I'm not sure why you posted the original note. I quit using SAS in 1991 and
haven't needed it yet.
Frank
RogerJDeAngelis wrote
> Sorry about the double post. But I keep getting 'post' rejections, so I
> resubmitted about an hour later.
-
Frank Harrell
Department of Biostatistics, Vanderb
I appreciate all the feedback on this. I ended up using this line to solve my
problem, just because I stumbled upon it first...
> alldata <- alldata[alldata$REC.TYPE == "SAO " | alldata$REC.TYPE ==
> "FM-15",,drop=FALSE]
But I think Jim's solution would work equally as well. I was a bit confus
Did you try using MLE to approximate the marginal?
On Mar 3, 2013, at 5:26 AM, Ali A. Bromideh wrote:
> Dear Nicole,
>
> First of all, my sincere gratitude goes to your kind reply. As I told to Mr.
> Gunter, this is a part of my research and differs from homework. However, I
> am going to cla
On 13-03-03 7:08 PM, Jerrod Parker wrote:
Help,
I've been having a difficult time trying to create 3d partial dependence
plots using "rgl". It looks like this question has been asked a couple
times, but I'm unable to find a clear answer googling. I've tried creating
x, y, and z variables by ex
Help,
I've been having a difficult time trying to create 3d partial dependence
plots using "rgl". It looks like this question has been asked a couple
times, but I'm unable to find a clear answer googling. I've tried creating
x, y, and z variables by extracting them from the partialPlot output to
That the most common formula, but not the only one. See
Kvålseth, T. (1985). Cautionary note about $R^2$. *American Statistician*,
*39*(4), 279285.
Traditionally, the symbol 'R' is used for the Pearson correlation
coefficient and one way to calculate R^2 is... R^2.
Max
On Sun, Mar 3, 2013 a
On 13-03-03 3:39 PM, Oleguer Plana Ripoll wrote:
Dear Milan and other users,
Thank you for your help, it worked. The problem is that the function "do.call"
is not ready for vectors and I need it in order to integrate it afterwards.
do.call() is fine, it's the argument list that needs fixing.
Dear Milan and other users,
Thank you for your help, it worked. The problem is that the function "do.call"
is not ready for vectors and I need it in order to integrate it afterwards.
With the pweibull, I can write:
pweibull(1,shape=1)
pweibull(2,shape=1)
pweibull(1:2,shape=1)
When I do the same
I was under the impression that in PLS analysis, R2 was calculated by 1-
(Residual sum of squares) / (Sum of squares). Is this still what you are
referring to? I am aware of the linear R2 which is how well two variables
are correlated but the prior equation seems different to me. Could you
expla
Le dimanche 03 mars 2013 à 19:49 +0100, Oleguer Plana Ripoll a écrit :
> Hello everyone,
>
> I have a quick question but I am stuck with it and I do not know how
> to solve it.
>
> Imagine I need the distribution function of a Weibull(1,1) at t=3,
> then I will write pweibull(3,1,1).
>
> I want
Hello everyone,
I have a quick question but I am stuck with it and I do not know how to solve
it.
Imagine I need the distribution function of a Weibull(1,1) at t=3, then I will
write pweibull(3,1,1).
I want to keep the shape and scale parameters in a list (or a vector or
whatever). Then I hav
Thank you for your response Jim! I will give this one a try! But a couple
followup questions...
In my search for a solution, I had seen something stating match() is much more
efficient than subset() and will cut down significantly on computing time. Is
there any truth to that?
Also, I found th
HI,
You could also use ?data.table()
n<- 30
set.seed(51)
mat1<- as.matrix(data.frame(REC.TYPE=
sample(c("SAO","FAO","FL-1","FL-2","FL-15"),n,replace=TRUE),Col2=rnorm(n),Col3=runif(n),stringsAsFactors=FALSE))
dat1<- as.data.frame(mat1,stringsAsFactors=FALSE)
table(mat1[,1])
#
# FAO FL-1
On 03-Mar-2013 16:29:05 Angelo Scozzarella Tiscali wrote:
> For example, I want to simulate different populations with same mean and
> standard deviation but different distribution.
>
> Il giorno 03/mar/2013, alle ore 17:14, Angelo Scozzarella Tiscali ha scritto:
>> Dear R friends,
>>
>> I'd lik
If you are using matrices, then here is several ways of doing it for
size 300,000. You can determine if the difference of 0.1 seconds is
important in terms of the performance you are after. It is taking you
more time to type in the statements than it is taking them to execute:
> n <- 30
> te
For example, I want to simulate different populations with same mean and
standard deviation but different distribution.
Il giorno 03/mar/2013, alle ore 17:14, Angelo Scozzarella Tiscali ha scritto:
> Dear R friends,
>
> I'd like to generate random sample (variable size and range) without a
>
there are way "more efficient" ways of doing many of the operations , but you
probably won't see any differences unless you have very large objects (several
hunfred thousand entries), or have to do it a lot of times. My background is
in computer performance and for the most part I have found th
Angelo Scozzarella Tiscali tiscali.it> writes:
>
> Dear R friends,
>
> I'd like to generate random sample (variable size and range) without a
specified distribution but with
> given mean and standard deviation.
>
> Could you help me?
>
The problem is underspecified, so no, we can't.
A
Dear R friends,
I'd like to generate random sample (variable size and range) without a
specified distribution but with given mean and standard deviation.
Could you help me?
thanks in advance
Angelo
__
R-help@r-project.org mailing list
https://stat.e
Hi,
Try this:
set.seed(51)
mat1<- as.matrix(data.frame(REC.TYPE=
sample(c("SAO","FAO","FL-1","FL-2","FL-15"),20,replace=TRUE),Col2=rnorm(20),Col3=runif(20),stringsAsFactors=FALSE))
dat1<- as.data.frame(mat1,stringsAsFactors=FALSE)
dat1[grepl("SAO|FL-15",dat1$REC.TYPE),]
# REC.TYPE Col2
On 03/03/2013 09:58, Rani Elkon wrote:
Dear all,
I calculate the test statistic for the KS test outside R, and wish to use R
only to calculate the corresponding p-value.
There is no public way to do this in R. But you can read the code of
ks.test and see how it does it, and extract the cod
Dear Cedric,
If I understand correctly what you want to do, and if you're willing to
assume that the variables are normally distributed, then you should be able
to use any of the latent-variable structural-equation-modeling packages in
R, such as sem, OpenMX, or lavaan.
Here's an artificial exam
Thank you for your response Max. Is there some literature that you make
that statement? I am confused as I have seen many publications that
contain R^2 and Q^2 following PLSDA analysis. The analysis usually is to
discriminate groups (ie. classification). Are these papers incorrect in
using thes
Hello,
You can compute the p-value from the test statistic if you know the
samples' sizes. R calls functions written in C for the several cases,
for the two samples case, this is the code (edited)
n.x <- 100 # length of 1st sample
n.y <- 100 # length of 2nd sample
STATISTIC <- 1.23
PVAL <-
Try this:
dataset <- subset(dataset, grepl("(SAO |FL-15)", REC.TYPE))
On Sun, Mar 3, 2013 at 1:11 AM, Matt Borkowski wrote:
> Let me start by saying I am rather new to R and generally consider myself to
> be a novice programmer...so don't assume I know what I'm doing :)
>
> I have a large matr
Dear all,
I calculate the test statistic for the KS test outside R, and wish to use R
only to calculate the corresponding p-value.
Is there a way for doing this? (as far as I see, ks.test() requires raw
data as input). Alternatively, is there a way to provide the ks.test() the
two CDFs (two
Let me start by saying I am rather new to R and generally consider myself to be
a novice programmer...so don't assume I know what I'm doing :)
I have a large matrix, approximately 300,000 x 14. It's essentially a 20-year
dataset of 15-minute data. However, I only need the rows where the column I
Dear Nicole,
First of all, my sincere gratitude goes to your kind reply. As I told to Mr.
Gunter, this is a part of my research and differs from homework. However, I
am going to clarify the problem. Suppose we have received an observation
from a Poisson distr. i.e. Y_1~Pois(Lam_1), where Lam_1~Gam
I forgot to say:
Also do not depend on equality in this situation.
You want to test equality with a tolerance.
See Circle 1 of 'The R Inferno':
http://www.burns-stat.com/documents/books/the-r-inferno/
I also see that 't' is a vector unlike what I was
thinking before, thus you want to use 'ifels
On 03-03-2013, at 00:57, Louise Stevenson
wrote:
> Hi,
>
> I'm trying to set up R to run a simulation of two populations in which every
> 3.5 days, the initial value of one of the populations is reset to 1.5. I'm
> simulation an experiment we did in which we fed Daphnia populations twice a
I think the command you want is:
if(t %in% feed_days) C_A <- 1.5 else C_A <- 0
Do not confuse `%in%` (which is essentially
"are the left-hand values in the right-hand
vector)
with
"in" of the `for` loop.
By the way,
if(t == TRUE)
is redundant -- better is:
if(t)
Pat
On 02/03/2013 23:57
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