It's standard form to cc the list so that replies and experience get pooled.
This sounds like a problem with your data set to be honest. Can you
post a link to it? If it's not publicly available, there's probably
nothing that we (having no access to it) can do to help interpret.
If you know the s
Sorry, that last line should read:
FUN=function(z){
lz <- length(z)
SDF(z,method="lag window",
window=taper(type="parzen",n.sample=lz,cutoff= 2*sqrt(lz)), npad=2*lz)
}
On Sun, Jan 29, 2012 at 11:29 PM, R. Michael Weylandt
wrote:
> It's customary to keep the list cc'd.
>
> I can't run your co
It's customary to keep the list cc'd.
I can't run your code without the data, but it does seem to me that
your problem is in the FUN argument, as you guess.
You have:
FUN=function(z) SDF(adezoo,method="lag window",
window=taper(type="parzen",n.sample=n.d,cutoff=(2*sqrt(n.d))),
npad=2*n.d)
But t
I believe you need to use the scale_fill_brewer since fill is the
color of the bars while color is the outside of the bars in
ggplot2-speak:
E.g., with built-in data (it's polite to provide yours so that your
minimal working example is working):
data(diamonds)
ggplot(diamonds, aes(clarity)) + geo
Perhaps I'm misunderstanding you, but it doesn't sound like this is
much of an R question at all: what is "the code book"? If it's an
actual (dead tree) book, I don't think there's much you can do in R to
automate identifications; if it's an online API, you *might* be able
to rig a matching algorit
Reread ?lm and note that the lhs can be a matrix. I believe this is
exactly what you want.
-- Bert
On Sun, Jan 29, 2012 at 2:05 PM, Martin Batholdy
wrote:
> Hi,
>
>
> I would like to fit lm-models to a matrix with 'samples' of a dependent
> variable (each row represents one sample of the depend
If it's a simple one variable OLS regression and you only need
regression coefficients, you'll probably get best performance by
hard-coding the closed form solutions. apply() might help a little
(since it's a very good loop) but ultimately you'll be best served by
deciding exactly what you want and
Hi,
I would like to fit lm-models to a matrix with 'samples' of a dependent
variable (each row represents one sample of the dependent variable).
The independent variable is a vector that stays the same:
y <- c(1:10)
x <- matrix(rnorm(5*10,0,1), 5, 10)
now I would like to avoid looping over
If v is your original data,
v <- c(-20, rep(0,98), 20)
why not use
mean( -20 < v & v < 2)
as your estimate of the probability that v is in (-20,2)?
Estimating a density is like taking the derivative
of a smooth of the empirical distribution function,
so why not eliminate the middleman instea
Hi,
I have what I suppose is the same problem as this. I am using the
linear mixed model function lme, and this does not seems to take the
attribute "model=TRUE" at the end of the function.
Is there a more general way of solving this problem?
Is my description of the problem below correct (f
Try this,
plot(1:10)
img <- grid::grid.cap()
# grid.raster(img)
stream <- col2rgb(img)
write.table(stream, file="dam.txt", row.names = FALSE,
col.names = FALSE)
(you'll have to restore the dimensions of the matrix once you've read
the rgb values for each pixel)
HTH,
baptiste
On 30
I can write a plot to a files of a given format using this:
x<-sample(c(1:100),10)
bmp("c:/mygraph.bmp")
plot(x)
dev.off()
and then show the image file in another application. This application can also
display the image from the stream of numbers that define the image.
How I can get the plot as
Thank you so very much for your help. That worked perfectly. Yes, it's
homework, but hopefully what I'm learning with the homework I'll be able to
apply to my own research in the future, so I'd really like to learn the
program. And by the by, the "someone" was a classmate. Thanks again!
Crystal
?ifelse
Type that in and you'll be in good shape.
On 1/29/12, C_Crown wrote:
> Hi all, I am very new to R. I am taking a course and am trying to complete
> my first assignment.
> For the assignment I have to get the program to generate different color
> combinations possible for a sample size o
CD gmail.com> writes:
>
> Hello,
> I'm trying to analyse the spatial organization of different fields planted
> with different varieties (each field has only one variety), but I have
> problems trying to understand the results of the test I did.
> Can somebody help me understanding what rea
On 29-01-2012, at 18:18, C_Crown wrote:
> Hi all, I am very new to R. I am taking a course and am trying to complete
> my first assignment.
> For the assignment I have to get the program to generate different color
> combinations possible for a sample size of 55 with the probabilities of each
>
Hello,
I have a dataset which I am calling RB09.
I am trying to match the questions in the code book with variable codes.
It is not very intuitive.
example:
names(RB09)
[1] "ea1""eaf1" "eaf1a" "eaf2" "eaf2_7"
[6] "eaf3" "eafimpun" "eafunpun" "evi
Thanks, Carlos,
your solution looks nice, however, there are 4 bars instead of 2 i.e. bars are
not stacked.
Meanwhile I got a solution for that task.
So thanks for sharing your version of it!
Good luck,
Mario
Von: Carlos Ortega
Cc: "r-help@r-project.org"
On Sun, Jan 29, 2012 at 11:36 AM, Ajay Askoolum wrote:
> Thank you. I need some clarification.
>
> dput(AirPassengers)
>
> gives:
>
> structure(c(112, 118, 132, 129, 121, 135, 148, 148, 136, 119,
> 104, 118, 115, 126, 141, 135, 125, 149, 170, 170, 158, 133, 114,
> 140, 145, 150, 178, 163, 172, 178
Take a look at
?sample
HTH,
Jorge.-
On Sun, Jan 29, 2012 at 11:21 AM, Ron Michael <> wrote:
> Dear all, here I need to draw all possible samples of size 2, from a
> population, which is characterized by c(1,4,56, 3). Sampling will be done
> with replacement. Is there any direct R function to f
On 01/29/2012 08:12 AM, Kevin Ummel wrote:
Sorry, guys. I'm not active on the listserve, so my last post was held by the
moderator until after Dirk's solution was posted.
Excellent stuff.
Is 'diff' really the bottleneck in your calculations? I would have said
diff was in the class of 'fast'
Hello, R friends,
I'm trying to change colors of my horizontal bars so that they show a sequence.
I chose the ColorBrewer palette "Blues". However the resulting plot doesn't
show any changes to the default.
I tried several places of "+ scale_colour_brewer(type="seq", pal = "Blues")"
with no eff
Hi all, I am very new to R. I am taking a course and am trying to complete
my first assignment.
For the assignment I have to get the program to generate different color
combinations possible for a sample size of 55 with the probabilities of each
color being chosen as: .24, .24, .16, .20, .13, .14
On Jan 29, 2012, at 11:21 AM, Ron Michael wrote:
Dear all, here I need to draw all possible samples of size 2, from a
population, which is characterized by c(1,4,56, 3).
> combn(c(1,4,56, 3), 2) # an enumeration of possible tuples
[,1] [,2] [,3] [,4] [,5] [,6]
[1,]1114
On Jan 29, 2012, at 11:36 AM, Ajay Askoolum wrote:
Thank you. I need some clarification.
dput(AirPassengers)
gives:
structure(c(112, 118, 132, 129, 121, 135, 148, 148, 136, 119,
104, 118, 115, 126, 141, 135, 125, 149, 170, 170, 158, 133, 114,
140, 145, 150, 178, 163, 172, 178, 199, 199, 184,
Dear all, here I need to draw all possible samples of size 2, from a
population, which is characterized by c(1,4,56, 3). Sampling will be done with
replacement. Is there any direct R function to faciliate this darwing?
Thanks for your help
__
R-help@
You said something about extracting January to June -- that's not so
possible with the default ts (best I understand it): ts() only allows
regular time-series so you can't go jan to jun at 1 month intervals
and then jump to jan again (6 months). If you want irregular time
series, check out the zoo
Sorry about that -- Gmail threaded things by arrival in my mailbox,
not by timestamp (surprisingly)
Rcpp really is one of the coolest new things in the R ecosystem --
hope it works well for you.
M
On Sun, Jan 29, 2012 at 11:12 AM, Kevin Ummel wrote:
> Sorry, guys. I'm not active on the listserv
Thank you. I need some clarification.
dput(AirPassengers)
gives:
structure(c(112, 118, 132, 129, 121, 135, 148, 148, 136, 119,
104, 118, 115, 126, 141, 135, 125, 149, 170, 170, 158, 133, 114,
140, 145, 150, 178, 163, 172, 178, 199, 199, 184, 162, 146, 166,
171, 180, 193, 181, 183, 218, 230, 2
Sorry, guys. I'm not active on the listserve, so my last post was held by the
moderator until after Dirk's solution was posted.
Excellent stuff.
thanks,
kevin
On Jan 29, 2012, at 8:37 AM, R. Michael Weylandt wrote:
> Have you not followed your own thread? Dirk is Mr. Rcpp himself and he
> give
On Jan 29, 2012, at 7:38 AM, Christopher Kelvin wrote:
Please, Help me,
How do I generate data from the weibull distribution if the data
contain both failure and interval censored,
For example, I want to generate n=100, shape=2 and scale =4 with 30%
interval censored.
What about right cen
see dump() or dput().
On Sun, Jan 29, 2012 at 10:56 AM, Ajay Askoolum wrote:
> Given:
>
> data(AirPassengers)
>
> I get a ts data structure AirPassengers in the workspace.
>
> How can I generate the code that can create that structure? That is, given an
> example of a data structure, is there a
dput()
Michael
On Sun, Jan 29, 2012 at 10:56 AM, Ajay Askoolum wrote:
> Given:
>
> data(AirPassengers)
>
> I get a ts data structure AirPassengers in the workspace.
>
> How can I generate the code that can create that structure? That is, given an
> example of a data structure, is there a way to
Given:
data(AirPassengers)
I get a ts data structure AirPassengers in the workspace.
How can I generate the code that can create that structure? That is, given an
example of a data structure, is there a way to generate the code that can
greate that structure?
Alternatively, is there a refere
Have you not followed your own thread? Dirk is Mr. Rcpp himself and he
gives an implementation that gives you 25x improvement here as well as
tips for getting even more out of it:
http://tolstoy.newcastle.edu.au/R/e17/help/12/01/2471.html
Michael
On Sat, Jan 28, 2012 at 12:28 PM, Kevin Ummel wr
Please, Help me,
How do I generate data from the weibull distribution if the data contain both
failure and interval censored,
For example, I want to generate n=100, shape=2 and scale =4 with 30% interval
censored.
What about right censoring
Thank you
[[alternative HTML version deleted]
On 27.01.2012 19:51, Ondřej Mikula wrote:
Dear r-helpers,
I have a trouble with a package downloaded from sourceforge.net
(namely the package 'kopls'). I installed it from the local zip file
with the expected result
utils:::menuInstallLocal()
package ‘kopls’ successfully unpacked and MD5 sum
On Jan 28, 2012, at 6:44 PM, h...@komarac.biologija.unios.hr wrote:
Thank you very much for help. But would you be kind to show me on this
simple example how to obtain x, y and z axis. Namely, it is not
problem to
avoid box with "par.box = list(col="transparent")" but to keep axis?
library(
Thanks.
I guess it may be due to the software conflict in my computers as there are the
same software packages which have been installed in these two laptops. Anyway,
I can still use my desktop with Windows XP system.
Jim
--- On Sat, 28/1/12, Uwe Ligges wrote:
> From: Uwe Ligges
> Subjec
On Jan 29, 2012, at 12:17 , Christopher Kelvin wrote:
> Hello,
> If i write a function as below using log of weibull distribution i do not get
> the required
>
> results in estimating the parameters what do i do, please
Presumably find and fix the error in your likelihood function!
> z2 <- f
>
>
> I am new to this group - joined today. I am in because i have an assignment
> to do using R => microarray analysis using affy (bioconductor).
>
> while reading some tutorials, i came across this and i am stuck. i want to
> understand it and would appreciate if anyone can tell me.
>
>
Hello,
If i write a function as below using log of weibull distribution i do not get
the required
results in estimating the parameters what do i do, please
a/b * (t/b)^a-1 * exp(-t/b)^a
n=500
x<-rweibull(n,2,2)
z<-function(p) {(-n*log(p[1])+n*log(p[2])-
(p[1]-1)*sum(log(x))+(p[1]-1)*log(p[2])
Hello,
I'm trying to analyse the spatial organization of different fields planted
with different varieties (each field has only one variety), but I have
problems trying to understand the results of the test I did.
To do this, I created different neighbourhood matrix. For example, for the
first matr
On Jan 29, 2012, at 02:42 , Ben Bolker wrote:
>>
>
> (quoting removed to make Gmane happy)
>
> AFAICS, this is a bug.
>
>
> I think so too, although I haven't got my head around it yet.
>
> Chuck, are you willing to post a summary of this to r-devel
> for discussion ... and/or post a bug r
On Sat, Jan 28, 2012 at 10:25:47AM -0800, Melrose2012 wrote:
> Hi Again,
>
> I am writing a 'for loop' to create a matrix of randomly sampled colors.
> I've written this loop in matlab and it works fine. I then tried to do it
> in R and apparently there is something wrong with my syntax b/c ever
Dear everyone,
I have met a problem when using the R function arima.
The following is the coding.
> www <- "http://www.massey.ac.nz/~pscowper/ts/pounds_nz.dat";
> x <- read.table(www, header = T)
> x.ts <- ts(x, st = 1991, fr = 4)
> x.ma <- arima(x.ts, order = c(0, 0, 1))
> x.ma
> acf(x.ma$res[-
Hello,
>
> Every time the loop blows up, I get back this error:
>
> Error in table.combos[i, ] <- table(combos) :
> number of items to replace is not a multiple of replacement length
>
This is because not all colors are present in that one sample that breaks
the code.
>
> There is no appar
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