i tried using your suggestion but still can't manipulate the dimensions such
that the chart fills the whole window. i've attached the picture i get when
i tried your suggested method. as you can see, there's still a lot of white
space around it.
is there anyway i can:
a) output to png ONLY the 3d
You should also check out the functions head() and tail(). For
instance, if you type head(bmtrend), the first few rows of your data
will be displayed.
Also, if you type tail(bmtrend), then the last few rows will be displayed.
--
John A. Ramey, M.S.
Ph.D. Candidate
Department of Statistical Scie
Hello,
I want to solve: x*(3^x)*log(4)-x*log(4/3)-(3^x)+1=0 for x. I used the
following code,
uniroot(function(x) x*(3^x)*log(4)-x*log(4/3)-(3^x)+1, lower = -2, upper = 2,
tol = 0.001 )
While using this I am getting the following error. Can anyone please help me
out.
Error in uniroot(function
I am trying to install R2.11 on RHEL5.3. The main code and base
packages compile fine and get installed to \usr\local\R211 (which I
set using --prefix). I would like to install a set of contrib
packages (car, ggplot2, etc.) also to the same location. Once R is
fired up, I can do install.packages()
Dear All,
I want to create a surface plot from the data. My data set is consists of x,
y and z data.
I plotted in very easy way by Excel worksheet as shown in the attached
picture.
I did some steps in R, but I cannot have the same plot as in Excel
worksheet's figure.
the R code is
x <- c(-
Need way to resize an existing graphics window.
This should be applicable across platforms (as part of a package).
Context: function1() draws main plot (I'm using grid), function2() adds smaller
plot
above main plot, but this one can sometimes overflow the original graphics
window
area.
Thanks
On Sat, May 1, 2010 at 11:55 AM, Tal Galili wrote:
> Hi Mohan,
> Check:
>
> dim(bmtrend)
>
> If the output is like the dimension of your data, then it would appear you
> succeeded in reading in the data :)
>
Yes . You are right. It reads all the data.
Thanks you.
Mohan L
[[alternativ
Hi Mohan,
Check:
dim(bmtrend)
If the output is like the dimension of your data, then it would appear you
succeeded in reading in the data :)
Best,
Tal
Contact
Details:---
Contact me: tal.gal...@gmail.com | 972-52-7275845
Rea
Hi All,
I am new to R and the mailing list.
I have a data file (.xls) format with little bit large (16 column,35000 rows
) data file. I am trying to read this file for calculation. I have
converted the file into .csv format and read like that :
> bmtrend <- read.csv(file="simple.csv",head=TRUE,
On Sat, May 1, 2010 at 11:02 AM, Mohan L wrote:
> Hi All,
>
> I am new to R and the mailing list.
>
> I have a data file (.xls) format with little bit large (16 column,35000
> rows ) data file. I am trying to read this file for calculation. I have
> converted the file into .csv format and read l
Dear R lists,
Would anyone help me with setting data frames according to the date?
X.RIC Date.G. Time.G. GMT.Offset Type Price
VolumeTime
1 QAN.AX 01-DEC-2008 00:00:28.611 11 Trade 2.28105 2008-12-01
11:00:28.611
2 QAN.AX 01-DEC-2008 00:00:43.155
On 04/30/2010 05:45 PM, Marc Schwartz wrote:
On Apr 30, 2010, at 5:13 PM, Max Gunther wrote:
Dear "R" list,
Our statisticians usually give us results back in a PDF format. I would like
to be able to copy and past tables from "R" output directly into a Microsoft
Word table since this will save
On 30/04/2010 9:01 PM, Mandy Xu wrote:
i tried using your suggestion but still can't manipulate the dimensions such
that the chart fills the whole window. i've attached the picture i get when
i tried your suggested method. as you can see, there's still a lot of white
space around it.
is there an
Greg Snow wrote:
When I work with clients who want to cut and paste to word or
powerpoint I usually use the odfWeave package, set up a template file
with the tables and graphs (possibly other output), then I run that
through odfWeave and then use openoffice to save the results as a
word file that
Hello R users,
I am trying to read files from a website that requires authentication.
I have been working with url() to to open a connection in Windows.
For example:
setInternet2(TRUE)
con <-
url(description="http://myusername:mypassw...@www.somesite.com/myfile.txt";,
open="r")
open(con)
read.
Greg Snow wrote:
When I work with clients who want to cut and paste to word or
powerpoint I usually use the odfWeave package, set up a template file
with the tables and graphs (possibly other output), then I run that
through odfWeave and then use openoffice to save the results as a
word file that
i tried using your suggestion but still can't manipulate the dimensions such
that the chart fills the whole window. i've attached the picture i get when
i tried your suggested method. as you can see, there's still a lot of white
space around it.
is there anyway i can:
a) output to png ONLY the 3d
Sorted!
Thanks everybody!
2010/4/30 Marc Schwartz :
> Dang ityou are right Greg.
>
> names <- c("ABC", "ABCD", "ABCDE", "ABCDEF")
>
>> substr(names, 1, nchar(names) - 1)
> [1] "AB" "ABC" "ABCD" "ABCDE"
>
>
> and...that is faster than using gsub().
>
>
> Thanks for catching thatnow i
Note the "diag" argument of upper.tri(). Thus:
x <- A[upper.tri(A,diag=TRUE)]
Then to recreate A from x:
A <- matrix(0, 4, 4)
A[upper.tri(A,dia=TRUE)] <- X
A <- A+t(A)
diag(A) <- diag(A)/2
(I grant that the division may cause slight floating point precision error)
Bert Gunter
Genentech
The yags package has QIC. Try
library(yags)
example(yags)
SPRUJ
On Fri, Apr 30, 2010 at 6:34 PM, Sachi Ito wrote:
> Hi,
>
> I'm using 'geepack' to run Generalized Estimating Equations. I'm aware that
> I can use anova to compare two models, but would it be possible to test QIC
> on R? It se
I did not understand enough of the rest of your question to give any better
response than others have given.
Looking back at your previous posts, there is one suggestion that I can make
that may help. You can use the approx or approxfun functions to approximate an
inverse, just generate a bunc
Dang ityou are right Greg.
names <- c("ABC", "ABCD", "ABCDE", "ABCDEF")
> substr(names, 1, nchar(names) - 1)
[1] "AB""ABC" "ABCD" "ABCDE"
and...that is faster than using gsub().
Thanks for catching thatnow it's time for cawfee
Marc
On Apr 30, 2010, at 6:28 PM, Greg Sno
No, substr is vectorized, you just have a typo, you are using a different
vector for nchar than you are subsetting.
--
Gregory (Greg) L. Snow Ph.D.
Statistical Data Center
Intermountain Healthcare
greg.s...@imail.org
801.408.8111
> -Original Message-
> From: r-help-boun...@r-project.or
On 2010-04-30 15:48, Prasenjit Kapat wrote:
Hello:
[Kindly Cc when replying]
The question in a nutshell is this: Is there a more robust alternative
to Ecdf ()?
The details:
I've used Ecdf () _a lot_ over the past few years and I have learned
to live with its warnings. But I am running short o
I have already learned a lot from the list, both technical and not,
and cannot thank enough for those valuable suggestions. In fact, as
said in my previous posts, I got really critical help and advices,
which really addresses the issues I have.
By the way, there is one point or two in your post I
You have the correct general idea, but it looks like lp may have already been
transformed to be in the range 0-1 rather than number of days, if that is the
case then you don't need to dived by 29. Again.
--
Gregory (Greg) L. Snow Ph.D.
Statistical Data Center
Intermountain Healthcare
greg.s...@
> -Original Message-
> From: r-help-boun...@r-project.org
> [mailto:r-help-boun...@r-project.org] On Behalf Of Greg Snow
> Sent: Friday, April 30, 2010 4:04 PM
> To: mailing-l...@gmx.net; r-help@r-project.org
> Subject: Re: [R] as.environment Error
>
> I don't know what is specifically ca
So does each person have multiple rows and you want to sample the set of rows?
The usual approach that I take is to split them into a list, sample from the
list, the put the list back together, for example:
tmp1 <- split(as.data.frame(state.x77), state.division)
tmp2 <- sample(tmp1, replace=TRU
On 2010-04-30 15:26, Nicholas Andrews wrote:
Here's an easy question: I'd like to convert a symmetric matrix to a
vector composed of the upper.tri() part of it plus the diagonal, and
convert it back again. What's the best way to achieve this? I'm
wondering if there are some built in functions t
I don't know what is specifically causing the error, but I think that you will
be happier in the long run (and probably short run) if you abandon the use of
assign and <<- and instead use lists.
--
Gregory (Greg) L. Snow Ph.D.
Statistical Data Center
Intermountain Healthcare
greg.s...@imail.org
odfWeave might be a less daunting option here, as it can output results in
some .doc formats. I have no idea how well tables would survive the
translations, however.
-- Bert
Bert Gunter
Genentech Nonclinical Statistics
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help
On Apr 30, 2010, at 5:44 PM, Sebastian Kruk wrote:
> Hi, i have a vector filled with names:
>
> [1] Alvaro Adela ...
> [25] Beatriz Berta ...
> ...
> [10] ...
>
> I would like to drop last character in every name.
>
> I use the next program:
>
> for (i in 1:10) {
>
When I work with clients who want to cut and paste to word or powerpoint I
usually use the odfWeave package, set up a template file with the tables and
graphs (possibly other output), then I run that through odfWeave and then use
openoffice to save the results as a word file that I can send to t
The nchar and substring functions are both vectorized, you can do something
like:
> substring(state.name, 1, nchar(state.name)-1)
And it should be much faster.
--
Gregory (Greg) L. Snow Ph.D.
Statistical Data Center
Intermountain Healthcare
greg.s...@imail.org
801.408.8111
> -Original M
On Apr 30, 2010, at 5:13 PM, Max Gunther wrote:
> Dear "R" list,
>
> Our statisticians usually give us results back in a PDF format. I would like
> to be able to copy and past tables from "R" output directly into a Microsoft
> Word table since this will save us tons of time, be more accurate to
>
On Apr 30, 2010, at 5:26 PM, Nicholas Andrews wrote:
Here's an easy question: I'd like to convert a symmetric matrix to a
vector composed of the upper.tri() part of it plus the diagonal, and
convert it back again. What's the best way to achieve this? I'm
wondering if there are some built in f
Hi, i have a vector filled with names:
[1] Alvaro Adela ...
[25] Beatriz Berta ...
...
[10] ...
I would like to drop last character in every name.
I use the next program:
for (i in 1:10) {
largo <- nchar(names[i]-1)
names[i] <- substri
On 04/30/2010 05:13 PM, Max Gunther wrote:
Dear "R" list,
Our statisticians usually give us results back in a PDF format. I would like
to be able to copy and past tables from "R" output directly into a Microsoft
Word table since this will save us tons of time, be more accurate to
minimize human
On Apr 30, 2010, at 4:57 PM, Erik Iverson wrote:
>
>> I'm sure it's not a bug, but could someone point to a thread or offer some
>> gentle advice on what's happening? I think it's related to:
>> test <- data.frame(name1 = 1:5, name2 = 6:10, test = 11:15)
>> eval(expression(test[c("name1", "name
On Fri, Apr 30, 2010 at 11:13 PM, Max Gunther
wrote:
> Dear "R" list,
>
> Our statisticians usually give us results back in a PDF format. I would like
> to be able to copy and past tables from "R" output directly into a Microsoft
> Word table since this will save us tons of time, be more accurate
Hi,
I'm using 'geepack' to run Generalized Estimating Equations. I'm aware that
I can use anova to compare two models, but would it be possible to test QIC
on R? It seems that there were similar questions a couple of years ago, but
the question has not been answered yet.
I'd appreciate if someo
> -Original Message-
> From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-
> project.org] On Behalf Of Kyeong Soo (Joseph) Kim
> Sent: Friday, April 30, 2010 4:10 AM
> To: kMan
> Cc: r-help@r-project.org
> Subject: Re: [R] Curve Fitting/Regression with Multiple Observations
[snip
On Apr 30, 2010, at 6:13 PM, Max Gunther wrote:
Dear "R" list,
Our statisticians usually give us results back in a PDF format. I
would like
to be able to copy and past tables from "R" output directly into a
Microsoft
Word table since this will save us tons of time, be more accurate to
min
Dear "R" list,
Our statisticians usually give us results back in a PDF format. I would like
to be able to copy and past tables from "R" output directly into a Microsoft
Word table since this will save us tons of time, be more accurate to
minimize human copying errors and help us update data in our
Hi,
I am looking for an more automated way to fit functions of the form
y~a*x^b+c
to given data, for example
x<-c(0.36,0.63,0.90,1.166,1.43, 1.70, 1.96, 2.23)
y<-c(9.0,9.62,10.11,10.53,10.9, 11.25, 11.56, 11.86)
I tried for example with nls, which did only work with really good initial
guessed
You can use nls2 to try many starting values. It works just like nls but:
- if you give it a two row data frame as the start value it will
create a grid between the upper and lower values of each parameter and
then run an optimization starting at each such point on the grid
returning the best
- i
I'm sure it's not a bug, but could someone point to a thread or offer
some gentle advice on what's happening? I think it's related to:
test <- data.frame(name1 = 1:5, name2 = 6:10, test = 11:15)
eval(expression(test[c("name1", "name2")]))
eval(expression(interco[c("name1", "test")]))
scra
Hello!
I'm reading through a logistic regression book and using R to replicate
the results. Although my question is not directly related to this, it's
the context I discovered it in, so here we go.
Consider these data:
interco <- structure(list(white = c(1, 1, 0, 0), male = c(1, 0, 1, 0),
Thomas,
I think the issue of having reasonable starting values is inherent in all
nonlinear optimization problems (unless they have some additional
properties like convexity, for example). Using a different algorithm may
or may not help. In fact, a vast majority of existing algorithms
guaran
Hello:
[Kindly Cc when replying]
The question in a nutshell is this: Is there a more robust alternative
to Ecdf ()?
The details:
I've used Ecdf () _a lot_ over the past few years and I have learned
to live with its warnings. But I am running short on time and patience
now [*] Here is a reproduc
I am trying to read and write database tables that have a date field in
them.
I am constructing a data.frame, then using dbWriteTable to create the
table and dbReadTable to read it.
>datatbl<-data.frame( dates, trialnum, fooddel, ethdel, trialtime,
trialtype, deliveries, food, ethanol, fcumr
Here's an easy question: I'd like to convert a symmetric matrix to a
vector composed of the upper.tri() part of it plus the diagonal, and
convert it back again. What's the best way to achieve this? I'm
wondering if there are some built in functions to do this easily. I
can encode fine:
v <- c(d
Hello David,
On Apr 30, 2010, at 11:00 PM, David Winsemius wrote:
> Note: Loops may be just as fast or faster than apply calls.
>
How come!? is this true also for other similar functions: lapply, tapply and
sapply?
Then the only advantage of these above is only syntactic sugar?
>>
>> indices
On Apr 30, 2010, at 4:52 PM, Giovanni Azua wrote:
Hello,
I have a bootstrap implementation loop that I would like to replace
by a faster "batch operator".
The loop is as follows:
for (b in 1:B) {
indices <- sample(1:N, size=N, replace=TRUE) # sample n elements
with replacement
theta_s
Hello,
I have a bootstrap implementation loop that I would like to replace by a faster
"batch operator".
The loop is as follows:
for (b in 1:B) {
indices <- sample(1:N, size=N, replace=TRUE) # sample n elements with
replacement
theta_star[b,] = statistic(data,indices) # exe
I am having troubles in fitting functions of the form
y~a*x^b+c
to data, for example
x<-c(0.1,0.36,0.63,0.90,1.166,1.43, 1.70, 1.96, 2.23)
y<-c(8.09,9.0,9.62,10.11,10.53,10.9, 11.25, 11.56, 11.86)
I tried for example with nls, which did only work with really good initial
guessed values.
Any su
Dear all,
I have a time series data X and want to calculate the sum of squared residuals
(SSR) at each time point.
SSR at time point m consists of two parts. The first part is the sum of squared
residuals for all time points before m, and the second part is the sum of
squared residuals for all
You could use pch='.' (assuming pairs, don't know if you used that or splom or
something else).
For lots of points you may want to look at the hexbin package (bioconductor).
--
Gregory (Greg) L. Snow Ph.D.
Statistical Data Center
Intermountain Healthcare
greg.s...@imail.org
801.408.8111
> ---
On 2010-04-30 12:42, jh556 wrote:
Some quick googling suggests that they are the same thing. Thanks for the
help!
And note that profile likelihood CIs are produced by default on
glm objects, i.e. R uses MASS's confint.glm for glm objects.
confint.default(your model) let's you compare with Wa
Some quick googling suggests that they are the same thing. Thanks for the
help!
--
View this message in context:
http://r.789695.n4.nabble.com/Likelihood-ratio-based-confidence-intervals-for-logistic-regression-tp2077303p2077354.html
Sent from the R help mailing list archive at Nabble.com.
___
require(randomForest)
rf.pred<-predict(fit, valid, type="prob")
> rf.pred[1:20, ]
0 1
16 0. 1.
23 0.3158 0.6842
43 0.3030 0.6970
52 0.0886 0.9114
55 0.1216 0.8784
75 0.0920 0.9080
82 0.4332 0.5668
120 0.2302 0.7698
128 0.1336 0.8664
147 0.4272 0.5728
148 0.0490 0.9510
Dear Gabor,
Thank you! I already suspected that this works differently in R because even
simple examples gave results that were, to me, unexpected. Strapply is a very
useful function.
Cheers!!
Albert-Jan
~~
All right, but
jh556 wrote:
I'm applying logistic regression to a moderate sized data set for which I
believe Wald based confidence intervals on B coefficients are too
conservative. Some of the literature recommends using confidence intervals
based on the likelihood ratio in such cases, but I'm having diffic
Thanks for this, Kjetil. I see that I was getting interpretation and
estimation confused.
My apologies on not including the paper title. For the benefit of
those who read this post after me, it is:
Ai, Chunrong, and Edward C. Norton. 2003. "Interaction Terms in Logit
and Probit Models." E
The Predict.Plot and TkPredict functions in the TeachingDemos package can also
help with visualizing what the model means and the effect of the different
terms.
--
Gregory (Greg) L. Snow Ph.D.
Statistical Data Center
Intermountain Healthcare
greg.s...@imail.org
801.408.8111
> -Original Me
What do you care if its optimal or not. If is fast enough for your
application that is all that matters. If its not then you will likely
have to code it yourself.
On Fri, Apr 30, 2010 at 2:09 PM, Richard R. Liu wrote:
> Gabor,
>
> Thanks for the hint to use charmatch. It is faster than what I
I'm applying logistic regression to a moderate sized data set for which I
believe Wald based confidence intervals on B coefficients are too
conservative. Some of the literature recommends using confidence intervals
based on the likelihood ratio in such cases, but I'm having difficulty
locating a
Gabor,
Thanks for the hint to use charmatch. It is faster than what I was
doing, but I suspect it might still be suboptimal. I have a list of
words, and I want to know which of those exactly or partially match
one word. By the very description of charmatch's arguments, the
second being
I would recommend the following :
a) create new factor variables out of the original factor variables
(Condition_1, Condition_2, ..., Condition_n) to
reflect the analyses that you want to do. The builtin function "interaction"
will do this for any number of factors
b) use the "summary"
It works fine for us, hence you need to tell us where you have not
followed the manual "R Installation and Administration" step by step.
It is also helpful to see how you called "CHECK" and id your package
installs at all.
Uwe Ligges
On 30.04.2010 17:53, John Lande wrote:
dear R user,
I
Hello,
I have just ordered the "ggplot2: Elegant Graphics for Data Analysis (Use R)"
but while it arrives :) can anyone please show me how to setup and add a simple
legend to a ggplot?
This is my use case, I need a legend showing CI "Classic", "Own bootstrap", "R
bootstrap":
library(ggplot2)
There exists a set of batchfiles for windows (they have been mentioned several
times on this list before, searching the archives should give you their
location) that will move or copy the installed packages from the folder for an
old R installation to the folder for your new installation without
This is a Frequently Asked Question, so frequently in fact that it is found in
the FAQ along with a couple of answers and discussion of their merits.
--
Gregory (Greg) L. Snow Ph.D.
Statistical Data Center
Intermountain Healthcare
greg.s...@imail.org
801.408.8111
> -Original Message-
>
Here is a workaround:
for(i in 1:nrow(x)) x[i, x[i, ] > 0] <- 0
On Fri, Apr 30, 2010 at 11:10 AM, Abiel X Reinhart
wrote:
> When using the tis time series package (v1.9), I cannot select or alter a
> subset of a time series when the time series is created from a matrix and the
> matrix contain
On 30/04/2010 1:17 PM, Matthias Rieber wrote:
Hi,
I've some problems with the new R version converting date to year-week:
R version 2.11.0 (2010-04-22)
Copyright (C) 2010 The R Foundation for Statistical Computing
ISBN 3-900051-07-0
> load("test.data.R")
> str(test.data)
Class 'Date' num [1
On Thu, Apr 29, 2010 at 6:15 PM, Santosh wrote:
> Richard,
> Thanks for your email. I am not looking for that kind of plot as you had
> suggested. I would like to see overlaid boxplots. Deepayan had earlier shown
> the method of overlay of boxplots (using panel.groups=panel.bwplot)...
>
> I have s
Hi all,
I have a large dataset that has >10k entries. The dataset is stored in a
dataframe with the headers:
SubID Condition1 Condition2 Result1 Result2
There are multiple entries for a given SubID(Subject ID). Condition 1 has 3
levels and condition2 has 2 levels (therefore there are 6 possible
* On Thu 05:53PM -0700, 29 Apr 2010, Seth (sjmy...@syr.edu) wrote:
>
> Hi,
>
> I've searched online, in a few books, and in the archives, but haven't seen
> this. I believe that xerror is scaled to rel error on the first split.
> After fitting an rpart object, is it possible with a little math to
dear R user,
I al rying to compile an R package with some C code. everithing work fine as
long as I work in Linux or MaCOX. unluckyly I can cant compile it under
Windows, here is the error, but I cant figure it out what is the problem.
can you help me?
* using log directory 'C://MyPKG.Rcheck'
* u
When using the tis time series package (v1.9), I cannot select or alter a
subset of a time series when the time series is created from a matrix and the
matrix contains NA values.
Example:
x<-tis(t(c(1:10,NA)), start=c(2000,1), freq=12)
x[x>0]<-0
The second line yields "Error in if (any(i > nro
Dear R community,
I have a problem with assign:
for ( iii in 1:dim(ref_df)[2] )
{
ret <<-
ref_df[,iii][names(single_string[ii])]*(single_count/sum(ref_df[,iii]))
assign(paste("expected_sing_ref",iii,sep=""),c(get(paste("expec
Hi,
I've some problems with the new R version converting date to year-week:
R version 2.11.0 (2010-04-22)
Copyright (C) 2010 The R Foundation for Statistical Computing
ISBN 3-900051-07-0
> load("test.data.R")
> str(test.data)
Class 'Date' num [1:3599546] 13888 14166 14188 14189 14189 ...
> res
Hi all,
I have a large dataset that has >10k entries. The dataset is stored in a
dataframe with the headers:
SubID Condition1 Condition2 Result1 Result2
There are multiple entries for a given SubID(Subject ID). Condition 1 has 3
levels and condition2 has 2 levels (therefore there are 6 possible
Dear Keith,
I will keep that in mind in my future posting.
Again, thanks for your time and advice!
Regards,
Joseph
On Fri, Apr 30, 2010 at 3:54 PM, kMan wrote:
> Dear Joseph,
>
> I have had a similar experience to replies. Andy's assessment about signal to
> noise on the list is, I believe, qu
Dear Andy,
You're the "kind soul" I mentioned in my previous e-mail!
Certainly yours is the kind of response I've been looking for, and now
I can start with that, especially "splinefun()" with "monoH.FC"
method.
As for my simulation data, your understanding is correct; there are
multiple y value
On 30/04/2010 12:07 PM, D Sonderegger wrote:
dump and dput but have the same behavior but dump has an extra option
'evaluate'. From the documentation, 'evaluate=TRUE' looks like it should
force promises to be evaluated, ie convert 1:6 to be c(1,2,3,4,5,6).
Both 1:6 and c(1,2,3,4,5,6) are e
Glad you got it working!
By "true value" do you mean the means? If so:
library(ggplot2)
data(mtcars)
p <- ggplot(mtcars, aes(x=cyl, y=mpg))
p + stat_summary(fun.data = "mean_cl_boot", colour = "red", geom =
"errorbar") + stat_summary(fun.data = "mean_cl_normal", colour =
"blue", geom = "errorb
On Apr 30, 2010, at 11:20 AM, Trevor Hastie wrote:
> I installed
> R version 2.11.0 (2010-04-22)
> on may macbook (snow leopard)
> and run R from within emacs
>
> Now when I try to get help, I get
>> ?lm
>
> (in the new "help" window)
>
>
> Error in help("lm", htmlhelp = FALSE) :
> unused a
Hello,
After installing gfortran from http://r.research.att.com/gfortran-4.2.3.dmg it
finally works! see below.
Thank you all.
@Ista Zahn: Looks fantastic! :) thank you so much! ... is there a way to have a
small circle on the true value?
Best regards,
Giovanni
> install.packages("Hmisc", d
HI, Dear R community,
I have 1 and 0 outcome in data set, I used the SVM (e1071). what is the
meaning of 1/0 in the decision.values?
if the decision.values>0 then it will class to 1, and if decision.values<0,
then it will classify to 0, is this right?
> attributes(svm.pred)$decision.values
I installed
R version 2.11.0 (2010-04-22)
on may macbook (snow leopard)
and run R from within emacs
Now when I try to get help, I get
> ?lm
(in the new "help" window)
Error in help("lm", htmlhelp = FALSE) :
unused argument(s) (htmlhelp = FALSE)
Help!
p.s. I am running:
This is GNU Emacs
UNIX grep selects out lines in a file and R grep similarly selects out
components of a vector of strings.On the other hand re.findall
extracts substrings from strings. These are different concepts so
there is no logical reason to expect that these two sets of commands
behave the same. Instead,
dump and dput but have the same behavior but dump has an extra option
'evaluate'. From the documentation, 'evaluate=TRUE' looks like it should
force promises to be evaluated, ie convert 1:6 to be c(1,2,3,4,5,6).
I think that either my understanding of what a 'promise' is and what it
means to be
Hello David,
On Apr 30, 2010, at 6:00 PM, David Winsemius wrote:
> Looks like you do not have the RTools bundle and perhaps not the XCode
> framework either?
>
> I am not suggesting that you do so, since it appears you are not conversant
> with compiling source code packages. If I am wrong abo
Hi:
On Fri, Apr 30, 2010 at 6:08 AM, arnaud Gaboury wrote:
> Dear group,
>
> I am losing my mind with a simple question. Sorry if obvious, but I maybe
> start to be confused after days and days of reading documentations.
>
> Df :
>
>
> df <-
> structure(list(a = 1:3, b = 4:6, c = structure(c(1L,
Hi,
The regular expression (grep) below does not behave at all like the equivalent
in Python. Also, I would be happy if somebody could tell me what the R
equivalent for Python's re.findall is. The regex filters out any numbers not
enclosed by square brackets, including fractions (with either co
I think you are Googling the wrong "reference." Note in ?acf the following:
References:
Venables, W. N. and Ripley, B. D. (2002) _Modern Applied
Statistics with S_. Fourth Edition. Springer-Verlag.
(This contains the exact definitions used.)
On Fri, Apr 30, 2010 at 10:42 AM, z
Hi Vivek,
On Thu, Apr 29, 2010 at 8:37 PM, Vivek Ayer wrote:
> Hi David,
>
> Thanks for the help. It's working! I still find it to be a new
> concept. I haven't encountered storing loops in objects in any other
> languages. Can it even be done in other languages? Very novel, quite
> intriguing.
See ?apply. The apply function works on *matrices*.
Actually arrays, and matrices are arrays with 2 dimensions.
characters. This is all explained in the first paragraph of ?apply.
Also see ?as.matrix
__
R-help@r-project.org mailing list
htt
D Sonderegger wrote:
> What I was expecting was for the second dump was:
> foo <-
> structure(c(1,2,3,4,5,6), .Dim = c(2L, 3L))
>
> That is, I was expecting dump to expand the 1:6 and 2:3 into the actual
> vectors.
Well,
> dput(c(1,2,3))
c(1, 2, 3)
> dput(c(1L,2L,3L))
1:3
And dump() does lik
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