> R Michael Weylandt
> on Mon, 22 Oct 2012 09:06:35 +0100 writes:
> On Monday, October 22, 2012, Christian Hoffmann wrote:
>> Hi,
>>
>> Each > help.start() generates a new tree of the R help
>> system, somewhere in 127.0.0.1:xxx, each xxx being
>> difeerent. T
On Oct 22, 2012, at 3:28 PM, Thomas Hoffmann wrote:
> Dear all,
>
> I am doing log-transformed bootstrap regression using:
>
> x=c(0.038,0.054,1.030,1.250,10.240,52.000,228.100,240.000,758.000,1502.600,0.140,9.170,280.000,1.000,0.150,0.388,20)
> y=c(3961.5,25987.5,526557,321094.5,642,235927
Copy the table data (leave out the column headings and the total line at the
bottom) into the clipboard. Then
> Dta <- read.delim("clipboard", header=FALSE, stringsAsFactors=FALSE)
> colnames(Dta) <- c("No", "State", "Abbrev", "Population", "Area.km",
"PopDensity.km", "Area.mi", "PopDensity.
On 23/10/12 16:17, David Arnold wrote:
All,
A friend of mine would like to use this data with his stats class:
http://www.worldatlas.com/aatlas/populations/usapoptable.htm
I can't figure a way of capturing this data due to the mysql commands in the
source code.
Any thoughts?
Copying and pas
All,
A friend of mine would like to use this data with his stats class:
http://www.worldatlas.com/aatlas/populations/usapoptable.htm
I can't figure a way of capturing this data due to the mysql commands in the
source code.
Any thoughts?
David.
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Thank u so much for helping! Now I got it ;)
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https:
On Oct 22, 2012, at 5:14 PM, nserdar wrote:
> I can not find any packages for polynomial optimisation in R.
>
> Please let me know any packages for that.
I can think of code written in base R that might satisfy that request. Perhaps
you should a) say what you understand by that phrase, and b)
Hi,
?ctab() also gets the same result.
library(catspec)
dat2<-within(dat1,{Gender<-factor(Gender);Dosage<-factor(Dosage)})
ctab(dat2$Gender,dat2$Dosage)
# a b
#
#f 4 4
#m 4 4
A.K.
- Original Message -
From: David Winsemius
To: arun
Cc: asafwe ; R help
Sent: Monday, October
[Apologies if you receive multiple copies of this Call For Papers.]
CALL FOR PAPERS
=
The 4th International Conference on Information and Communication Systems,
ICICS 2013
Organized by Jordan University of Science and Technology
April 23-25 2013, Irbid, Jordan
http://www.icics.info
I can not find any packages for polynomial optimisation in R.
Please let me know any packages for that.
Regards,
Ser
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I would start by making sure your R installation and all packages are
up-to-date.
update.packages() will be helpful once you've upgraded R itself.
Sarah
On Monday, October 22, 2012, Vinicius Campos wrote:
> Dear Gabor
>
> You got solution for your problem?
>
> "I have just got a laptop with win
You need to load the dataset.
First, run
data(CO2)
Then it should work.
Jeremy
[[alternative HTML version deleted]]
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PLEASE do read the posting guide http://www
Hello:
I am new on R, this is the first time I work with it,
I am trying to run some R example to get learn of them,
but, I have a problem when running the example below:
http://stat.ethz.ch/R-manual/R-patched/library/datasets/html/zCO2.html
CO2.Qn1 <- CO2[CO2$Plant == "Qn1", ]
SSasympOff(CO2.Q
Inline.
On Mon, Oct 22, 2012 at 1:10 PM, CMB123 wrote:
> I apologize if the term "submatrix" was confusing - I am basically 2 rows and
> 2 columns from a r x c matrix to construct 2x2 matricies. Thus, the
> choose(r,2) * choose(c,2) possible combinations.
>
> For each matrix [(a,b), (c,d)],
What
Dear all,
I am doing log-transformed bootstrap regression using:
x=c(0.038,0.054,1.030,1.250,10.240,52.000,228.100,240.000,758.000,1502.600,0.140,9.170,280.000,1.000,0.150,0.388,20)
y=c(3961.5,25987.5,526557,321094.5,642,23592715.5,4050,
228853521.1,320980392,42900,58435.5,13525240.5
Thank you all; David -- this is, in fact, exactly what I need!
Asaf
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Hi
Sorry, that should be ...
(i) To start a new page, use plot.new() (if you are trying to mix 'grid'
and base graphics)
(ii) Watch out for 'grid' and base graphics fighting each other over the
clipping region (you might need to use grid.clip() after a call to
plot() or your 'grid' output m
Hi,
May be this helps:
psidp<-function(x){
p <- c(1/10, 2/5, 2/5, 2/5, 2/5, 1/10, 1/10, 1/10, 1/10, 1/10)
ifelse(x>0 & x<=length(p),sum(p[seq(1,x,1)]),NA)}
psidp(0)
#[1] NA
psidp(5)
#[1] 1.7
psidp(10)
#[1] 2.2
A.K.
- Original Message -
From: Rlotus
To: r-help@r-project.org
Cc:
Sen
I apologize if the term "submatrix" was confusing - I am basically 2 rows and
2 columns from a r x c matrix to construct 2x2 matricies. Thus, the
choose(r,2) * choose(c,2) possible combinations.
For each matrix [(a,b), (c,d)], I am testing a > b, b > d, d > c, and c > a.
For the sake of simplicity
Hi
Here are a couple of suggestions:
(i) To start a new page, use plot.new() FIRST, followed by
grid.newpage() (if you are trying to mix 'grid' and base graphics)
(ii) Watch out for 'grid' and base graphics fighting each other over the
clipping region (you might need to use grid.clip() after
Hi,
I'm playing with moving some of my lattice graphics into ggplot2, and I'd
like to ask how to achieve a couple of things, both of which are fully
illustrated in self-contained code (and mostly minimal, although that left
quite a bit) following this written description.
1. I quite often like to
Dear Gabor
You got solution for your problem?
"I have just got a laptop with win 7 64 bit installed on it. I
installed R and quite a few packages I use. When I try to start
BiodiversityR the library loads without any problems but when starting
the GUI it gives me the following error message:
So
I think this is what you want.
p <- c(1/10, 2/5, 2/5, 2/5, 2/5, 1/10, 1/10, 1/10, 1/10, 1/10)
psidp <- function(x){
if (x>0&&x<=10)
{
return(sum(p[1:x]))
}
else{
return("Input integer between 1 and 10")
}}
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Dear Paul,
Thank you very much for helping, works perfectly fine.
I now see the scaling when looking at grid.show.layout(), but I would have never
come up with looking there in the first place :-)
Cheers,
Marius
Paul Murrell writes:
> Hi
>
> Apologies for the slow response.
>
> This part o
On Oct 22, 2012, at 7:48 AM, arun wrote:
> HI,
> Another way:
> dat1<-read.table(text="
> Observation Gender Dosage Alertness
> 11 m a8
> 22 m a12
> 33 m a13
> 44 m a12
> 5
On Oct 22, 2012, at 1:51 AM, PIKAL Petr wrote:
> Hi
>
>> -Original Message-
>> From: r-help-boun...@r-project.org [mailto:r-help-bounces@r-
>> project.org] On Behalf Of David Arnold
>> Sent: Monday, October 22, 2012 4:26 AM
>> To: r-help@r-project.org
>> Subject: [R] Margin annotation
>>
On Fri, Oct 19, 2012 at 6:28 PM, Ali Tofigh wrote:
> On Wed, Oct 17, 2012 at 4:08 PM, ilai wrote:
> > On Wed, Oct 17, 2012 at 11:10 AM, Ali Tofigh
> wrote:
>
> ## this works as intended with a mix of plot.new() and grid.newpage
> pdf("test3.pdf")
> plot.new(); my.plot(); grid.newpage(); my.plot
Hello,
Try ?cumsum.
cumsum(p)
Hope this helps,
Rui Barradas
Em 22-10-2012 19:41, Rlotus escreveu:
I have an array of probabilitiesit is p. So if user types x=1 then
probability is p1=1/10.
If user types x=2 it means that p2= p1+p2
if user types x=3 it means that p3=p1+p2+p3and so on.
Hello,
If your matrix is in the order of 300x300, the problem of extracting all
possible submatrices and applying a function will allways be a large
one, but the use of ?combn may reduce it a bit if the order of
rows/columns in the submatrices doesn't matter. It can reduce it from
300^4 = 8.1
This looks like homework to me.
Here's a hint, though:
p<-(1/10 2/5 2/5 2/5 2/5 1/10 1/10 1/10 1/10 1/10)
This is not how you create a vector in R.
for (i in 1 to x)
This is not how you create a for loop in R.
Sarah
On Mon, Oct 22, 2012 at 2:41 PM, Rlotus wrote:
> I have an array of probabil
300x300 isn't terribly large; looping should work just fine. But I'm
confused about a 2x2 submatrix:
I would have thought that a submatrix would be adjacent elements, like
x[1:2, 1:2]
or
x[13:14, 296:297]
but your loop compares all possible sets of four elements, so the
matrix position doesn't matt
Dheeraj,
You need to give us some more hints of what
you get and what you don't get.
My guess is that what has happened is that the
optimization algorithm didn't converge.
The R-sig-finance mailing list would be a more
appropriate place for this discussion (you have
to subscribe before you can
I have an array of probabilitiesit is p. So if user types x=1 then
probability is p1=1/10.
If user types x=2 it means that p2= p1+p2
if user types x=3 it means that p3=p1+p2+p3and so on.
So i created a code. but it doesnt work properly. Help me plz to fix it)
Thank u in advance.
psidp
thank you so much for helping! Now i got how to do it ;)
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thank you, that looks good
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Hello,
Another usefull link is
http://www.r-statistics.com/2010/02/post-hoc-analysis-for-friedmans-test-r-code/
Hope this helps,
Rui Barradas
Em 22-10-2012 20:18, arun escreveu:
HI,
Have u tried the example in the link
(http://stat.ethz.ch/R-manual/R-patched/library/stats/html/friedman.test.
Hi all,
I'm working with a large data set (on the order of 300X300) and trying to
apply a function which compares the elements of all possible 2x2
submatrices. There are rc(r-1)(c-1) such submatrices, so obviously the naive
method of looping through the rows and columns is computationally unfeasib
Hi there,
I am exporting plm results (from the plm package) to Latex using the
texreg function (from the texreg package).
How can I force the exported results to appear in scientific format?
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HI,
Another way:
dat1<-read.table(text="
Observation Gender Dosage Alertness
1 1 m a 8
2 2 m a 12
3 3 m a 13
4 4 m a 12
5 5 m b 6
6 6 m b
Thanks Rui, your solution works great and is so fast!
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Hi
Apologies for the slow response.
This part of the problem is I think just a bit of a misunderstanding.
The diagram drawn by grid.show.viewport() is drawn within a *subset* of
the current page (or viewport), to allow room for labelling, which is
why your subsequent "real" viewports do not a
WARNING: Use with caution!
There is a way to effect the catenation of factors: The data.frame
method for rbind() does this. E.g.
set.seed(42)
f1 <- factor(sample(letters[1:3],42,TRUE))
f2 <- factor(sample(letters[1:4],66,TRUE))
d1 <- data.frame(f=f1)
d2 <- data.frame(f=f2)
dd <- rbind(d1,d2)
HI,
Have u tried the example in the link
(http://stat.ethz.ch/R-manual/R-patched/library/stats/html/friedman.test.html).
It is working for me.
wb <- aggregate(warpbreaks$breaks,
by = list(w = warpbreaks$wool,
t = warpbreaks$tension),
FUN = mean)
friedman.test(x ~ w | t, data = wb)
# Fried
On Mon, Oct 22, 2012 at 3:35 PM, sagarnikam123 wrote:
> I want to connect R with HTML/PHP pages to take input from user,do some
> statistical processing on it & show results to HTML page again.
> I search on net,i got Rserve package,but examples are mainly for java
> langaure & not for PHP
> i
Incorrect. The code is 90% compatible. Look at
http://biostat.mc.vanderbilt.edu/Rrms for differences.
Frank
Joost Kremers wrote
> On Mon, Oct 22 2012, R. Michael Weylandt <
> michael.weylandt@
> > wrote:
>> Design has been superceeded by rms -- give that a shot.
>
> I know, but it doesn't l
Your attachments did not make it through to the official R-help mailing list.
Nabble R is not R-help.
I downloaded the files from the links to Nabble.
See further down for my comments.
On 22-10-2012, at 14:24, paulfjbrowne wrote:
> I haven't determined the root cause of the numerical accuracy bu
Then what about:
fun<-function(a,b,c,data) {
ifelse(data > c, (a*(data-c)^0.5)+(b*(data-c)), 0)
}
y=c(100,210,320,130,170,120,220,90,55,45)
fun(1,0.2,150,data=y)
> fun(1,0.2,150,data=y)
[1] 0.00 77471.67 130418.05 0.00 44725.36 0.00 83680.00
[8] 0.00 0
Try this -- this uses the vectorization which you need to read up on
in the Intro to R:
> #set the function
> fun<-function(a,b,c,data)
+ {
+ N <- numeric(length(data)) # initialize to zero
+ indx <- data > c # determine which data is greater than c
+ N[indx] <- (a * (data[indx] - c) ^ 0.5) + (b
example(jitter)
library('ggplot2')
example(geom_jitter)
RMW
On Monday, October 22, 2012, swertie wrote:
> Thank you for your comment. I would be very interested, but I did not know
> about jitter. Would you have a small example?
>
>
>
>
> --
> View this message in context:
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Hello,
Your error message means that the arrays have different dim attributes.
check with
dim(M1) == dim(M2)
They must be the same.
If they are, the result is not 0/1 but FALSE/TRUE, which you can treat
as or coerce to integer.
Hope this helps,
Rui Barradas
Em 22-10-2012 16:44, shahab esc
Hi Rui,
Thanks for the quick reply! It was my mistake not to notice $country in
thr fourth line of your code. I went back and changed it to $name, and
got the following output when I mapped the borders:
http://i.imgur.com/DQ3IB.png
Here is the output of your function:
1> exmpl <- sub[, c(
Hello,
I'm not the greatest expert in ggplto2 but you're overcomplicating (if
the word exists).
1. In a black&white plot don't use argument color nor scale_color_manual().
2. your second plot object, p0, is completely unnecessary.
3. if you want it to look like the graph in the link, it's geom
Hi,
I would like to xor (bitwise) two matrices filled with binary values
(0,1). The result of such XOR is expected to be 0,1.
But apparently neither of xor nor bitXor is working in this case.
I got ": binary operation on non-conformable arrays" error message
when I used xor (M1,M2) .
The proble
Greetings
I went back over the examples in the R Graphics book on the grid
webpage and saw an example which I have adapted and seems to work:
pushViewport(viewport(layout=grid.layout(nrow=26,ncol=2)))
pushViewport(viewport(layout.pos.col=1,layout.pos.row=1))
plot_gene_map(dna_segs=list(segs[
Hello,
In your original post, there was a column named 'country', it now seems
to be 'name', therefore my function shouldn't work. To see the output of
head(9 is helpfull but the better way is dput(). Try the following:
exmpl <- sub[, c("name", "idxy", "ix", iy")]
dput( head(exmpl, 30) ) # p
thank you. yes I want to loop over the elements of the data. I want the
output to be corresponded to the input by unchanging index. at the same
time, the data that is more than c value should follow the
function (a*(data-c)^0.5)+(b*(data-c), and the rest (same or less than c)
should return a zero.
Hello,
I'm using VAR models in R in order to obtain impulse responses of stock
market shock on US economy.
I have series of quarterly changes in real gdp, S&P 500 and quarterly level
of unemployment for 1985 - 2012 period.
My series are stationary. So I did all the steps below. However I don't
und
Please stop posting Mac-specific questions on R-help. You have been
asked not to do so before
There _is_ a quartz.save() function in R.app, whose FAQ you are reading.
The RMacOSX-FAQ is not part of R: it is part of the separate Mac-GUI
project. You really do need to contact the authors
Rui,
Thanks a lot for your help. Unfortunately this doesn't work though:
1> is.border <- function(idx, DF){
1+ i1 <- DF$ix %in% (DF$ix[idx] + c(-1, 1)) & DF$iy == DF$iy[idx]
1+ i2 <- DF$iy %in% (DF$iy[idx] + c(-1, 1)) & DF$ix == DF$ix[idx]
1+ any(DF$country[idx] != DF$country[i1 | i
Thank you for your comment. I would be very interested, but I did not know
about jitter. Would you have a small example?
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Thank you very much. I managed to do it with a similar approach and it looks
like what I expected.
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Hi,
try this:
dat1<-read.table(text="
Observation Gender Dosage Alertness
1 1 m a 8
2 2 m a 12
3 3 m a 13
4 4 m a 12
5 5 m b 6
6 6 m b
Hi,
You can also use ?merge()
dat$ind_dat<-TRUE
bad$ind_bad<-TRUE
res<-merge(dat,bad,all=TRUE)
res1<-res[is.na(res$ind_bad),][,1:3]
res1
# id year age
#5 2 2 Adult
#6 2 2
#7 2 3
#8 2 2
#9 2 2 Adult
#10 3 2
#11 3 3 Adult
#12 3 4
A.K.
- Ori
I want to connect R with HTML/PHP pages to take input from user,do some
statistical processing on it & show results to HTML page again.
I search on net,i got Rserve package,but examples are mainly for java
langaure & not for PHP
i am wondering how to connect it to PHP-Apache-MySQL
Is there any g
On Mon, Oct 22 2012, R. Michael Weylandt wrote:
> Design has been superceeded by rms -- give that a shot.
I know, but it doesn't look like it's a dorp-in replacement for Design,
so I suspect the code in the book won't work with it...
Joost
> On Mon, Oct 22, 2012 at 10:51 AM, Joost Kremers
>
Dear Simon & R-Users.I have applied your solution also for larger buffer
window,and it's working well.
But I'm getting the problem harder.I have two vectors: x<-
c(0,0,0,0,1,0,0,0,0,1,0,0,0) y<- c(0,-10,0,-10,0,0,0,0,-10,0,0,0,0)
And I want to know where the value -1 in x have at least 2 values
Hello everyone,
I have been trying to perform a Friedman test for a while with no success.
I have tried the command: > friedman.test (y~A|B, data = mydata), but it
doesn't work .
I searched youtube for help, but still could not find much. Can anyone help
me with that!
Thank you very much,
Pedro
I haven't determined the root cause of the numerical accuracy bug yet (I
haven't determined a way to compare the values used when the code is run
from R as opposed to from a Fortran test), but I did have another issue with
retrieving the matrix from the Fortran code that I hoped you might be able
t
Hello, I have the following sample dataset.
sex <- as.factor(c(rep(0,12),rep(1,12)))
char <-
c("x","x","x","x","y","y","y","y","z","z","z","z","x","x","x","x","y","y","y","y","z","z","z","z")
a <- runif(24,0,2)
b <- runif(24,4,10)
data <- data.frame(cbind(sex,char,a,b))
I would like to generate a
Thank you very much for your answer; my Gooogleing was not good engouh.
Claire
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Dear Dr. Murrell
Sorry to bother you at home, so to speak, but my emails to r-help@r-project.org
are falling into the void.
I am trying to make a multiple panel plot with 26 rows and 2 columns
with plots from a tool (genoPlotR) that is layered on grid. My code
and session info are below.
N=as.vector()
This doesn't mean anything in R: you can read the help using
?as.vector
to see what the required arguments are.
Do you want to loop over the elements of data? That's not what your
loop is doing. Instead it is looping over each element of the sequence
from the min to the max *value*
Hi all,
I'm trying to create a function where it can process a vector and also give
a vector output accordingly
#input: a,b anc c are constants, data is the vector
#set the function
fun<-function(a,b,c,data)
{
N=as.vector()
for (i in min(data):max(data)){
if(i>c){
N<-(a*(i-c)^0.5)+(b*(i-c))}
else
Hi,
In the FAQ for Mac http://cran.r-project.org/bin/macosx/RMacOSX-FAQ.html
there is
7.2 Saving the image
You can can (???) the content of the quartz device window into a PDF file.
This is not explained!
This is a very simple way to export high quality graphics from R into
other applications
Hi
> -Original Message-
> From: Bert Gunter [mailto:gunter.ber...@gene.com]
> Sent: Monday, October 22, 2012 4:00 PM
> To: PIKAL Petr
> Cc: noobmin; r-help@r-project.org
> Subject: Re: [R] How to use tapply with more than one variables grouped
>
> Inline.
>
> On Mon, Oct 22, 2012 at 6:55
Inline.
On Mon, Oct 22, 2012 at 6:55 AM, PIKAL Petr wrote:
> Hi
>
>
>> -Original Message-
>> From: r-help-boun...@r-project.org [mailto:r-help-bounces@r-
>> project.org] On Behalf Of noobmin
>> Sent: Monday, October 22, 2012 12:31 PM
>> To: r-help@r-project.org
>> Subject: Re: [R] How to
Tal:
There was a recent discussion on this list about this (Sam Steingold
was the OP IIRC).
The issue is ?c . In particular:
"c is sometimes used for its side effect of removing attributes except
names, for example to turn an array into a vector."
Hence, the factor attribute is removed and you
Hi
> -Original Message-
> From: r-help-boun...@r-project.org [mailto:r-help-bounces@r-
> project.org] On Behalf Of noobmin
> Sent: Monday, October 22, 2012 12:31 PM
> To: r-help@r-project.org
> Subject: Re: [R] How to use tapply with more than one variables grouped
>
> interTable <-data.
Hello all,
Please review the following simple code:
# make a factor:
x <- factor(c("one", "two"))
# what should be the output to the following expression?
c(x, "3")# <===
# I expected it to be as the output of:
c(as.character(x), "3")
# But in fact, the output is wh
Shortly ago I did something similar to what you intend to.
The data frame was called 'fore'. There were eight different variables and I
wanted to plot each one by column, and depict the frequencies in terms of the
size of the dots. The problem I ran into is that the plots got extremely big
for
I think it's more generally encouraged to "jitter" the points in this
circumstance.
RMW
On Mon, Oct 22, 2012 at 9:34 AM, swertie wrote:
> Hello! I would like to make a scatterplot of my data, but the problem is that
> several points have the same x and y values and are represented as only one
>
Hi Tammy,
I'm afraid this is pretty obviously homework, so we can't really do
much to help you. It's not a personal thing: just the considered
opinion of this list to believe that giving you answers (or even hefty
hints) may undermine whatever intent your teacher has in assigning the
problem. Ther
Design has been superceeded by rms -- give that a shot.
Cheers,
Michael
On Mon, Oct 22, 2012 at 10:51 AM, Joost Kremers
wrote:
> Hi all,
>
> I'm planning to work through the book "Analyzing Linguistic Data" by
> R.H. Baayen, which is an introduction to R used for, well, what the
> title says. ;-
You need to install netcdf, including its development files. The notes
you quote are pretty specific about that. Googling for that leads to
http://packages.debian.org/source/squeeze/netcdf
and it seems you need libnetcdf-dev and libnetcdf6 installed.
On 22/10/2012 11:39, claire1234 wrote:
H
Therneau doesn't know the answer either.
The predictions are positively correlated since they all depend on the same beta-hat from
the original model. The same will be true for any model: logistic, poisson, linear, ...
Terry T
On 10/20/2012 06:06 PM, Omar De la Cruz C. wrote:
I have a follow
Not in ggplot2 as far as I know. As Jim Lemon says the plotrix package should
handle this.
An alternative that probably gives better data interpretation might be to use
facet.grid() or perhaps facet_wrap() in ggplot2.
Crude example
library(ggplot2)
xx <- data.frame(aa = c(10, 12, 15, 55,
Hello,
If I understand it well,
idx <- !dat$id %in% bad$id
dat[idx, ]
Also, to create bad you are complicating, this would do:
bad <- data.frame(id = c(1,4))
Hope this helps,
Rui Barradas
Em 22-10-2012 12:04, penguins escreveu:
Hi, I am trying to remove a series of records from a large dat
Sent from my iPhone
Begin forwarded message:
> From: Michael Muratet
> Date: October 21, 2012 12:08:55 PM CDT
> To: r-help@r-project.org
> Subject: Navigating a grid.
>
> Greetings
>
> I am trying to make a multiple panel plot from a tool (genoPlotR) that is
> layered on grid. My code and s
On 12-10-22 3:24 AM, Christian Hoffmann wrote:
Hi,
Each > help.start() generates a new tree of the R help system, somewhere
in 127.0.0.1:xxx, each xxx being difeerent. This tree disappears when
exiting R. How can the current help tree copied to a permanent place for
reference outside a running R
Hi, I am trying to remove a series of records from a large dataframe. The
script I have written works fine but takes a long time to run. Can anyone
suggest a quicker way to do this?
Here is an example of the code I've written. The end result of this bit of
code would be a dataframe with any record
This is perilously close to a straightforward homework question which is not
usually answwered here but you get brownie points for saying you are a student.
Also you did not follow the posting guidelines so we have no idea what are
doing, what package(s) you may be using etc. Have a look at
Thanks, that has worked like a dream!
--
View this message in context:
http://r.789695.n4.nabble.com/conditional-value-assignment-tp4646945p4646988.html
Sent from the R help mailing list archive at Nabble.com.
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Hi,
Is there any Unicode Support for JSON that helps to read Latin letters?
Thanks
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Hello,
I am a new user of R and Linux (debian squeeze), and I am trying to install
the ncdf library.
First I don't know wich is the best to use (ncdf or ncdf4), But not matter
the one I tried to install, it always failled.
I try to find a solution to this common problem on the web, but I never fin
interTable <-data.frame (Tapply ($ contb_receipt_amt date, list ($ cand_nm
date, $ contbr_st date), sum))
I got create a table with the sum total contribution (contb_receipt_amt) of
each presidential candidate (cand_nm) in each state (contbr_st). How could
from interTable create a table of states
Hi,
I have a problem about the Latin letters in R. I try to get data from
facebook to oracle and while R is reading the data, it cannot write some
letters like "ü,Å,Ä" etc and just mark as "?". I know these are the Latin
letters and R cannot read it but is there any tool or function to convert
t
Hi
I was trying to implement Egarch (1,1) with Student t distribution using
rugarch. But I was not getting any value.
Following were the commands that I was using:
library(rugarch)
spec=ugarchspec(variance.model=list(model="eGARCH", garchOrder=c(1,1)),
mean.model=list(armaOrder=c(1,1), arfima=FA
Hello everyone,
I am developing a web application and I would like to perform two kind of
statistical/modeling operations.
(1) Batch analysis from data stored in the backend of my app (HBase
cluster). Typically, this operation needs to be performed on regular basis,
say every night. The size of
Hi all,
I'm planning to work through the book "Analyzing Linguistic Data" by
R.H. Baayen, which is an introduction to R used for, well, what the
title says. ;-) On the first page of the book, Baayen says that in order
to work with the book, R needs to download and install a number of
packages from
Hello! I would like to make a scatterplot of my data, but the problem is that
several points have the same x and y values and are represented as only one
point. I wonder if there is a way to represent the data points, but with
point size representing the frequency of the depicted x-y value?
Thank
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