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On 12/05/13, 16:11 , Sarah Goslee wrote:
> Adding the argument all.x=TRUE to merge() will retain the NA
> values, but the only reliable way I've found to preserve order with
> NA values in a merge is to add an index column to x, merge the
> data, sort
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On 12/05/13, 16:37 , arun wrote:
> Hi, Try ?join()
>
> library(plyr)
Well - what would we do without Hadley ...
He solved many problems we didn't know we would have soon...
Cheers,
Rainer
> y$ref <- y$id
>> join(x,y,by="ref")
> ref id val 1 N
Hi all,
I am working with data.table objects within nested foreach loops and I am
having trouble creating the results object the way I would prefer.
Code below with sample data:
library(iterators)
library(data.table)
library(foreach)
#generate dummy data
set.seed(1212)
sample1 <- data.frame(par
The following question is inspired by Jun's problem, which resembles some
of my own problems, but goes off on a tangent about applying plot3D from
Karline Soetart.
On Thu, Dec 5, 2013 at 11:52 PM, Bert Gunter wrote:
>
> Your comment that:
>
> " I can see the critical point here is to find a rig
Hello everyone,
I'm trying to generate a sequence that consists of random numbers and the
following algorithm works well
###
a <- 0.08
b <- 0.01
T <- 90
t <- 0:T
alpha <- 1
e <- rnorm(T, mean = 0, sd = 0.1)
d <- c( runif(1,0, a*T), rep(0, T-1) )
for (i in 2:T)
{
d[i] <- alpha * d[i-1] +
I am using the step() function to select a model using backward
elimination, with AIC as the selection criterion. The full regression
model contains three predictors, plus all the second order terms and
two-way interactions. The full model is fit via lm() using two different
model formulae. One
Uh, no. You are forgetting to take the square root of 10, and to divide
by the square root of 12.
The variance of Y is (exactly) (56^2 - 1)/12, so the variance of Y-bar
is this quantity over 10,
so the standard deviation of Y-bar is sqrt((56^2 - 1)/12)/sqrt(10).
Which is approximately
(ig
The latest code I have put together is this. Could you point out what is
missing here ?
#Reference values plotted on x-axis. These are constant.
#These values could be time of day. So every day at the same
#time we could collect other measurements
referenceset <- data.frame(c(5,10,15,20,25,30
The model you are fitting is a random-effects model and does not include any
potential moderators/covariates. Therefore, the estimated intercept of that
model is *the* estimated/predicted (average) effect and it applies to each
study. That is why the predict function also just gives you that val
One more thing ... You used the command:
iat_result = rma(yi=Mean, vi=Variance_rounded, ni=N, sei=Std_error,
slab=Study_Name, subset=(Country == "AUT"), data=cma_iat, method="HS")
This probably does not do what you want it to do. First of all, if you specify
vi, there is no need to specify sei
Hej all,
actually i try to tune a SVM in R and use the package "e1071" wich works
pretty well.
I do some gridsearch in the parameters and get the best possible parameters
for classification.
Here is my sample code
type<-sample(c(-1,1) , 20, replace = TRUE )
weight<-sample
What value do you want d to take on if it is outside that interval?
Here is an example where if d is outside the interval, it is assigned to be
one of the interval endpoints.
minx <- 0
for(i in 2:T) {
x <- alpha * d[i-1] + e[i]
maxx <- a*(T-i)
if(x < minx) {
d[i] <- minx
} else {
if(x > maxx)
At 21:05 05/12/2013, Alma Wilflinger wrote:
Hi,
I am struggling a bit with creating a forest plot containing the
predicted effect size. As seen in other studies these effect sizes
are shown per study usually as a light grey diamond - which is what
I want to achieve.
The calls I use are:
iat
Aya Anas feps.edu.eg> writes:
> Hello all,
>
> I need to perform the following integration where the integrand is the
> product of three functions:
> f(x)g(y)z(x,y)
>
> the limits of x are(0,inf) and the limits of y are(-inf,inf).
>
> Could this be done using R?
There is a saying: Don't ask C
Hi,
May be this helps:
dat1 <- read.table(text="
a a b b c c
x y x y x y
12 34 256 25 5 32
5 45 23 452 21
45",sep="",header=TRUE,stringsAsFactors=FALSE,check.names=FALSE)
mat1 <- matrix(0,5,5,dimnames=list(NULL,c("x",letters[1:4]))
I am having trouble understanding how to use sapply (or similar
functions) with a user defined function with multiple parameters.
I have the following functions defined
q1.ans <- function(x)
{
retVal = 0
if (x == 1) {
retVal = 1
} else if (x ==2) {
Hi
The warning is due to fact that "if" takes only single scalar value not an
entire vector.
Maybe you shall explain more clearly what result do you expect.
I bet that there is vectorised solution to your problem but I am lost in your
ifs and cannot follow what shall be the output.
Please use
Thank you for your response!
I am attempting to determine a preference from the answers to three
binomial questions;
q.1) 1 or 2q.2) 1 or 3q.3) 2 or 3
However, the questions are coded with either a 1 or 2 (though no answer
is also possible) and the first three functions (q#.ans) conv
Hi
So first step is over. Anyway, is there any problem with using dput as I
suggested?
Instead of using your date I need to generate my own.
A<-sample(0:2, 10, replace=T)
B<-sample(0:2, 10, replace=T)
C<-sample(0:2, 10, replace=T)
df<-data.frame(A,B,C)
df[df[,2]==2,2]<-3
df$C<-as.numeric(as.ch
> I have been researching and it appears that I should be using the sapply
> function to apply the evaluate.question function above to each row in
> the data frame like this
Read the documentation more closely: sapply(dataFrame, func)
applies func() to each column, not row, of dataFrame.
> prefer
Hey everyone,
I have a list of genes for which I would like to get Gene Ontology profiles
(i.e. what are the most common GO terms). First I had a look at topGO, but
since that compares two data sets, which I don’t have, it wasn’t right for this
purpose. I then found goProfiles, which seems to d
Hi,
I want to set up a mixed model ANCOVA but cannot find a way to do it.
There is:
* 1 subject factor (random, between subjects) called Subject
* 3 categorical within subjects factors called Emotion, Sex, Race
* 1 continuous covariate (**WITHIN subjects**) called Score
and
* a continuous depend
Dear,
I need to calculate the following equation
tr(Sigma^-1 %*% D.Sigma)
I know only Sigma (positive definite) and D.Sigma (derivative of Sigma), a
naive code is
sum(diag(solve(Sigma,D.Sigma)))
but these matrices are dense and big dimension (1 x 1), and I need
to evaluate this equatio
Does anyone know if the error being generated when trying to predict
test set data in the Easy Uplift Tree package is something fixable by
the user or is this a bug in the program making the package essentially
non-operable?
This is from the package documentation and fails on the last step of
a
A fast computation I use is based on the following:
A <- matrix(rnorm(16), ncol = 4)
B <- matrix(rnorm(16), ncol = 4)
C <- A %*% B
sum(diag(C))
### This is less expensive to compute when the matrix multiplication is
expensive
sum(A * t(B))
So, it just uses the elementwise calculations and sums
On 12/06/2013 10:43 AM, William Dunlap wrote:
I have been researching and it appears that I should be using the sapply
function to apply the evaluate.question function above to each row in
the data frame like this
Read the documentation more closely: sapply(dataFrame, func)
applies func() to eac
laurie bayet gmail.com> writes:
>
> Hi,
>
> I want to set up a mixed model ANCOVA but cannot find a way to do it.
>
> There is:
>
> * 1 subject factor (random, between subjects) called Subject
> * 3 categorical within subjects factors called Emotion, Sex, Race
> * 1 continuous covariate (**WI
Hi Chris,
May be this helps.
#Suppose the working directory is `FirstLevel`
D <- dir(recursive=TRUE)
D
#[1] "S1/S1data.txt" "S2/S2data.txt" "S3/S3data.txt"
sapply(D,function(x) nrow(read.table(x,sep="",header=TRUE)))
#S1/S1data.txt S2/S2data.txt S3/S3data.txt
# 20 20
Robert Lynch gmail.com> writes:
> I am modeling grade as a function of membership in
> various cohorts. There
> are four "cohorts". (NONE, ISE07,ISE08,ISE09) and two times of cohorts
> coded as ISE = TRUE (ISE0#) or FALSE (NONE). There is clear co-linearity
> but that is to be expected.
>
>
Rosario Garcia Gil slu.se> writes:
>
> Hello
>
> I have run an anova analysis for
> the fallowing model: H_obs=mu+REGION+MANAGEMENT + e
>
> When I run it in ASRelm I get the p-value for mu, and,
> of course also for the two dependent variables (REGION
> and MANAGEMENT)
>
> When I run it in R
> I understand that such for loops aren't 'best practice' in R and am
> trying to learn its approach.
sapply() is an encapsulated loop and loops have their place in R.
'Best practice' is a nebulous term, but explicit loops can make
code that is hard to understand (by a compiler or by a human)
and
Karen,
Look at the help for the drop1() function.
?drop1
There you will see, "The hierarchy is respected when considering terms to
be added or dropped: all main effects contained in a second-order
interaction must remain, and so on."
So, for fit2, the step() function will only consider dropp
What is a "good" way to create quantiles with approximately the same
number of data points within each quantile?
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/p
Thanks again!
Can the ifelse statement be nested like
ifelse(condition1,
ifelse(condition2,yes,no)
ifelse(condition3,yes,no)
)
?
On 12/06/2013 12:23 PM, William Dunlap wrote:
I understand that such for loops aren't 'best practice' in R and am
trying to learn its approach.
sapply() is
By default I believe. See
http://en.wikipedia.org/wiki/Quantile
Others more erudite may correct me.
On Dec 6, 2013, at 11:47 AM, Anika Masters wrote:
> What is a "good" way to create quantiles with approximately the same
> number of data points within each quantile?
>
> _
Hello,
Use function ?quantile.
See this example, each group has exactly, not approximately, 25 elements.
x <- rnorm(100)
qnt <- quantile(x)
tapply(x, findInterval(x, qnt, rightmost.closed = TRUE), length)
Hope this helps,
Rui Barradas
Em 06-12-2013 19:47, Anika Masters escreveu:
What is a "
On Dec 5, 2013, at 9:46 PM, A Xi Ma wrote:
> The following question is inspired by Jun's problem, which resembles some
> of my own problems, but goes off on a tangent about applying plot3D from
> Karline Soetart.
>
>
> On Thu, Dec 5, 2013 at 11:52 PM, Bert Gunter wrote:
>
>>
>> Your comment
The rms package has had several updates in version 4.1-0:
* Fixed orm.fit to not create penalty matrix if not needed
(penalties are not yet implemented anyway)
* Added yscale argument to plot.Predict
* Added Wald test simulation to orm help file
* Added example in help file for plot
On Dec 4, 2013, at 7:49 AM, kevinod wrote:
> I have a concern about the survAUC package option AUC.cd.
>
So shouldn't you be sending this to the package authors? They may or may not be
regular readers of R help. It's apackage I have never heard of.
> I am exploring package functionality, sp
I have a data frame whose first colum contains the names of the variables
and whose second colum contains the values to assign to them:
: kkk <- data.frame(vars=c("var1", "var2", "var3"),
vals=c(10, 20, 30), stringsAsFactors=F)
If I do
: assign(kkk$vars[1], kkk$vals
Hi,
Try
vec1 <- 10958:10963
as.Date(vec1,origin="1960-01-01")
#[1] "1990-01-01" "1990-01-02" "1990-01-03" "1990-01-04" "1990-01-05"
#[6] "1990-01-06"
A.K.
I have imported a stata data into R and wanted to convert the date.
The format went OK, but the output doesn't represent my data. The he
On Dec 6, 2013, at 11:27 AM, Julio Sergio Santana wrote:
> I have a data frame whose first colum contains the names of the variables
> and whose second colum contains the values to assign to them:
>
> : kkk <- data.frame(vars=c("var1", "var2", "var3"),
> vals=c(10, 20, 30
> -Original Message-
> From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org]
> On Behalf Of arun
> Sent: Friday, December 06, 2013 3:11 PM
> To: R help
> Subject: Re: [R] Wrong date fromat?
>
> Hi,
> Try
> vec1 <- 10958:10963
> as.Date(vec1,origin="1960-01-01")
> #[1] "
Hi Lishu,
I run into the similar large-scale problems recently. I used a parallel
SGD k-means described in this paper for my problem:
http://www.eecs.tufts.edu/~dsculley/papers/fastkmeans.pdf
Let n be the samples, k be the number of clusters, and m be the number of
nodes,
1. First, each node r
Hi Uwe,
It looks SVM in e1071 and Kernlab does not support feature selection, but
you can take a look at package penalizedSVM (
http://cran.r-project.org/web/packages/penalizedSVM/penalizedSVM.pdf).
Or you can implement a SVM-RFE (
http://axon.cs.byu.edu/Dan/778/papers/Feature%20Selection/guyon*.
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