somewhere I read that " !is.na(your_vector)" is better than "your_vector!=NA"
.
Thomas L Jones, PhD wrote:
>
> Difficulty handling NA's:
> Assume that I have a numeric vector y. For simplicity, assume that it has
> 10
> elements. Assume that the third element has the value NA. I give it the
>
Dear R-helper,
I am working with ROCR of Tobias Sing et. al. to compare the performances of
logistic and nnet models on a binary response.
I had the performance plots, but I have problem finding out other
performance statistics (eg. MSE/ASE, AUC). Any help on this?
Thanks
Ilham
[[altern
Good afternoon!
I'm trying to model a ts but unfortunately i'.m very new to this kind of
modeling so i 'll be very grateful if you have an advice.
This was my syntax:
t<-c(16115,17391,19011,20256,19034,18851,20016,18088,19166,21163,18463,19397,15800,16113,18879,20598,17252,19753,19110,19605,2
Hi,
I've been using the Matlab robustfit function for linear regressions
where I suspect some data points are outliers. Is there an equivalent
function in R?
Take care, Darren
PS, This is the Matlab help on robustfit:
>> help robustfit
ROBUSTFIT Robust linear regression
B = ROBUSTFIT(X,Y)
Dear Milton,
package distrEx has methods to compute skewness.
library(distrEx)
?skewness
To generate new distributions you could use package distr.
hth,
Matthias
Milton Cezar Ribeiro wrote:
> Dear all,
>
> How can I generate values which distrituions have left or rigth skewness.
> After that,
If its ok to use lattice try this:
library(lattice)
# no attach needed
# we want VC to be first level, not OJ
ToothGrowth$supp <- relevel(ToothGrowth$supp, "VC")
# rename levels
levels(ToothGrowth$supp) <- c("A", "B")
# plot
dotplot(len ~ supp | dose, ToothGrowth, layout = c(3, 1))
On Nov 20
You can do the following:
set1 <- Matrix1[,1]
set2 <- Matrix2[,1]
common <- intersect(set1,set2)
ind1 <- which(set1 %in% common)
ind2 <- which(set2 %in% common)
A1 <- Matrix1[ind1,-1]
A2 <- Matrix2[ind2,-1]
and if you wish,
rownames(A1) <- common
rownames(A2) <- common
--- "C.S. Liu" <[EMAIL P
Hi all,
I I need to manipulate the x axis in a stripchart.
I will use one of the data sets included in R to
explain what I need to do.
attach(ToothGrowth)
stripchart(len[supp=='VC']~dose[supp=='VC'],
vertical=TRUE, group.names=c('A','A','A'))
stripchart(len[supp=='OJ']~dose[supp=='OJ'], add=
Dear list,
I have two martix like these, each rows with name in the matrix.
I try to select rows with identical name from two matrix and calculate
their correlation.
Matrix1:
2-Sep9 5.51 7.18 10.08 139.45
2-Sep7 4.915 8.22 61.68 9 123.
I wrote the original rowSums (in S-PLUS).
There, rowSums() does not coerce integer to double.
However, one advantage of coercion is to avoid integer overflow.
Tim Hesterberg
>... So, why does rowSums() coerce to double (behaviour
>that is undesirable for me)?
__
Hi,
I am a new student to R and I was hoping someone could help me with
the error message I keep getting when I try to use the nb2listw()
function.
I have 45 sites and I want to get an autocorrelation value for sites
within 1000m of each other.
coords<-cbind(EAST,NORTH)
coords<-as.matrix(co
On Tue, 2007-11-20 at 17:13 -0800, Anup Nandialath wrote:
> Dear Friends,
>
> My objective is to do element wise multiplication of two vectors. For example
> suppose I have
>
> a <- (1,1,1)
> b <- (2,4)
>
> My output should be (2,4,2,4,2,4). I managed to write it down with loops as
> follows
Dear Friends,
My objective is to do element wise multiplication of two vectors. For example
suppose I have
a <- (1,1,1)
b <- (2,4)
My output should be (2,4,2,4,2,4). I managed to write it down with loops as
follows
r <- c(1,1,1)
l <- c(2,4)
x <- 1
for (j in 1:3)
{
for (i in 1:2)
{
Let x be the input vector and cx be the cumulative running sum of it.
Then seq_along(cx) - match(cx, cx) gives increasing sequences
starting at 0 and for those after the leading zeros we start them
at 1 by adding cummax(x).
x <- c(0, 1, 0, 1, 0, 1, 0, 0, 1, 1, 1, 0, 0, 0) # input
cx <- cumsum(x)
Like always, there is much to be learned from the R-help list! Another
message had a much simpler approach.
plot(xy.coords(reizstaerke, kennlinie1, log="x"), log="x")
Julian
Julian Burgos wrote:
> Hey John,
>
> You can do simply
>
> plot(reizstaerke, kennlinie1, ylim=c(0, max(kennlinie1,
>
Hey John,
You can do simply
plot(reizstaerke, kennlinie1, ylim=c(0, max(kennlinie1,
kennlinie2)),log="x")
but if you want to fine tune where the tick marks are, you can do it "by
hand".
kennlinie1 <- c(8.0746909, 3.9916973, 9.9789444, 19.962869);
kennlinie2 <- c(6.0994206, 8.9661081, 19.924
Of course. I was just putting it inline to be as minimalist as possible...
I'm in a minimal mood today...
-- Bert
Bert Gunter
Nonclinical Statistics
7-7374
-Original Message-
From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] On
Behalf Of Henrik Bengtsson
Sent: Tuesday, November 20, 200
Hi,
I cannot find a 'vectorized' solution to this 'for loop' kind of problem.
Do you see a vectorized, fast-running solution?
Objective:
Take the value of X at each timepoint and calculate the corresponding value
of Y. Leading 0's and all 1's for X are assigned to Y; otherwise Y is
incremented b
Thanks guys, that's exactly what I was looking for!
Pieter
--
This message was sent on behalf of [EMAIL PROTECTED] at openSubscriber.com
http://www.opensubscriber.com/message/r-help@r-project.org/8030546.html
__
R-help@r-project.org mailing list
https:
So why not suggest
pathname <- file.choose()
z <- read.table(pathname, dec=".", header=TRUE)
instead.
/Henrik
On Nov 20, 2007 2:58 PM, Bert Gunter <[EMAIL PROTECTED]> wrote:
> z=read.table(chosen <- file.choose(),dec=".",header=TRUE)
>
> will assign the return value of file.choose (the path nam
Dear all,
How can I generate values which distrituions have left or rigth skewness.
After that, which function compute the skewness of each distributions?
Kind regards.
miltinho
para armazenamento!
[[alternative HTML version deleted]]
__
z=read.table(chosen <- file.choose(),dec=".",header=TRUE)
will assign the return value of file.choose (the path names of the file as a
character string) to chosen.
This sort of code is seductive, but I find it quite difficult to read. I
have also gotten myself into trouble from time to time doin
Hello everybody,
I'm sorry for asking a dumb question, but I don't find how to loos it.
I need to display the name of the choosing file using this command :
z=read.table(file.choose(),dec=".",header=TRUE)
Thanks for any help.
--
View this message in context:
http://www.nabble.com/display-bas
Hi and thank you
Discussion is surely tough via the web
And I have trouble with "tone" also, and English
My only point was that Epselon <[EMAIL PROTECTED]> received education
but not much encouragement IMO and
With the method of Bolker, it might be possible to achieve both for newbies,
ie both e
On Tuesday 20 November 2007, Deepayan Sarkar wrote:
> On 11/20/07, Dylan Beaudette <[EMAIL PROTECTED]> wrote:
> > On Tuesday 20 November 2007, Deepayan Sarkar wrote:
> > > On 11/20/07, Dylan Beaudette <[EMAIL PROTECTED]> wrote:
> > >
> > > [...]
> > >
> > > > Example:
> > > > library(lattice)
> > >
Thomas L Jones, PhD wrote:
> Difficulty handling NA's:
> Assume that I have a numeric vector y. For simplicity, assume that it has 10
> elements. Assume that the third element has the value NA. I give it the
> following:
> NA_test <- function (){
> y <- numeric (10)
> y [3] <- NA
> if (y [3] != N
I did specify only the three columns of interest
c('bd.n','ht.n','dbh.n')
as the second (column) index, and I did superficially check the results
which turned out as I had hoped. Only the three columns listed were
changed and existing values _in_ those columns were retained.
So, I appear to be on m
On 11/20/07, Dylan Beaudette <[EMAIL PROTECTED]> wrote:
> On Tuesday 20 November 2007, Deepayan Sarkar wrote:
> > On 11/20/07, Dylan Beaudette <[EMAIL PROTECTED]> wrote:
> >
> > [...]
> >
> > > Example:
> > > library(lattice)
> > >
> > > # generate some data:
> > > resp <- rnorm(100)
> > > pred <-
On Tuesday 20 November 2007, Deepayan Sarkar wrote:
> On 11/20/07, Dylan Beaudette <[EMAIL PROTECTED]> wrote:
>
> [...]
>
> > Example:
> > library(lattice)
> >
> > # generate some data:
> > resp <- rnorm(100)
> > pred <- resp*1.5 + rnorm(100)
> > d <- data.frame(resp=resp, pred=pred)
> >
> > # add
The problem with that solution is that if any of the columns in a row
has an NA in it, then all the columns of that row will be set to zero.
Is the what you want, or do you only want the specific entries set to
0?
On Nov 20, 2007 4:07 PM, Thompson, David (MNR)
<[EMAIL PROTECTED]> wrote:
> Jim,
>
On Tue, 20 Nov 2007, Geertje Van der Heijden wrote:
> Hi,
>
> I'd like some advice on bootstrapping in R.
>
> I have a species x with 20 individuals and a factor containing 0 and 1's
> (in this case 5 zeros and 15 ones). I want to compare the frequency of
> the occurrence of 1 with a probability v
Jim,
What I'm getting at is my lack of understanding when manipulating
data frame indices. I have tried something like your matrix example
with variants of: Bout[ is.na(Bout[,c('bd.n','ht.n','dbh.n')]), ] <- 0
only to receive the following error message:
Error in `[<-.data.frame`(`*tmp*`, is.na(B
On 11/20/07, Dylan Beaudette <[EMAIL PROTECTED]> wrote:
[...]
> Example:
> library(lattice)
>
> # generate some data:
> resp <- rnorm(100)
> pred <- resp*1.5 + rnorm(100)
> d <- data.frame(resp=resp, pred=pred)
>
> # add a grouping factor:
> d$grp <- gl(4, 25, labels=letters[1:4])
>
> # plot: loo
?which
On Tue, 20 Nov 2007, Barb, Jennifer (NIH/CIT) [E] wrote:
> Does anyone know which function (if any) will return the index of the
> true locations in a Boolean vector?
>
>
>
> For instance:
>
> A=c(1,3,5,7,4);
>
> B=c(2,4,77,3,3);
>
> X=A
>
>
> So X is:
>
>> X
>
> [1] TRUE TRUE TRUE FA
try:
which(X)
Best,
Dimitris
Dimitris Rizopoulos
Ph.D. Student
Biostatistical Centre
School of Public Health
Catholic University of Leuven
Address: Kapucijnenvoer 35, Leuven, Belgium
Tel: +32/(0)16/336899
Fax: +32/(0)16/337015
Web: http://med.kuleuven.be/biostat/
http://www.student.
which(A < B)
On Nov 20, 2007, at 3:06 PM, Barb, Jennifer (NIH/CIT) [E] wrote:
> Does anyone know which function (if any) will return the index of the
> true locations in a Boolean vector?
>
>
>
> For instance:
>
> A=c(1,3,5,7,4);
>
> B=c(2,4,77,3,3);
>
> X=A
>
>
> So X is:
>
>> X
>
> [1] TRUE
?which
On 21/11/2007, at 9:06 AM, Barb, Jennifer (NIH/CIT) [E] wrote:
> Does anyone know which function (if any) will return the index of the
> true locations in a Boolean vector?
>
>
>
> For instance:
>
> A=c(1,3,5,7,4);
>
> B=c(2,4,77,3,3);
>
> X=A
>
>
> So X is:
>
>> X
>
> [1] TRUE TRUE TRU
?which
> which(X)
[1] 1 2 3
Max
-Original Message-
From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED]
On Behalf Of Barb, Jennifer (NIH/CIT) [E]
Sent: Tuesday, November 20, 2007 3:07 PM
To: [EMAIL PROTECTED]
Subject: [R] Loc function in R??
Does anyone know which function (if any) will r
On Tuesday 20 November 2007, Deepayan Sarkar wrote:
> On 11/20/07, Dylan Beaudette <[EMAIL PROTECTED]> wrote:
> > On Tuesday 20 November 2007, Deepayan Sarkar wrote:
> > > On Nov 20, 2007 11:14 AM, David Afshartous <[EMAIL PROTECTED]>
> >
> > wrote:
> > > > All,
> > > >
> > > > I'm trying to make a
Does anyone know which function (if any) will return the index of the
true locations in a Boolean vector?
For instance:
A=c(1,3,5,7,4);
B=c(2,4,77,3,3);
X=A X
[1] TRUE TRUE TRUE FALSE FALSE
I'd like to know if there is a function that will tell me the locations
of where the TRUE valu
I have a graph which looks like hyperbole. I'd like to fit a
straight line to the lower segment going to infinity, approaching the
X axis -- I'm interested in the angle. If I'd do it manually, I'd
cut off a certain initial part of the range, [0..x_min], and then do
an lm with the rest. Y
On 21/11/2007, at 8:40 AM, Deepayan Sarkar wrote:
> There's a way to do almost everything, but we need a reproducible
> example (preferably minimal) and a clear statement of what you want to
> do to help.
Is this a candidate for fortune() ?
cheers,
On 2007-11-20, jim holtman <[EMAIL PROTECTED]> wrote:
> Does something like this work for you? You can vary the mgp parameter
> for placement of the label.
>
You can also use mtext to place the axis label separately:
barplot(rev(modelledprofile),horiz=TRUE,xlim=c(0,1),col="cornflowerblue",names.
On 11/20/07, Dylan Beaudette <[EMAIL PROTECTED]> wrote:
> On Tuesday 20 November 2007, Deepayan Sarkar wrote:
> > On Nov 20, 2007 11:14 AM, David Afshartous <[EMAIL PROTECTED]>
> wrote:
> > > All,
> > >
> > > I'm trying to make a basic plot: data points superimposed upon the a line
> > > connecting
On Tuesday 20 November 2007, Deepayan Sarkar wrote:
> On Nov 20, 2007 11:14 AM, David Afshartous <[EMAIL PROTECTED]>
wrote:
> > All,
> >
> > I'm trying to make a basic plot: data points superimposed upon the a line
> > connecting the points w/ a different color. Example below doesn't work
> > as
Try this:
xyplot(y ~ Hour, xlab = list("Hour", font=2, cex=2), ylab= list("U_x * V",
font=2, cex=2), type = 'l', scales = list( x= list(tick.number=5, cex=2,
limits=c(0,4.5)), y=list(cex=0)), lines=TRUE, lwd = 2)
trellis.focus("panel", 1, 1, highlight = FALSE)
print(panel.points(Hour, y, pch = 16
It really depends on what you want to set the values that contain NAs
to for the various type of the columns. Do you always want numerics
=0, characters ="", and factors =whatever? Do you want to do this for
all the columns in a dataframe? If you want to it for all the columns
in a matrix, it is
According to the Octave manual (page 146 of Edition 3 for Octave
version 2.9.16) Octave can read Matlab version 7 files. It can also
output with similar options to Matlab. As there are no issues or
licencing problems when running Octave and R on the one machine. Note
that there is a new native W
On Nov 20, 2007 11:14 AM, David Afshartous <[EMAIL PROTECTED]> wrote:
>
> All,
>
> I'm trying to make a basic plot: data points superimposed upon the a line
> connecting the points w/ a different color. Example below doesn't work as
> the first xyplot call doesn't remain. Suggestions?
xyplot(y ~
Hi Loren,
It wasn't my intention to sound arrogant, cruel or "off putting".
English is not my first language, and perhaps my message had a tone I
did not intended. The person posting the message I responded was
literally asking for somebody on the list to do his/hers homework (there
are even refe
All,
I'm trying to make a basic plot: data points superimposed upon the a line
connecting the points w/ a different color. Example below doesn't work as
the first xyplot call doesn't remain. Suggestions?
David
Hour = c(NA,1,2,3,4)
y = c(2,2,3,2,1.5)
xyplot(y ~ Hour, xlab = list("Hour", font
On Tue, 20 Nov 2007, Duncan Murdoch wrote:
> On 11/20/2007 12:56 PM, Michael Grishin wrote:
>> I am running R on the linux platform. Is there any way to connect to
>> UDB(DB2) dbms? How?
>
> I don't know that dbms at all, but if it supports ODBC, then the RODBC
> package should be able to talk t
See:
http://finzi.psych.upenn.edu/R/Rhelp02a/archive/57255.html
On Nov 20, 2007 1:21 PM, John Wiedenhoeft <[EMAIL PROTECTED]> wrote:
> Hi there,
>
> I guess this must be a standard issue, but I'm starting to go crazy with it. I
> simply want a plot with the x axis being logarithmic, having labels
On Tue, 2007-11-20 at 13:30 -0500, Duncan Murdoch wrote:
> On 11/20/2007 12:56 PM, Michael Grishin wrote:
> > I am running R on the linux platform. Is there any way to connect to
> > UDB(DB2) dbms? How?
>
> I don't know that dbms at all, but if it supports ODBC, then the RODBC
> package shoul
Does something like this work for you? You can vary the mgp parameter
for placement of the label.
modelledprofile <- c(7.072289e-01, 2.625354e-01, 2.932886e-02,
8.986474e-04, 8.155270e-06, 2.537943e-08, 3.137954e-11, 1.729522e-14,
4.579875e-18, 6.069698e-22, 4.100828e-26, 1.423359e-30, 1.272139e
On 11/20/2007 12:56 PM, Michael Grishin wrote:
> I am running R on the linux platform. Is there any way to connect to
> UDB(DB2) dbms? How?
I don't know that dbms at all, but if it supports ODBC, then the RODBC
package should be able to talk to it.
Duncan Murdoch
Hi there,
I guess this must be a standard issue, but I'm starting to go crazy with it. I
simply want a plot with the x axis being logarithmic, having labels 1, 10,
100..., and ten unlabelled ticks between each of them - just as they
introduce logarithmic axis at school. I've played around a bit
Hi there,
I guess this must be a standard issue, but I'm starting to go crazy with it. I
simply want a plot with the x axis being logarithmic, having labels 1, 10,
100..., and ten unlabelled ticks between each of them - just as they
introduce logarithmic axis at school. I've played around a bit
I am running R on the linux platform. Is there any way to connect to
UDB(DB2) dbms? How?
Thanks,
Michael Grishin
-
DTCC DISCLAIMER: This email and any files transmitted with it are
confidential a
On Nov 20, 2007 10:29 AM, Gabor Grothendieck <[EMAIL PROTECTED]> wrote:
> update already has formula and packageStatus methods so its
> being extended beyond models already even within core packages.
Also, update methods are extensively used for the "other kind" of
graphs in the lattice package.
On 11/20/2007 11:22 AM, Alexy Khrabrov wrote:
> xy.coords can have a log="xy" parameter which then plot interprets to
> use log scale.
> I wonder whether plot can be instructed in a similar way to use log10
> scale instead of natural logs.
I still don't see the problem. When I do
> x <- 1:1
Hello,
How do I do the following more concisely?
Bout[is.na(Bout$bd.n), 'bd.n'] <- 0
Bout[is.na(Bout$ht.n), 'ht.n'] <- 0
Bout[is.na(Bout$dbh.n), 'dbh.n'] <- 0
Would the form of such a command be different
between numeric, character and factor columns?
. . . between data.f
Here's an example of my problem:
modelledprofile <- c(7.072289e-01, 2.625354e-01, 2.932886e-02, 8.986474e-04,
8.155270e-06, 2.537943e-08, 3.137954e-11, 1.729522e-14, 4.579875e-18,
6.069698e-22, 4.100828e-26, 1.423359e-30, 1.272139e-35, 5.449925e-46,
1.431925e-57, 1.629660e-70)
depthnames <-
c
Have a look at scale_x_log10() and scale_y_log10() from the ggplot2
package. The helpfile with some examples is at
http://had.co.nz/ggplot2/scale_continuous.html
HTH,
Thierry
ir. Thierry Onkelinx
Instituut voor natu
Hi, Marion,
I believe the package FlexMix provides a more generalized version of
finite mixture modeling than is found in mclust/mclust02.
Please see:
http://cran.r-project.org/doc/vignettes/flexmix/flexmix-intro.pdf
Karen
---
Karen M. Green, Ph.D.
[EMAIL PROTECTED]
Research Investigator
> "LRC" == Luis Ridao Cruz <[EMAIL PROTECTED]>
> on Tue, 20 Nov 2007 15:57:07 + writes:
LRC> R-help,
LRC> Sorry if this question has been discussed/posted before
LRC> but I can't just find it myself.
LRC> How can I print the comment character (") ?
Note that ' "
update already has formula and packageStatus methods so its
being extended beyond models already even within core packages.
On Nov 20, 2007 8:54 AM, Luke Tierney <[EMAIL PROTECTED]> wrote:
> On Tue, 20 Nov 2007, Prof Brian Ripley wrote:
>
> > update() is generic, so the recommended approach would
On Tue, 2007-11-20 at 11:07 -0500, Duncan Murdoch wrote:
> On 11/20/2007 10:50 AM, Ken Fullish wrote:
> > > as.character(seq(-.25,.95,.1))
> > [1] "-0.25" "-0.15" "-0.05" "0.05" "0.15" "0.25" "0.35" "0.45"
> > "0.55" "0.65" "0.75" "0.85" "0.95"
> >
> > > as.character(seq(-.35,.95,.
xy.coords can have a log="xy" parameter which then plot interprets to
use log scale.
I wonder whether plot can be instructed in a similar way to use log10
scale instead of natural logs.
Cheers,
Alexy
On Nov 20, 2007, at 7:01 PM, Duncan Murdoch wrote:
> On 11/20/2007 10:41 AM, Alexy Khrabrov
On Tue, 2007-11-20 at 16:32 +0100, Søren Højsgaard wrote:
> Thanks, Luke Tierney - the documentation of update was exactly what
> prompted my question. Yet, I conclude that writing an update.myobject
> method is OK.
I think Luke's point was that whilst it would be OK if you wrote your
own method i
Hello,
I'm trying to find out how I can move the y axis label of a horizontal barplot
further away from the axis, in order to avoid overlap between the names (which
I have rotated using las=1) and the label. I have been fiddling with the
graphical parameters but nothing seems to work. Maybe a g
Hi
Apologies for this question.
I have read the help for GDD and font mapping, and it tells me that the
config file, /usr/lib/R/library/GDD/fonts/basefont.mapping is
responsible for mapping fonts to the GDD library.
However:
> .GDD.font()
[1] "/usr/X11R6/lib/X11/fonts/Type1/l048013t.afm"
[2] "/
> -Original Message-
> From: [EMAIL PROTECTED]
> [mailto:[EMAIL PROTECTED] On Behalf Of Animesh Acharjee
> Sent: Tuesday, November 20, 2007 7:47 AM
> To: [EMAIL PROTECTED]
> Subject: [R] Canonical Correlation Analysis
>
> Dear Sir/Madam,
> This is Animesh work
On 11/20/2007 10:50 AM, Ken Fullish wrote:
> > as.character(seq(-.25,.95,.1))
> [1] "-0.25" "-0.15" "-0.05" "0.05" "0.15" "0.25" "0.35" "0.45"
> "0.55" "0.65" "0.75" "0.85" "0.95"
>
> > as.character(seq(-.35,.95,.1))
> [1] "-0.35" "-0.25"
> "-0.15"
see FAQ 7.31
On Nov 20, 2007 10:50 AM, Ken Fullish <[EMAIL PROTECTED]> wrote:
>
> > as.character(seq(-.25,.95,.1))
> [1] "-0.25" "-0.15" "-0.05" "0.05" "0.15" "0.25" "0.35" "0.45"
> "0.55" "0.65" "0.75" "0.85" "0.95"
>
> > as.character(seq(-.35,.95,.1))
> [1] "-0.35" "-0.
Hi everybody! I just joined, I started using R a few weeks ago and it
seems to be a really versatile system. Last weekend I
half-accidentally installed all R packages to my computer. Independent
of this (at least I hope), I managed to break my Kubuntu's KDE and so
far haven't been able to repair it
R-help,
Sorry if this question has been discussed/posted before
but I can't just find it myself.
How can I print the comment character (") ?
Thanks in advance
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PLEASE do
On 11/20/2007 10:41 AM, Alexy Khrabrov wrote:
> Is there a way to teach xy.coords, when given log="xy", or just "x"
> or "y" separately, to do a decimal log10 instead of the natural log?
xy.coords doesn't do any transformation other than setting non-positive
values to NA. So your question doe
> as.character(seq(-.25,.95,.1))
[1] "-0.25" "-0.15" "-0.05" "0.05" "0.15" "0.25" "0.35" "0.45"
"0.55" "0.65" "0.75" "0.85" "0.95"
> as.character(seq(-.35,.95,.1))
[1] "-0.35" "-0.25"
"-0.15" "-0.0499" "0.05"
[6] "0.15"
Dear Sir/Madam,
This is Animesh working on Canonical
Correlation Analysis ( CCA ) and I am using CCA package in R. I have a doubt
and it may be very basic question. Hope someone will get back to me with
answer. I am trying to follow the documentation of CCA but could n
Is there a way to teach xy.coords, when given log="xy", or just "x"
or "y" separately, to do a decimal log10 instead of the natural log?
Cheers,
Alexy
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Hi all!
Is it possible to model a multiple regression in which the response becomes
zero when one of the two covariates is zero?
lm(y~ x1+x2) and y=0 if x1=0.
However, when x1=0, y=x2+1(intercept).
Does this mean I cannot have a second covariate and intercept or should I
eliminate only the
Thanks, Luke Tierney - the documentation of update was exactly what prompted my
question. Yet, I conclude that writing an update.myobject method is OK.
Regards
Søren
Fra: Luke Tierney [mailto:[EMAIL PROTECTED]
Sendt: ti 20-11-2007 14:54
Til: Prof Brian Ripl
Simplest would be to use the select.list() R command:
xnames <- select.list(colnames(iris))
For a prettier and more functional approach the gWidgets package provides
a layer over a number of lower level packages for creating GUI front
ends in R.
http://wiener.math.csi.cuny.edu/pmg/gWidgets
Other
Good morning and thank you
One correction please
I was not the poster
I am [EMAIL PROTECTED]
Poster was Epselon <[EMAIL PROTECTED]>
What you say is most certainly true. But then the newbie corrections might
have the style displayed by Mr. Bolker to both educate and further promote
participation.
Selon Rajarshi Guha <[EMAIL PROTECTED]>:
> Hi, I'm writing code to generate a plot, in which I draw a series of
> rectangles. So my code is of the form
>
> plot.new()
> plot.window( ... )
>
> draw rectangle
> draw rectangle
> ...
>
> Is there a way for me to insert a PNG or PDF graphic at a specif
On Tue, 20 Nov 2007, S Ellison wrote:
> Also check the asp= parameter in plot.default and plot.window; this sets
> the aspect ratio so that 1 unit in x is the same physicla length as 1
> unit in y. I don;t know whether it is respeced by your particular
> biplot, though.
That's not how biplot.defa
Thanks to your help, I managed to link R and a Mysql Database, send queries,
plot the results and put everything in a pdf document (with Sweave).
My co-workers find the job not bad but they would like to have a friendly
interface to send queries and see the graphs (the pdf document is not
necessa
I've tried tetxplot() but it fills the whole frame. I could either
use mfrow but it would devote unnecessarily large space to the text.
In fact I want it in a corner of my current plot, not occupied by the
curves. Here's what I ended up with -- it works, but I have to
manually move top t
stephen,
Thanks for your help. I want to draw a plot which shows separate
survival curves for each category of X on the same plot(same set of
axes). Your code produces a separate curve for each combination of X
and Y but I don't want curves for combinations of X and Y since Y has
many levels and a
Check out this thread:
http://tolstoy.newcastle.edu.au/R/e2/help/07/04/14273.html
On Nov 20, 2007 9:26 AM, Rajarshi Guha <[EMAIL PROTECTED]> wrote:
> Hi, I'm writing code to generate a plot, in which I draw a series of
> rectangles. So my code is of the form
>
> plot.new()
> plot.window( ... )
>
Hi, I'm writing code to generate a plot, in which I draw a series of
rectangles. So my code is of the form
plot.new()
plot.window( ... )
draw rectangle
draw rectangle
...
Is there a way for me to insert a PNG or PDF graphic at a specific
position in the plot (ideally in plot coordinates)? I
To Walter,
I am building a model of voting in the EU Council of Ministers. A voting
scenario assumes the probabilities of yes votes, possible different for
each country, and correlation coefficient between them.
The country variables (data) reflect the background characteristics that
are releva
Hi All,
I'm stumped on something that must be trivial. I created a correlation
matrix on 4 variables (6 correlations) using Hmisc's rcorr function. I
wanted to correct the P-value matrix for the number of tests done, so I
ran it through the p.adjust function. That function adjusted for the 12
p-va
On Tue, 20 Nov 2007, Prof Brian Ripley wrote:
update() is generic, so the recommended approach would be to write a method
for your objects.
Creating your own function update() in a package would probably not break too
much, as namespaces would protect most functions using the generic in stats
you can use the newdata statement in the predict function, as in
n<-50
x<-sample(40:70,n,rep=T)
y<-.7*x+rnorm(n,sd=5)
plot(x,y,xlim=c(20,90),ylim=c(0,80))
mylm<-lm(y~x)
abline(mylm,col="red")
newx<-seq(20,90)
prd<-predict(mylm,newdata=data.frame(x=newx),interval = c("confidence"),
level = 0.90,ty
Dear All,
I have several socio-economic and geographic variables for the 27 EU
countries. I would to use these data to derive a correlation matrix between
groups of countries (for a different application).
I thought of using kmeans to cluster the groups, and then calibrate between
group correlat
update() is generic, so the recommended approach would be to write a
method for your objects.
Creating your own function update() in a package would probably not break
too much, as namespaces would protect most functions using the generic in
stats. But it could be very confusing to users.
O
Hello,
I am trying to generate a confidence interval (90 or 95%) of a regression
line. This is primarily just for illustration on a scatter plot (i.e. I am
trying to make this
http://www.ast.cam.ac.uk/~rgm/scratch/statsbook/graphics/anima4.gif).
I have been trying to use the predict.lm function,
Thank you Gabor for your fast answer (received as I was reading R News 4/1)
Ptit Bleu.
library(chron)
chron(paste("1/1/", year, sep = "")) + julday + hour/24
That produces a chron, i.e. dates/times class datetime. If you want
it as a string use format(...above expression...)See the help de
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