data frame, so the m2qldf statement that follows
> is completely unnecessary.
>
> 2. Please read ?read.table carefully, and especially the bit about the
> stringsAsFactors argument. The problem is that
You can learn a lot from the help files. Check out the help files for the
lm() and summary.lm() functions
?lm
?summary.lm
You can extract the beta values in a few different ways.
These two will give you just the estimates in a vector:
coef(result)
result$coef
These two will give you the estim
I think it has been moved to the scales package but I've never used it so I
don't know the syntax.
John Kane
Kingston ON Canada
> -Original Message-
> From: fjpcaball...@gmail.com
> Sent: Thu, 26 Jul 2012 07:41:58 -0400
> To: r-help@r-project.org
> Subject: [R] scale_y_logit not present
It is really not clear what you want without some idea of what the variables
and data look like.
However if the data is in a couple of vectors you could try something like this
postal <- c(2134, 54, 12)
superb <- c(2,5,54433)
cities <- c("City1", "City2", "City3")
hds <- c("Postal.Code",
I'm currently developing several tools in R that I'd like to deploy for use
by multiple analysts in my research group. Many of the analysts have no
background in using R (but have plenty of experience with SAS), so part of
my effort will be in training them to use the new tools. Some of the
analyse
Dear Henrik
Thank you so much for the clarification.
Best regards
waheed
On Thu, Jul 26, 2012 at 9:32 PM, Henrik Singmann [via R] <
ml-node+s789695n463791...@n4.nabble.com> wrote:
> Dear Waheed,
>
> As you correctly inferred, these are just warnings and dont need to bother
> you now. The maint
Hi,
I have problem to read hdf4 files in R, I would be very grateful if
somebody can tell me how to deal with hdf4 files.
I can read hdf5 using "hdf5" and "h5r" packages, but these packages does
not work for hdf4 files.
I was trying to open some MODIS data files those are in hdf4 format.
Th
Hey,
I want to estimate a spatial linear mixed model y=X\beta+Zv+e with e\sim
N(0,4) and v\sim N(0,G).
G is a covariance matrix of a simultaneously autoregressive model (SAR) and
is given by
G=\sigma_v^2((I-pW)(I-pW^T))^{-1} where p is a spatial correlation parameter
and W is a matrix which desc
Hello all,
My problem is similar to Sergey's.
I could not try what Mario suggested, since doMPI is not available for
windows.
I have tried the Cedrick's commands, i.e.
cl.tmp = makeCluster(rep("localhost",2), type="SOCK")
registerDoSNOW(cl.tmp)
but this is instead increasing the time taken. Fo
Hello,
I'm attempting to plot error bars side by side rather than stacked on top
of each other with ggplot2. Here is the sample code I am using:
#Code
#Data
spd<-c("s","f","f","s","f","s","s","s","f","f","s","f")
r<-c(4.9,3.2,2.1,.2,3.8,6.4,7.5,1.7,3.4,4.1,2.2,5)
#Turn spd into a factor
spd.f<
Dear R-help list,
apparently lda from the MASS package can be used in situations with
collinear variables. It only produces a warning then but at least it
defines a classification rule and produces results.
However, I can't find on the help page how exactly it does this. I have a
suspicion (
Hi all,
I am using R package 'ordinal' to fit a cumulative logit ordinal model with
random effects. Does someone know
1) The optimization method used in estimating the parameters from the
marginal likelihood?
2) In Adaptive Gauss-Hermite Quadrautre, let f() be the function to be
intergrated. The
Thanks Gabor for your invaluable help! I learned a lot.
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___
You can try this one...
#
library(RCurl)
library(rjson)
ids <- c("tt0110074", "tt0096184", "tt0081568", "tt0448134", "tt0079367")
titles <- data.frame()
for ( i in 1:length(ids)) {
req <- paste("http://www.imdbapi.com/?i=";, ids[i] , "&tomatoes=TRUE",
sep="")
u <- getURL(req)
j <- fromJSON(u
Dear all,
I have been trying to plot hazard function in R for survival data, but in
vain.
Can anybody help me out in plotting hazard function in R?
Dr Suman Kumar
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Sent fro
HI,
I guess this should be the one:
dat1<-read.table(text="
Postal Code | Superb
City1 | 2134 | 2
City2 | 254 | 5
City3 | 12 | 54433
",sep="|",header=TRUE)
write.table(dat1,"dat7.txt",sep="|",quote=FALSE)
#contents of dat7.txt
Postal.Code|
thanks! the line command worked well.
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__
R-help@r-project.org mail
Hi,
I am trying to optimize a complex model calibration using a genetic algorithm.
I choose rgenoud package because it allows for easy parallelization through
snow package. I am on a Mac Pro with 2 x 2.66 GHz 6-Core Intel Xeon machine,
i.e. I have 12 CPUs available.
So I set the cluster optio
Ok, here a simple example. The file
http://r.789695.n4.nabble.com/file/n4637924/test.csv test.csv has 400 lines
containing 20 columns (1. column is class label, the other 19 are the
features).
So what I'm doing is
/
data <- read.csv(file="test.csv", head=F, sep=",")
names(data) <- c("Class","V1
Check str(lmobj). You can see the underlying structure of lm object. It is
actually a list. You can access its individual components with $ operator.
Bye
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Hi Friends,
I have a data frame X, and I want to add “%” & “$” in row 4 and 5
respectively. when I’m trying using below logic, I’m getting warning
message.
Can anyone help me out on this.
X:
Summary G Y R T
Accts582644 0 1226
AcctCov 230
Something as simple as dat1[2,2] <- "3%" where your data is in a data.frame
called dat will change 3 to 3% it but it changes everything in that column to
character if it was numeric.
str(dat1)
John Kane
Kingston ON Canada
> -Original Message-
> From: saileshchowd...@gmail.com
> S
Hi.
I don't know how did you create data frame X but if you check str(X) you
will see that you have one or more factors inside.
Try using stringsAsFactors=FALSE options while creating data frame.
Hope this helps.
Andrija
On Thu, Jul 26, 2012 at 3:23 PM, namit wrote:
> Hi Friends,
>
>
> I have
Is this what you mean?
dat1 <- data.frame( spd = c("s","f","f","s","f","s","s","s","f","f","s","f"),
r = c(4.9,3.2,2.1,.2,3.8,6.4,7.5,1.7,3.4,4.1,2.2,5))
myplot<-ggplot(dat1, aes(spd, r, colour = spd)) +
geom_errorbar(aes(ymin=3, ymax=5), width=.1) +
geom_point() + coord_flip
On Thu, Jul 26, 2012 at 7:59 AM, Carrie Wager wrote:
> I'm currently developing several tools in R that I'd like to deploy for use
> by multiple analysts in my research group. Many of the analysts have no
> background in using R (but have plenty of experience with SAS), so part of
> my effort will
It's sad, but not an impossible result with synchronization overhead
(though I wouldn't have guessed it would be that bad) -- can you try
it on a more reasonable benchmark?
Also, that advice might be somewhat out of date -- why not use the
tools provided in
library(parallel)
available for recen
You can make different lm objects by adding all predictors and compare them
with anova(lm1,lm2,lm3...). See if p value is not significant, the more
complex model is not appropriate.
Dr Suman Kumar
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HI,
I posted reply in nabble.
One more comment regarding your code. If your dataset is X. I wonder how it
changed to Z16. Probably, you have to use X[5,2:5]
A.K.
- Original Message -
From: namit
To: r-help@r-project.org
Cc:
Sent: Thursday, July 26, 2012 9:23 AM
Subject: [R] Get
Hi all,
I've fit a mixed linear model to some longitudinal data. I'm interested in the
differences in patterns of decrease in the dependent variable according to
group status, and my hypothesis particularly predicts a difference between the
groups in trajectory of change at between specific a
Hello,
I am running classification trees fro the purpose of predicting dividends.
I have training and test data sets, but am running to issues when
evaluating the prediction accuracy of the tree as it isn't a simple
'predict' formulation as the determinant variable is not a simple set of
classifyi
Hi,
I'm trying to using pspline in bic.surv{BMA}.
#
library(BMA)
library(survival)
data(veteran)
test.bic.surv<- bic.surv(Surv(time,status) ~
karno+pspline(age,df=3)+diagtime+prior, data = veteran, factor.type = TRUE)
summary(test.bic.surv, conditional=FALSE, digits=2
1. Your post is unacceptable (imho, of course). Read the bottom of
this message (re: "posting guide") and re-post properly.
2. This is not an R-help question. Re-post on r-sig-mixed-models -- or
perhaps on a non-R statistical forum like stats.stackexchange.com, as
this appears to have little to do
Sadly, your commonly held belief is wrong (imho) -- p
values/statistical significance are not a legitimate decision criteria
for model "appropriateness," especially scientific appropriateness.
That requires more careful consideration of a relevant "utility
function" (to use Frank Harrell's phrase),
It would be a useful additon to the help page to add
integrate(dnorm, lower = -1.96, upper = 1.96, mean = 2, sd = 1)
as an example.
Thanks,
Frank
Chicago
> Date: Mon, 23 Jul 2012 19:54:45 -0700
> From: ehl...@ucalgary.ca
> To: kri...@ymail.com
> CC: chicagobrownb...@hotmail.com; r-help@
Your post is unacceptable (imho, of course). Read the bottom of
this message (re: "posting guide") and re-post properly.
-- Bert
On Thu, Jul 26, 2012 at 9:30 AM, Orla Carey wrote:
> Hello,
>
> I am running classification trees fro the purpose of predicting dividends.
> I have training and test d
I might suggest:
integrate(dnorm, lower = -1.96 + 2 , upper = 1.96 + 2, mean = 2, sd = 1)
instead.
Incidentally, (and since I find this treatment of ... somewhat opaque)
I think anonymous first class functions are much easier:
integrate(function(x) dnorm(x, mean = 2, sd = 1), lower = -1.96, upp
"The following object(s) are masked from 'package:A'"
is equivalent to "The following object(s) from 'package:A' are masked"
and perhaps that might be a more universally understood phrasing.
I do find this better but I don't see any real need to change the status quo.
It becomes fairly obvious af
Hi,
You can try predict.rpart in rpart package for prediction on a test data.
?predict.rpart
Weidong
On Thu, Jul 26, 2012 at 12:30 PM, Orla Carey wrote:
> Hello,
>
> I am running classification trees fro the purpose of predicting dividends.
> I have training and test data sets, but am running t
Hello, All:
What references exist on how to link to C?
I'm familiar with sections 5.2 and 5.6 of the "Writing R
Extension" manual plus chapter 6 of Venables and Ripley (2000) S
Programming (Springer). From these, I get the following:
R storage mode C type
logical i
This will work:
model2007 <-
You can't start an identifier with a digit.
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
Behalf Of msherwood
Sent: Thursday, 26 July 2012 9:44a
To: r-help@r-project.org
Subject: [R] Package 'nlme' linear mixed
Hi there,
I would like to solve a simple equation in R
a^2 - a = 8.313
There is no real solution to this problem but I would like to get an
approximate numerical solution. Can someone suggest how I can set this up?
Thanks in advance,
Diviya
[[alternative HTML version deleted]]
___
Sorry it is important for me to constrain the value of 'a' between c(0,1)
On Thu, Jul 26, 2012 at 4:48 PM, Diviya Smith wrote:
> Hi there,
>
> I would like to solve a simple equation in R
>
> a^2 - a = 8.313
>
> There is no real solution to this problem but I would like to get an
> approximate nu
> -Original Message-
> From: r-help-boun...@r-project.org [mailto:r-help-bounces@r-
> project.org] On Behalf Of Diviya Smith
> Sent: Thursday, July 26, 2012 1:50 PM
> To: r-help@r-project.org
> Subject: Re: [R] Solving quadratic equation in R
>
> Sorry it is important for me to constrain t
On Thu, Jul 26, 2012 at 5:16 PM, Diviya Smith wrote:
> Thank you for pointing me to the uniroot function?
>
> Is there a way to constrain this solution so that it only gives me values
> of 'a' between c(0,1)?
>
> I tried using nlminb and for some reason it always estimates a = 0, even
> when I cha
I have two data frames. One with a matrix of months and the other with a
matrix of values. The two dataframes correspond to each other. I would like
to sum up all the values in by month.
What would be an efficient way to do this?
a=C(2,3,5,2,3,5)
b=c(2,6,3,2,6,3)
c=c(2,6,7,2,6,5)
months <- dat
Dear list
I have a data set involving binary responses (successes failures) for which
some explanatory variables result in a quasi complete separation problem.
To deal with the separation problem I tried to run a glm with "bayesglm" in
the arm package.
However when I try to compare different bay
Hi Arun,
Bymistake it kept X instead of Z16.
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Hi Dennis,
Part of my problem could be that I'm unsure how to nest another variable
withn spd.f. Perhaps if I give a better explanation of my goal things will
make more sense.
My intent is to calculate two sets of confidence intervals to show the
benefits of a DOE approach versus a Non-DOE appro
Hi all, is there a package for converting R code into C code?
Thanks.
Suman
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The values need not be a data frame as the number of unique months
could be different among the columns, right?. So you're going to have
to rethink your data structure. Probably as a list.
Once you get that straight,?tapply and friends should make it trivial,
if I understand you correctly.
-- Ber
I've searched hard in texts, email threads, FAQs etc. and cannot find out how
to successfully utilize sub-directories below my WorkingDirectory. I can
create them, and create R objects within the sub-directories but (a) the
objects() command lists ONLY the objects in the WorkingDirectory and no
Hello,
I am using names function to name an array.
It works first time when I use *as.numeric(names(myVar1)*
However, at a place later, when I tried to use a very similar line of code
*as.numeric(names(myVar2)*, it always returned 'numeric(0)' (or if I only
type 'names(myVar2), it gave me NULL'.
If you want a recommendation, why not use the one that comes with the manual
page for density():
?density
Under bw
"The default, "nrd0", has remained the default for historical and
compatibility reasons, rather than as a general recommendation, where e.g.,
"SJ" would rather fit, see also V&R (20
Hello,
Surely one of
f <- function(a) a^2 - a - 8.313
curve(f, from=0, to=1)
# zeros of f
root <- polyroot(c(-8.313, -1, 1))
ifelse(Im(root) == 0, Re(root), root)
# minimum of f
optimize(f, interval=c(0, 1))
Hope this helps,
Rui Barradas
Em 26-07-2012 22:13, Nordlund, Dan (DSHS/RDA) escreve
You asked why you could not get two discriminant functions and that question
was answered. The number of discriminant functions is one less than the
number of groups (assuming you have more variables than groups). Now you are
asking a different question. How to plot the discriminant boundary betwee
Hello,
About the cause of your problem, it's difficult to have an idea without
any code. But maybe it's helpfull to know that 'names' is an attribute,
so you can see its value with
attr(x, "names") # just 'names' attribute
attributes(x)# values of all attributes
Hope this helps,
Ru
Actually you probably want the full manual. It is here
http://cran.r-project.org/doc/manuals/fullrefman.pdf
--
David L Carlson
Associate Professor of Anthropology
Texas A&M University
College Station, TX 77843-4352
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