I am trying to resolve a problem I am having with running the rattle package on
two different Windows 7 x64 systems. It appears to be a problem with my two
specific systems, because others on Windows 7 x64 systems aren't complaining
about this problem. What I am looking for is a method for try
Hi,
Try the cubature package, and maybe play with the tolerance.
HTH,
b.
On 13 March 2012 18:39, Niroshan wrote:
> Dear R Members,
>
> I want to know a fast R function to do multidimensional integration. I used
> the function 'cuhre' in R2cuba library. But it takes painful time to get the
> a
Dear R Members,
I want to know a fast R function to do multidimensional integration. I used
the function 'cuhre' in R2cuba library. But it takes painful time to get the
answer.
I would be thankful if anyone could help me out to solve this problem
Thanks
Niroshan
--
View this message in co
Hi,
I have 6 variables and I want to do a PCA Kernel on the 6 variables. But I
want the scores from the from the PCA kernel method. for each subject. Does
anyone know how to do this?
--
Thanks,
Jim.
[[alternative HTML version deleted]]
__
R-h
On Mar 12, 2012, at 8:51 PM, Alex Chen wrote:
Hello,
I am new to the R software and attempting to create a time series
model with trend, seasonality, as well as cycles. I have already been
able to create a model that incorporates both the trend and
seasonality and my approach is to then use th
Hello,
I am new to the R software and attempting to create a time series
model with trend, seasonality, as well as cycles. I have already been
able to create a model that incorporates both the trend and
seasonality and my approach is to then use the fitted results from
this model in order to creat
On Mar 12, 2012, at 3:07 PM, Emmanuel Levy wrote:
Hi Jeff,
Thanks for your reply and the example.
I'm not sure if it could be applied to the problem I'm facing though,
for two reasons:
(i) my understanding is that the inverse will associate a new Y
coordinate given an absolute X coordinate.
On Tue, Mar 13, 2012 at 9:39 AM, Davy wrote:
> Thanks for the reply. Sorry to be a pain, but could perhaps explain what you
> mean by
> "you can center each SNP variable at its mean to make the interaction
> term uncorrelated with the main effects".
Suppose rs1 and rs2 are your SNPs
rs1cent <- r
Will this work for you:
> a <- 3 + 2
> print(a)
[1] 5
> cat(a, '\n')
5
On Mon, Mar 12, 2012 at 6:49 PM, mrzung wrote:
> hi,
> is there any way to erase the term "[1]"?
>
> for example,
>
>> a<-3+2
>> a
> *[1]* 5
>
> the term [1] in front of number 5.
>
> is there any way?
>
> --
> View this me
You can write your own print method or do it using cat:
cat(a, "\n")
Michael
On Mon, Mar 12, 2012 at 6:49 PM, mrzung wrote:
> hi,
> is there any way to erase the term "[1]"?
>
> for example,
>
>> a<-3+2
>> a
> *[1]* 5
>
> the term [1] in front of number 5.
>
> is there any way?
>
> --
> View t
hi,
is there any way to erase the term "[1]"?
for example,
> a<-3+2
> a
*[1]* 5
the term [1] in front of number 5.
is there any way?
--
View this message in context:
http://r.789695.n4.nabble.com/erasing-1-tp4467628p4467628.html
Sent from the R help mailing list archive at Nabble.com.
_
Is there a simple way to use Rserve/RSclient as a proxy to transparently
send requests from a local instance of R to a remote instance? It seems
like this would by doable by wrapping each call that doesn't refer to a
local path inside RSeval. Is this harder than it seems? Does this already
exis
Dear R users,
I have been trying to draw the following 3d graphs: The solid region bounded
above by the paraboloid z = 9 - x2 - y2 and below by the unit circle in the
xy-plane.
I wanted to visualize the solid region bounded by those two graphs. I could
draw those two 3d graphs separately, but I c
Hello again,
Thank you, vito for these functions: I've been able to look at the smoothing
spline matrix and its derivatives that I constucted with functions
"create.bspline.basis" and "smooth.basis"
Understanding from Hastie and Tishbirani that the S(lambda)-- the penalty matrix
25 февраля 201
--- On Mon, 3/12/12, aleksandr shfets wrote:
> From: aleksandr shfets
> Subject: Fwd: Re[2]: [R] B-spline/smooth.basis derivative matrices
> To: "Vassily Shvets"
> Received: Monday, March 12, 2012, 5:15 PM
>
>
>
> Пересылаемое сообщение
>
> От кого: aleksandr shfets
> Ко
--- On Mon, 3/12/12, aleksandr shfets wrote:
> From: aleksandr shfets
> Subject: Fwd: Re[2]: [R] B-spline/smooth.basis derivative matrices
> To: "Vassily Shvets"
> Received: Monday, March 12, 2012, 5:15 PM
>
>
>
> Пересылаемое сообщение
>
> От кого: aleksandr shfets
> Ко
You said:
> Does this make sense?
Not to me. I don't understand either of your points. It may be time for you to
provide a concrete example of your own (reproducible R code, or at least
specific input and output data).
---
J
Thanks for the reply. Sorry to be a pain, but could perhaps explain what
you mean by
"you can center each SNP variable at its mean to make the interaction
term uncorrelated with the main effects".
I'm coming from a web based programming background so often statistical
terminology doesn't click with
On Mon, Mar 12, 2012 at 5:55 PM, David A. Kim wrote:
> Does anyone happen to know what method cluster() for Stata's xtreg
> uses to calculate clustered SEs for panel data, and/or how this could
> be implemented equivalently in R? Any help would be much appreciated.
The Stata manuals usually say
On Tue, Mar 13, 2012 at 7:11 AM, Davy wrote:
> Dear useRs,
>
> First off, sorry about the long post. Figured it's better to give context
> to get good answers (I hope!). Some time ago I wrote an R function that
> will get all pairwise interactions of variables in a data frame. This
> worked fine a
As the error message show, the problem isn't in the subscripting but
rather in passing a NA to `:`
If you try to implement Dr. Winsemius' method you should be fine.
Also, send plain text email.
Michael
On Mon, Mar 12, 2012 at 11:58 AM, Alaios wrote:
> You will find here a reprodusible example,
You get the "good" behavior with
base + aes(x = cut) + facet_wrap(~ color, ncol = 5)
so this seems buggy to me.
If someone here doesn't step forward with more insight, I'd forward it
to the ggplot list to see if one of the developers there can give an
explanation or possibly make the official ca
Joshua,
Just confirming quickly that your method using cmpfun and your f function
below was fastest using my real data. Again, thank you for your help!
Ben
On Sat, Mar 10, 2012 at 1:21 PM, Joshua Wiley wrote:
> On Sat, Mar 10, 2012 at 12:11 PM, Ben quant wrote:
> > Very interesting. You are d
My code doesn't give that error so you must have done something
differently -- without seeing your code, I can't tell you what though.
Also, please cc the list on follow-up if you are seeking further help.
I'm not particularly knowledgeable about rpanel and others might be
able to give concrete he
Dear useRs,
First off, sorry about the long post. Figured it's better to give context
to get good answers (I hope!). Some time ago I wrote an R function that
will get all pairwise interactions of variables in a data frame. This
worked fine at the time, but now a colleague would like me to do this
Folks,
I have been working on VAR and VECM models and trying to simulate the results.
This is easy to do with a VAR model (in the "vars" package) as the covariance
matrix is easily extracted.
Unfortunately I can't figure out how to do this with a VECM fit.
I have used vec2var and created the
Hi Jeff,
Thanks for your reply and the example.
I'm not sure if it could be applied to the problem I'm facing though,
for two reasons:
(i) my understanding is that the inverse will associate a new Y
coordinate given an absolute X coordinate. However, in the case I'm
working on, the transformatio
Thanks both! That solves ! You've made a very happy newbie!
Simon
On 2012-03-12, at 2:52 PM, Sarah Goslee wrote:
> Hi Simon,
>
> On Mon, Mar 12, 2012 at 2:37 PM, Simon Kiss wrote:
>> Hi: I'm sure this is a very easy problem. I've consulted Data Manipulation
>> With R and the R Book and can't fi
Your function doesn't return the new data frame but rather the new
names. Note, e.g.
x <- 1:2
names(x) <- letters[1:2]
.Last.value # Not x!
Try this:
.xx<- lapply(.xx, function(x) {colnames(x)<-c('State', 'Year'); x})
or more explicitly
.xx<- lapply(.xx, function(x) {colnames(x)<-c('State', '
Hi Simon,
On Mon, Mar 12, 2012 at 2:37 PM, Simon Kiss wrote:
> Hi: I'm sure this is a very easy problem. I've consulted Data Manipulation
> With R and the R Book and can't find an answer.
>
> Sample list of data frames looks as follows:
>
> .xx<-list(df<-data.frame(Var1=rep('Alabama', 400), Var2
Hi,
On Mon, Mar 12, 2012 at 2:37 PM, Simon Kiss wrote:
> Hi: I'm sure this is a very easy problem. I've consulted Data Manipulation
> With R and the R Book and can't find an answer.
>
> Sample list of data frames looks as follows:
>
> .xx<-list(df<-data.frame(Var1=rep('Alabama', 400), Var2=rep(c
Hi: I'm sure this is a very easy problem. I've consulted Data Manipulation With
R and the R Book and can't find an answer.
Sample list of data frames looks as follows:
.xx<-list(df<-data.frame(Var1=rep('Alabama', 400), Var2=rep(c(2004, 2005, 2006,
2007), 400)), df2<-data.frame(Var1=rep('Tennes
The problem is: by default shouldn't it use "Huber's"?
And it should be convex problem no?
so when I do rlm(y~x) which is a single-beta fitting problem,
shouldn't it always converge?
Thanks!
Psi functions are supplied for the Huber, Hampel and Tukey bisquare
proposals as
Bill and Brian,
Thank you both for the finding my error. force() appears to do the trick.
Thanks again,
James
On Mon, Mar 12, 2012 at 1:09 PM, William Dunlap wrote:
> I think that adding a force(i) to your inner function should fix things up.
> > z <- mapply(assign,
> + c("half", "
I think that adding a force(i) to your inner function should fix things up.
> z <- mapply(assign,
+c("half", "third", "quarter"),
+lapply(2:4, function(i) {
+ force(i)
+ function(x) divide(x, i)}),
+pos = 1)
> half(10)
[1] 5
> third(
On Mar 12, 2012, at 1:19 PM, jianghong deng wrote:
Hello,
I have created two boxplots with following R code. There is one
outlier in
B group.
The outlier is 33. But the all other data are between 0 to 4.
How can I skip y-axis around 5 to 25, and expand 0-4 for this case.
Also I
want kee
On 3/12/2012 10:47 AM, J Toll wrote:
Hi,
I have a problem that I'm finding a bit tricky. I'm trying to use
mapply and assign to generate curried functions. For example, if I
have the function divide
divide<- function(x, y) {
x / y
}
And I want the end result to be functionally equivalent
Jianghong wrote:
> Hello,
>
> I have created two boxplots with following R code. There is one
outlier in
> B group.
> The outlier is 33. But the all other data are between 0 to 4.
>
> How can I skip y-axis around 5 to 25, and expand 0-4 for this case.
Also I
> want keep the outlier in my boxplot.
Hi,
I have a problem that I'm finding a bit tricky. I'm trying to use
mapply and assign to generate curried functions. For example, if I
have the function divide
divide <- function(x, y) {
x / y
}
And I want the end result to be functionally equivalent to:
half <- function(x) divide(x, 2)
th
Hello.
Such a procedure is not implemented in 'plm' but you can probably get
around to do what you want. One possible way (my preferred one) is to
demean the data by both dimensions as you would do for a two-way FE
model, and then estimate a random parameters model on the demeaned data.
This would
Hello,
I have created two boxplots with following R code. There is one outlier in
B group.
The outlier is 33. But the all other data are between 0 to 4.
How can I skip y-axis around 5 to 25, and expand 0-4 for this case. Also I
want keep the outlier in my boxplot.
I want my boxplot look like the
On Mon, Mar 12, 2012 at 09:14:43AM -0700, behave14 wrote:
> Dear community
>
> I have the following problem. I'd like to have the maximum of the remaining
> elements of a vector.
>
> ex.
> x<-c(1,2,3,4,1,2,3,1,2,1)
>
> f(x) = c(4,4,4,4,3,3,3,2,2,1)
>
> where the first element is: max(x[1:leng
> -Original Message-
> From: r-help-boun...@r-project.org [mailto:r-help-bounces@r-
> project.org] On Behalf Of behave14
> Sent: Monday, March 12, 2012 11:15 AM
> To: r-help@r-project.org
> Subject: [R] Maximum of remaining elements of vector
>
> Dear community
>
> I have the following pr
On 12 Mar 2012, at 12:47 , S Ellison wrote:
> Yes, to the extent that the default barplot plots the height of the bar so
> far as the sum of teh values so far, starting at teh first. For your first
> vector, no problem; for your second, the highest value is undefiuned, for the
> third, the sum
> -Original Message-
> Am I wrong that barplot is supposed to just skip NAs, and
> continue with the rest of the data in a matrix column? That's
> how I read various posts on the subject.
>
> But that's not what happens for me with R64.app (on a Mac,
> obviously). For example:
>
> d
You can use gsub() to delete all the characters after the first
semicolon by replacing them with "". If you provide the requested
reproducible example, I imagine someone will provide you with working
code, if reading the help for gsub() isn't enough to get you going.
Sarah
On Mon, Mar 12, 2012 at
Also, if it is not homework, why are you reinventing some functionality
already in R? R can easily calculate medians for you.
Kevin
On 03/12/2012 12:35 PM, Sarah Goslee wrote:
Did you try what I suggested?
If you go to www.rseek.org and search for "odd or even integers" you
will get various
Make A Profit with your own home businesshttp://tiz.co.kr/555cd.php?yppage=a9k2
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https://stat.ethz.ch/mailman/listinfo/r-help
PLEA
Hi,
I have a list of names (alphanumeric characters) and each row by itself
contains a different number of names seperated by a semicolon (";"). There
are also rows with no entry. I would like to have the name before the first
semicolon in each row, if there is any entry. Does anyone have an idea h
Dear community
I have the following problem. I'd like to have the maximum of the remaining
elements of a vector.
ex.
x<-c(1,2,3,4,1,2,3,1,2,1)
f(x) = c(4,4,4,4,3,3,3,2,2,1)
where the first element is: max(x[1:length(x])
the second one: max(x[2:length(x)])
the third one: max(x[3:length(x)]) an
Did you try what I suggested?
If you go to www.rseek.org and search for "odd or even integers" you
will get various solutions.
You could even write your own test by comparing x/2 with round(x/2).
Also, it's good form to copy the list on your replies, not just me.
Sarah
On Mon, Mar 12, 2012 at
Mark,
It is of great help. Thank you very much. I appreciate it.
Sincerely,
Jialin Huang
On Mon, Mar 12, 2012 at 11:27 AM, Mark Lamias wrote:
> cov(A) will compute a covariance matrix of the columns of a matrix A.
>
> cov(a, b) will compute a covariance of the vectors a and b.
>
> here are so
cov(A) will compute a covariance matrix of the columns of a matrix A.
Â
cov(a, b) will compute a covariance of the vectors a and b.
Â
here are some examples:
Â
#create random vectors
a=rnorm(10)
b=rnorm(10)
Â
Â
> a
 [1] -0.3110384 -2.6201009 -1.3078118 1.3384743 2.4032381 -0.7245201
 [7]
d2 <- as.matrix(c(2,NA,4))
barplot(d2,beside=T)
barplot(c(d2))
barplot(na.omit(d2))
d2[2,] <- 0
barplot(d2)
# So barplot is not "stopping" at the first NA (first 2 plots). But
what does stacking even mean when you have a missing group in the
middle ? you can't expect barplot to know... if yo
Hi Michael,
Thanks so much for your detailed reply!
I gained a better understanding of the read.fwf function, along with
ensuring I better note how these read-in functions convert variables, etc.
As well, your tip on removing "format" while converting the PRVDR_NUM
variable to numeric (from fac
Hi,
On Mon, Mar 12, 2012 at 12:10 PM, wrote:
> I've tried the if function? Its just defining when something is even or odd?
R's help system might be useful. www.rseek.org might be useful.
You haven't accomplished the second part of my response, and this
isn't a homework help list, so you'll ne
On Mar 12, 2012, at 12:05 PM, Prof Brian Ripley wrote:
On 12/03/2012 15:29, z2.0 wrote:
Trying to install RMySQL on 64-bit Windows 7.
Using R-2.14.2 with Rtools214 and MySQL Server 5.5.
Read through several step-by-steps of RMySQL source installation.
Troubleshooting:
- Copied libmysql.dll
Dear all
with image I can plot only one set of values in one plot.
Do somebody have any insight how to put those 2 matrices into one picture
so that in one cell in image picture are both values from mat[1,1] and
mat2[1,1].
mat<-matrix(1:4, 2,2)
mat2<-matrix(4:1,2,2)
x <-1:2
y <-1:2
image(x, y
You expected the sum to be 1 and it is:
> a <- hist(Y1, breaks=seq(0, 350, by=35), col="grey80", freq=FALSE)
> a$density*35
[1] 0.27923387 0.14717742 0.09979839 0.08064516 0.06955645 0.06350806
[7] 0.05241935 0.05342742 0.05544355 0.09879032
> sum(a$density*35)
[1] 1
Note that the first density
On 07.03.2012 18:31, Miklós Emri wrote:
On 03/07/2012 04:58 PM, Prof Brian Ripley wrote:
On 07/03/2012 15:45, Duncan Murdoch wrote:
On 12-03-07 7:59 AM, Miklós Emri wrote:
Dear experts,
I have to install ggplot2 packages for R 2.11 and 2.12 but this is
available for 2.14 only.
My question:
On 12/03/2012 15:29, z2.0 wrote:
Trying to install RMySQL on 64-bit Windows 7.
Using R-2.14.2 with Rtools214 and MySQL Server 5.5.
Read through several step-by-steps of RMySQL source installation.
Troubleshooting:
- Copied libmysql.dll to R-2.14.2/bin AND R-2.14.2/bin/i386.
- Copied libmys
Doran,
Thanks for your advice.
Sincerely,
Jialin Huang
On Mon, Mar 12, 2012 at 10:56 AM, Doran, Harold wrote:
> Always start with the help page. But, the best way to get help on this
> list is to construct some toy data. Even if imperfect, people will work to
> help you. Maybe show us some of
I'd start by using if().
If you can convince the list that this isn't homework, you might get more help.
Sarah
On Mon, Mar 12, 2012 at 7:15 AM, wrote:
> Hi just a quick question in which I must use the "function and return" part
> of R.
>
> If I was asked to define a function called "median"
You will find here a reprodusible example, and you will also see it "exploding"
myData<-seq(1:10)
windowList<-list(1,2,3,4)
totalLength<- length(myData)
windowSize<-30
for (i in 2:length(windowList)) {
myMean<-array()
indexes<-seq(1,totalLength,by=windowSize*windowList[[i]])
for
Dear all,
I have developed a loop but receive an error (error 1 below) because the
mcstoc function prohibits arrays. I tried creating an mcnode to represent
the data frame, but received Error 2 below (array size not compatible with
node dimension).
I'm still at a loss on how to apply the Monte-Ca
On Sat, Mar 10, 2012 at 04:44:00AM -0800, Alaios wrote:
> Dear all,
> I have a large vector (lets call it myVector) and I want to plot its value
> with the logic below
>
> yaxis<-myVector[1]
> yaxis<-c(xaxis,mean(myvector[2:3])
> yaxis<-c(xaxis,mean(myvector[4:8])
> yaxis yaxis
> this has to sto
Trying to install RMySQL on 64-bit Windows 7.
Using R-2.14.2 with Rtools214 and MySQL Server 5.5.
Read through several step-by-steps of RMySQL source installation.
Troubleshooting:
- Copied libmysql.dll to R-2.14.2/bin AND R-2.14.2/bin/i386.
- Copied libmysql.dll and libmysql.lib to MySQL Serv
Always start with the help page. But, the best way to get help on this list is
to construct some toy data. Even if imperfect, people will work to help you.
Maybe show us some of your actual data.
Of course, if a and b1 are really scalars, I donât think there is such a
thing as the covariance
Hi just a quick question in which I must use the "function and return" part of
R.
If I was asked to define a function called "median" where the function has a
single argument x and returns the median value of a vector. How would I go
about saying the median = element x in terms of n (say (n + 1
Hi,
Thanks for your response.
I tried to use cov() before, but got NA all over the place.
a and b1 are scalars in the example. So that is the problem. I will start
over from the formula.
Thanks again.
Sincerely,
Jialin Huang
On Mon, Mar 12, 2012 at 10:47 AM, Doran, Harold wrote:
> Can you
Can you provide sample data? This suggests that a and b1, for example, are
scalars. If that is what you actually have then you cannot produce a covariance
between two scalars.
> -Original Message-
> From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
> Behalf Of
On Mon, Mar 12, 2012 at 11:42 AM, huang jialin wrote:
> Hello,
>
> I have a vector {a, b1, b2, b3, b4}.
What does this mean? Is this a character vector giving names of
objects that exist elsewhere in the workspace? Else, how do you tell
a/b1/b2 apart in a vector?
You probably want ?cov but apply
?cov
HTH,
Jorge.-
On Mon, Mar 12, 2012 at 11:42 AM, huang jialin <> wrote:
> Hello,
>
> I have a vector {a, b1, b2, b3, b4}. How can I calculate the following
> matrix:
>
> var(a) cov(a, b1) cov(a, b2) cov(a, b3) cov(a, b4)
> cov(a, b1) var(b1) cov(a, b2) cov(a, b3) cov(a, b4)
> ...
> ..
Hello,
I have a vector {a, b1, b2, b3, b4}. How can I calculate the following
matrix:
var(a) cov(a, b1) cov(a, b2) cov(a, b3) cov(a, b4)
cov(a, b1) var(b1) cov(a, b2) cov(a, b3) cov(a, b4)
...
...
cov(a, b1) cov(a, b2) cov(a, b3) cov(a, b4) var(b4)
I would very appreciate your inputs.
It's not always true that going out of bounds in subscripting gives an
error (as you seem to assert in your original post)
x <- 1:3
x[4] # No error
and note that mean() has a na.rm argument.
Perhaps you should construct a *reproducible* example of what you
think will go wrong.
Michael
On Mon,
Yep I did some mistake while I was typing the example
What I want is this one
cbind(2^(0:(n-1))) and the problem is that at the last "window" this is going
to explode are there will be less elements than what the window asks for.
How should I deal with that?
Regards
Alex
_
Le lundi 12 mars 2012 à 14:48 +, aoife doherty a écrit :
> Hey guys,
>
> if i do a correspondance analysis, e.g.:
>
> table <- structure(c(4, 7, 0.2, 3, .1, 7, 222, 3, 10, 5, 11,
>8, 8, 10, 7), .Dim = c(5L, 3L), .Dimnames = list(c("gene1",
>"gene2", "gene3", "gene4", "gene5"), c("codo
Hey guys,
if i do a correspondance analysis, e.g.:
table <- structure(c(4, 7, 0.2, 3, .1, 7, 222, 3, 10, 5, 11,
8, 8, 10, 7), .Dim = c(5L, 3L), .Dimnames = list(c("gene1",
"gene2", "gene3", "gene4", "gene5"), c("codon1", "codon2",
"codon3")))
Library(ca)
plot(ca(table))
is there a way
Hello David.
Usually I'd ask for a reproducible example (see the posting guide), but
as I routinely check my results against Stata, this time I think I know
what happens already.
There are two issues here: one is cluster-robust covariance estimation,
which in Stata is done through 'vce(cluster )'
On Sun, Mar 11, 2012 at 9:21 PM, frauke wrote:
> Wow Jim, this is much more than I expected. Thank you!!
>
> It took me a while to figure out what exactly you are doing in that code.
> But I think I understand and it definitely runs. May I ask you two follow up
> questions?
>
> First, some of my f
Dear Jessica,
As you can see, it's a little hard to look at your commands because of the
lack of newlines, and you also don't give data, so one can't tell which
variables are latent and which observed, but I'm going to guess that this is
a latent-variable model with factors belonging, autonomy, op
> Peter Ehlers
> on Sat, 10 Mar 2012 11:04:06 -0800 writes:
> On 2012-03-10 08:35, sybil kennelly wrote:
>> Dear all.
>>
>> I wanted to read in a 20,000 row X 60 column matrix
>> (called "table") into R.
>>
>> i did this:
>>
>>> R table<- read.ta
Close the device anyways; test for an error as Sarah (and the help
pages) describe and use unlink() to delete the resultant (empty) file
if the plot call fails.
Michael
On Mon, Mar 12, 2012 at 8:30 AM, Alaios wrote:
> Thanks a lot for the answer..
> As you can see my chain is the following
> op
Ok.
On Mon, Mar 12, 2012 at 8:17 AM, Duncan Murdoch wrote:
> On 12-03-11 4:30 PM, Eduardo Mendes wrote:
>
>> Dear Prof. Ripley
>>
>> Many thanks. rtools 2.15 did the job.
>>
>> One note if I may - In the rtools site one reads
>>
>> rtools215.exe - r>2.14.1 to R 2.15.x No
>> rtoosl214.exe - r 2.13
Thanks a lot for the answer..
As you can see my chain is the following
open a pdf device.
plot and
close the device
pdf(file=filename, width=width)
try(plot_per_band_ (list(get(varlist[[k]])), i, j,meanPlot=TRUE))
dev.off()
I would like if plot fails (so try returns try-error) the file to not b
On Sat, 2012-03-10 at 12:00 +0100, r-help-requ...@r-project.org wrote:
> Hi All,
>
> s = Surv(outcome.[,1], outcome.[,2])
> survplot= (survfit(s ~ person.list[,1]))
>
> summary(survplot)
>
> This prints a summary of all the curves at specified
On Mon, Mar 12, 2012 at 4:39 AM, Alaios wrote:
> Thanks a lot.
> How I can use try though for plots. There are times where the plot function
> will fail
>
> pdf(file=filename, width=width)
> try(plot_per_band(list(get(varlist[[k]])), i, j, l,
> datalabels=seq(1:length(varlist
> dev.off()
>
On 12-03-11 4:30 PM, Eduardo Mendes wrote:
Dear Prof. Ripley
Many thanks. rtools 2.15 did the job.
One note if I may - In the rtools site one reads
rtools215.exe - r>2.14.1 to R 2.15.x No
rtoosl214.exe - r 2.13.x or R.2.14.x yes
which can mislead the reader. 2.14.x could mean 2.14.2
But as
I have encountered a problem with faceted bar plots. I have tried to
create something like the example explained in the ggplot2 book (see pp.
126-128):
library(ggplot2)
mpg4 <- subset(mpg, manufacturer %in%
c("audi", "volkswagen", "jeep"))
mpg4$manufacturer <- as.character(mpg4$manufacture
2012/3/12 David Winsemius :
>
> On Mar 12, 2012, at 4:29 AM, Mauricio Zambrano-Bigiarini wrote:
>
>> 2012/3/9 David Winsemius :
>>>
>>>
>>> On Mar 8, 2012, at 8:02 AM, Mauricio Zambrano-Bigiarini wrote:
>>>
Dear list members,
Within a loop, I need to create an xyplot with only a lege
Thank you Jeff, I took the short-cut and updated R to R-2.14.2 and it works
too.
A.
On 12 March 2012 06:43, Jeff Newmiller wrote:
> Works for me on Windows XP with R 2.14.1 (not patched).
>
> You may be too far out on the bleeding edge. Upgrading to 2.14.2 is
> probably the best first step. If
2012/3/9 David Winsemius :
>
> On Mar 8, 2012, at 8:02 AM, Mauricio Zambrano-Bigiarini wrote:
>
>> Dear list members,
>>
>> Within a loop, I need to create an xyplot with only a legend, not even
>> with the default external box drawn by lattice.
>>
>> I already managed to remove the axis labels and
2012/3/9 Greg Snow <538...@gmail.com>:
> Why do you want to do this? Lattice was not really designed to put
> just part of the graph up, but rather to create the entire graph using
> one command.
To make a long story short, you are right, I need to do this because
my knowledge with trellis graphi
On 12/03/12 21:44, jpm miao wrote:
Hello,
Thanks very much for your kind response. Yes, if I multiply by the
width "35", the area should be equal to one.
How can I plot the probability bars rather than density bars? That
is, I would like the height of the first bar to be 0.279, which i
On Mar 12, 2012, at 4:29 AM, Mauricio Zambrano-Bigiarini wrote:
2012/3/9 David Winsemius :
On Mar 8, 2012, at 8:02 AM, Mauricio Zambrano-Bigiarini wrote:
Dear list members,
Within a loop, I need to create an xyplot with only a legend, not
even
with the default external box drawn by latt
Hello,
Thanks very much for your kind response. Yes, if I multiply by the width
"35", the area should be equal to one.
How can I plot the probability bars rather than density bars? That is, I
would like the height of the first bar to be 0.279, which is the
probability that the variable fall
Thanks a lot.
How I can use try though for plots. There are times where the plot function
will fail
pdf(file=filename, width=width)
try(plot_per_band(list(get(varlist[[k]])), i, j, l,
datalabels=seq(1:length(varlist
dev.off()
the code above does not print anything.
B.R
Alex
__
Why do you say that the area is much lower than 1?
It is exactly equal to 1.
How did you calculate this area?
Your code seems extremely convoluted and confused.
You could construct your (rather bizarre) vector Y1 much
more simply as
Y1 <- rep((0.5 + 0:9)*35,c(277,146,99,80,69,63,52,53,55,98))
Am I wrong that barplot is supposed to just skip NAs, and continue with the
rest of the data in a matrix column? That's how I read various posts on the
subject.
But that's not what happens for me with R64.app (on a Mac, obviously). For
example:
d0 <- as.matrix(c(2,3,4))
d1 <- as.matrix(c(2,3,N
I'm trying to replicate a time-series cross-sectional analysis
(countries over years) with SEs clustered by country. The original
analysis was done in Stata 10 with: xtreg [DV] [IVs] fe
cluster(country).
Using plm() in R (cran.r-project.org/web/packages/plm/index.html),
I've replicated the coeffi
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