On Sun, 20 Apr 2008, Marko Milicic wrote:
Dear R helpers,
I'm trying to build logistic regression model large dataset 360 factors and
850 observations. All 360 factors are known to be good predictors of outcome
variable but I have to find best model with maximum 10 factors. I tried to
fit
Alex Reynolds wrote:
Is there a way to modify the choice of notch size [1] in R's boxplot
routine from outlining a 5% significance region, to say 1% or lower?
Not easily. If you look inside boxplots.stats you'll find the hardcoded
constant 1.58, and the documentation has the following
Alex Reynolds reynolda at u.washington.edu writes:
Is there a way to modify the choice of notch size [1] in R's boxplot
routine from outlining a 5% significance region, to say 1% or lower?
Not directly from boxplot, because it is hardwired to indirectly call fivenum,
not quantile.
Check
kedar nadkarni nadkarnikedar at gmail.com writes:
I have been trying to obtain confidence intervals for the fit after having
used lmer by using intervals(), but this does not work. intervals() is
associated with lme but not with lmer(). What is the equivalent for
intervals() in lmer()?
ci
David,
tkanks für your comment, the code and the link.
You are right: arbitrary is a better word than exact pair matching.
I took the term one-to-one exact matching from the paper MatchIt:
Nonparametric Preprocessing for Parametric Causal Inference (p. 6):
Hi List,
I trying an example from pumps{rbugs} with .Renviron in $HOME adjusted
for my box:
WINE=/usr/bin/wine,
BUGS=/usr/local/bin/WinBugs14/winbugs.exe
data(pumps)
pumps.data - list(t = pumps$t, x = pumps$x, N = nrow(pumps))
pumps.model - file.path(.path.package(rbugs), bugs/model,
On Sun, Apr 20, 2008 at 08:16:11PM +, David Winsemius wrote:
Gabor Csardi [EMAIL PROTECTED] wrote in
news:[EMAIL PROTECTED]:
Hmm, my understanding is different,
m - matrix(sample(10*10), ncol=10)
m2 - rbind( m[1:5,], 1:10, m[6:10,] )
m3 - cbind( m[,1:8], 1:10, m[,9:10] )
I
Suppose I write:
f1 - function(x) x + 1
f2 - function(x) 2 * f1(x)
f2(10)
# 22
f1 - function(x) x - 1
f2(10)
# 18
This is quite obvious. But is there any way to define f2
in such a way that we freeze the definition of f1?
f1 - function(x) x+1
f1frozen - f1
f2 - function(x)
[EMAIL PROTECTED] wrote:
Suppose I write:
f1 - function(x) x + 1
f2 - function(x) 2 * f1(x)
f2(10)
# 22
f1 - function(x) x - 1
f2(10)
# 18
This is quite obvious. But is there any way to define f2
in such a way that we freeze the definition of f1?
f1 - function(x) x+1
f1frozen
Dear all,
is it possible to set up RGUI or JGR on Windows PC to UTF-8 encoding?
I looked for it in mailing lists and in the documentation, but I
couldn't figure out it.
My problem is e.g. to split a given string containing German and
Russian words into characters.
example:
a - asdШas
Dear all,
I use the barchart-function (lattice) for plotting stacked barcharts.
The data is a summary table (data frame) of likert-scale-evaluations
(strongly agree, agree...strongly disagree) to different issues
constructed as follows (L1=precentage of strongly agree evaluations,
You didn't tell us your R version (or your locale). Windows has no UTF-8
locales, so a lot of work has had to be done to allow Unicode chars to be
handled on Windows.
Please look into 2.7.0 RC, and in particular its CHANGES file at
Hi Everyone,
I am running into a problem with matrices. I use R version 2.4.1 and
an older version.
The problem is this:
m-matrix(ncol=3,nrow=4)
m[,1:3]-runif(n=4)
That does what I expect; it fills up the rows of the matrix with the
data vector
m
[,1] [,2] [,3]
[1,]
On windows you can go to the graphics window; click history - recording. You
may also want to have a look at recordPlot() and replayPlot().
Regards
Søren
Fra: [EMAIL PROTECTED] på vegne af Norbert NEUWIRTH
Sendt: ma 21-04-2008 10:59
Til: r-help@r-project.org
On 21 Apr 2008, at 11:33, Prof Brian Ripley wrote:
You didn't tell us your R version (or your locale). Windows has no
UTF-8 locales, so a lot of work has had to be done to allow Unicode
chars to be handled on Windows.
It was more or less a general question on R running on Windows PCs.
On Mon, 21 Apr 2008, Hans-Joerg Bibiko wrote:
On 21 Apr 2008, at 11:33, Prof Brian Ripley wrote:
You didn't tell us your R version (or your locale). Windows has no UTF-8
locales, so a lot of work has had to be done to allow Unicode chars to be
handled on Windows.
It was more or less a
kathie wrote:
Dear R users,
I have 32 observations in data x. After sorting this, I want to compute
means and variances of 3 groups divided by nr.
Actually, the number of groups is flexible. Any suggestion will be greatly
appreciated.
Hi Kathryn,
One way (there are many others) is
Dear all,
I have several data frames for which I want to change the column names.
Example data:
data.1 - data.frame(x1 = rnorm(5))
data.2 - data.frame(x1 = rnorm(5))
.
.
What I want to achieve:
names(data.1) - y1
names(data.1) - y1
.
.
Is it possible to achieve this with a loop or any of the
Hi
[EMAIL PROTECTED] napsal dne 21.04.2008 09:03:30:
Dear R users,
I have 32 observations in data x. After sorting this, I want to compute
means and variances of 3 groups divided by nr.
Actually, the number of groups is flexible. Any suggestion will be
greatly
appreciated.
man4ish wrote:
I am trying to calculate the regression for the follwing input data stored in
'data.txt' file.I am reading this and storing it in the variable i .then i
am trying to get the predicted value using f1 as dependent and others
f2f10 as independent variables.It is giving the
hi ,
i am trying to predict the value of dependent variable using the independent
variable using R .
like y is dependent and x1,x2,x3 ...,xn are independent variables so how can
predict the value of y using x1,x2,x3 ...,xn .
y x1 x2 x3 x4 x5 x6
6 0 1 231 2
5 1 34
Is there a way to modify the choice of notch size [1] in R's boxplot
routine from outlining a 5% significance region, to say 1% or lower?
Yes, but it's not as simple as specifying the significance level. You'll
have to update the function boxplot.stats, specifically the line
conf - if
ermimi wrote:
Hello!!!
I have been read a much about as read data from Excel File, but I haven´t
found the necesary information to read the data.
Now, I can create a channel : channel - odbcConnectExcel(file.xls) but I
don´t know as read the data??
I hope that you could help me. Thank
On 4/21/2008 5:54 AM, William Simpson wrote:
Hi Everyone,
I am running into a problem with matrices. I use R version 2.4.1 and
an older version.
The problem is this:
m-matrix(ncol=3,nrow=4)
m[,1:3]-runif(n=4)
That does what I expect; it fills up the rows of the matrix with the
data
Doran, Harold wrote:
I'm curious if there are users of RPy on this list. I've recently
created a gui front end using Tkinter for some python scripts I've
written for some of our internal operations and I am quite pleased with
how this program works.
Currently, I can use py2exe to create a
Mallika,
I am not sure exactly what you mean by consensus approach. One easy
thing you can do is compute the Pareto front, which is the set of
non-dominated models. A model is dominated or covered if another
model exists which is unambiguously better according to the given
scores. So, this method
Prof. Paul, Prof. Frank.
Thank you very much for helping me out. The Design package did the
trick.
Here is how the anova table looks like without using the Design package:
anova(Fit1)
Analysis of Deviance Table
Cox model: response is Surv(Time, cancer)
Terms added sequentially (first to last)
Hi
Not sure what you want to do.
You can set dimensions to your vector.
vec-1:12
dim(vec)-c(3,4)
you can repeat your vector for n times
vec-rep(1:4,3)
dim(vec) - c(4,3)
or you can use byrow option
vec-1:12
matrix(vec, nrow=4, ncol=3, byrow=T)
Petr Pikal
[EMAIL PROTECTED]
724008364,
Generally lm and predict.lm will solve this kind of problems, see
?lm
?predict.lm
But with your given data there is nothing to predict, since you have 6
independent variables and 6 observations. So you have a complete System
of linear equations, which you can solve, see ?solve.
hth.
man4ish
Hi everyone,
I would like to do the following.
Given matrix m and matrix n, I would like to compute mn[i,,j]= m[i,,j]
+ n[i,,j] if either of these elements is 0. (In other words, whichever
number is nonzero.)
Else I want mn[i,,j]=(m[i,,j] + n[i,,j])/2
I need a fast method.
Thanks very much for
Dear R users,
I have 32 observations in data x. After sorting this, I want to compute
means and variances of 3 groups divided by nr.
Actually, the number of groups is flexible. Any suggestion will be greatly
appreciated.
Kathryn Lord
dear useRs and developeRs,
I am afraid it is a very basic question, but I did not find anything alike in
the literature.
The R standard graphics device shows the opportunity to activate the history of
plots drawn within the current session. Th user can scroll back and see the
last graphs (or
To help Kedar a bit:
Here is one way:
recall - c(10, 13, 13, 6, 8, 8, 11, 14, 14, 22, 23, 25, 16, 18, 20,
15, 17, 17, 1, 1, 4, 12, 15, 17, 9, 12, 12, 8, 9, 12)
fr - data.frame(rcl = recall, time = factor(rep(c(1, 2, 5), 10)),
subj = factor(rep(1:10, each = 3)))
(fr.lmer - lmer(rcl ~ time +
Hi all,
I've uploaded to CRAN a new version of the HardyWeinberg package. This
package has routines for performing graphical significance tests (based
on the ternary plot) for Hardy-Weinberg equilibrium of bi-allelic marker
data.
Jan.
--
On 4/21/08, Michael Kubovy [EMAIL PROTECTED] wrote:
To help Kedar a bit:
Here is one way:
recall - c(10, 13, 13, 6, 8, 8, 11, 14, 14, 22, 23, 25, 16, 18, 20,
15, 17, 17, 1, 1, 4, 12, 15, 17, 9, 12, 12, 8, 9, 12)
fr - data.frame(rcl = recall, time = factor(rep(c(1, 2, 5), 10)),
subj =
Dear, List members,
My student are creating some functions to implement the median polish
kriging (one of prediction method in geostatistic). She want to create
some dialog box (to input some data) and menu. For this she is using
winMenuAddItem and winDialogString commands in function. But
This may be a question to R-development but I'm not sure. Symbolic
differentiation is implemented in R (maybe not for extremely complex
expressions), but it proves that it can be done. I know that in C++ it can
be done (symbolic c++), do you think in R it can be programmed just using
the R
Hello, we are needing to generate optimal (Fractional) designs for
discrete choice applications, where we will be using logistic regression
or multinomial logit as the modeling technique.
It looks like optFederov, in the AlgDesign package may work, but not
sure if this algorithm works when the
Henrik Parn henrik.parn at bio.ntnu.no writes:
Dear all,
I have several data frames for which I want to change the column names.
Example data:
data.1 - data.frame(x1 = rnorm(5))
data.2 - data.frame(x1 = rnorm(5))
Use lists. I.e.:
data - list()
data[[1]] - data.frame(x1 = rnorm(5))
On Mon, 21 Apr 2008, Duncan Murdoch wrote:
On 21/04/2008 4:59 AM, Norbert NEUWIRTH wrote:
dear useRs and developeRs,
I am afraid it is a very basic question, but I did not find anything alike
in the literature.
The R standard graphics device shows the opportunity to activate the
history
yacas has symbolic integration and the Ryacas package interfaces to
it -- although that portion of yacas is not very mature. There are
examples in the vignette:
library(Ryacas)
vignette(Ryacas)
On Mon, Apr 21, 2008 at 7:09 AM, francogrex [EMAIL PROTECTED] wrote:
This may be a question to
On 21/04/2008 4:59 AM, Norbert NEUWIRTH wrote:
dear useRs and developeRs,
I am afraid it is a very basic question, but I did not find anything alike in
the literature.
The R standard graphics device shows the opportunity to activate the history
of plots drawn within the current session.
Hi Henrik,
afaIcs this should work:
for(v in sprintf(data.%d, 1:n)) {
f = get(v)
names(f) = whatever
assign(v, f)
}
--
Best wishes
Wolfgang
--
Wolfgang Huber EBI/EMBL Cambridge UK http://www.ebi.ac.uk/huber
R version 2.6.2 PowerBook G4
Hello R User,
I try to perform an ANCOVA using the glm function.
I have a dataset with continuous and categorical data (explanatory
variables) and my response variable is also binary categorical.
Fehler: NA/NaN/Inf in externem Funktionsaufruf (arg 4)
Zusätzlich:
Hi,
Given matrix m and matrix n, I would like to compute mn[i,,j]= m[i,,j]
+ n[i,,j] if either of these elements is 0. (In other words, whichever
number is nonzero.)
Else I want mn[i,,j]=(m[i,,j] + n[i,,j])/2
I need a fast method.
m - matrix(c(0,1,2,3,4,0,5,6,0),nrow=3,ncol=3)
n -
Hello,
is there a R package that provides a log rank trend test
for survival data in =3 treatment groups?
Or are there any comparable trend tests for survival data in R?
Thanks a lot
Markus
--
Dipl. Inf. Markus Kreuz
Universitaet Leipzig
Institut fuer medizinische Informatik, Statistik und
Hi folks:
I have a data set v1, v2, ... v10.
Can anyone tell me how to create a new data set where the entire row is
deleted if, say, v5, is missing, is NA on that row?
Thanks,
Charles
__
R-help@r-project.org mailing list
Dear Brigit,
My guess is that you forgot to specify the argument family=binomial in
the call to glm().
Had you included the commands that you used as well as the error that
was produced, it wouldn't be necessary to guess.
I hope this helps,
John
On Mon, 21 Apr 2008 14:23:13 +0200
Birgit
Gabor Csardi [EMAIL PROTECTED] wrote in
news:[EMAIL PROTECTED]:
On Sun, Apr 20, 2008 at 08:16:11PM +, David Winsemius wrote:
Gabor Csardi [EMAIL PROTECTED] wrote in
news:[EMAIL PROTECTED]:
Hmm, my understanding is different,
m - matrix(sample(10*10), ncol=10)
m2 - rbind(
Hello dear R users!
I know this question is not strictly R-help, yet, maybe some of the guru's
in statistics can help me out.
I have a sample of data all from the same population. Say my regression
equation is now this:
m1 - lm(y ~ x1 + x2 + x3)
I also regress on
m2 - lm(y ~ x1 +
On 4/21/2008 8:16 AM, Prof Brian Ripley wrote:
On Mon, 21 Apr 2008, Duncan Murdoch wrote:
On 21/04/2008 4:59 AM, Norbert NEUWIRTH wrote:
dear useRs and developeRs,
I am afraid it is a very basic question, but I did not find anything alike
in the literature.
The R standard graphics
Hello John,
I am really sorry about that. I wanted to include the code but I
forgot and you are completely right, I forgot the family-argument.
Thanks for the help.
B.
Am 21.04.2008 um 14:50 schrieb John Fox:
Dear Brigit,
My guess is that you forgot to specify the argument
?na.omit
Weidong Gu,
Department of Medicine
University of Alabama, Birmingham
1900 University Blvd., Birmingham, Alabama 35294
Email: [EMAIL PROTECTED]
PH: (205)-975-9053
-Original Message-
From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED]
On Behalf Of Charles Vetterli
Sent: Monday,
On Mon, Apr 21, 2008 at 12:50:08PM +, David Winsemius wrote:
[...]
Am I correct in assuming that after the creation of m by way of a
temporary matrix that the temporary matrix would then be available for
garbage collection, whereas if both m and m2 were created, there would
be more
?complete.cases
On Mon, Apr 21, 2008 at 8:43 AM, Charles Vetterli
[EMAIL PROTECTED] wrote:
Hi folks:
I have a data set v1, v2, ... v10.
Can anyone tell me how to create a new data set where the entire row is
deleted if, say, v5, is missing, is NA on that row?
Thanks,
Charles
Douglas Bates bates at stat.wisc.edu writes:
If you want to examine the three means then you should fit the model as
lmer(rcl ~ time - 1 + (1 | subj), fr)
True, but for the notorious error bars in plots that reviewers always request
the 0.35 is probable more relevant than the 1.87. Which I
Dear R users,
I am trying to fully understand the difference between estimating
overdispersion with glm.nb() from MASS compared to glm(..., family =
quasipoisson).
It seems that (i) the coefficient estimates are different and also (ii) the
summary() method for glm.nb suggests that overdispersion
Thanks Doug,
You write: If you want to examine the three means then you should fit
the model as lmer(rcl ~ time - 1 + (1 | subj), fr)
I do just that (which is what Dieter just sent). But the CIs are much
too big compared to the CIs for differences between means (which
should be bigger than
Hello R users!
I got again an error message.
I used this code:
with (FemMal85_Sex, {
ModelFemMal85-
glm
(Sex~outLatTep_like_other*outLatTep_like_conduplicate*outLatTep_keeled_w
Tianxu [EMAIL PROTECTED] wrote in
news:[EMAIL PROTECTED]:
I am wondering how to do survival analysis with time-related IVs in
R. For example,
See section 4 of Fox's contribution:
http://cran.r-project.org/doc/contrib/Fox-Companion/appendix-cox-regression.pdf
Hello,
How can I label a secondary axis in R? At the moment it's labelled as
c(-100,200). Obviously I would like it to be more sensible.
Here is the code I am using
newx = -100+37.5*((1:9)-1)
axis(4,at=newx,labels=(newx+100)/3750)
Thanks,
Rob
--
View this message in context:
Sorry, I meant to say: For the moment I wonder if the solution is not
to use CIs based on the two low SEs produced by the ~ time model, and
to treat them as least-significant difference intervals.
_
Professor Michael Kubovy
University of Virginia
Department of
Hello to all R users,
up to my knowledge, neither garch(tseries) nor garchFit(fGarch)
support including external regressors in the regression, which, for
example, arima(stats) can do by setting xreg. Is there a package or
any other way that can do this?
To be precise, I want to estimate a
On 4/21/2008 9:02 AM, Nakamura wrote:
Hello,
How can I label a secondary axis in R? At the moment it's labelled as
c(-100,200). Obviously I would like it to be more sensible.
Here is the code I am using
newx = -100+37.5*((1:9)-1)
axis(4,at=newx,labels=(newx+100)/3750)
I don't
Hello everyone,
for some exploratory analysis I would like to compare the distribution of an
observable WERT pairwise between several samples identified by STICHPROBE
(which differ in size).
str(stichproben_o1o4_20080327ff[c(STICHPROBE, WERT)])
'data.frame': 6087 obs. of 2 variables:
$
Apologies for the private mail, Nabble has not yet updated the thread so I
can write another post in it. I think I have confused things. I don't mean
the labels are incorrect. They are fine. What I am referring to is a title
for the secondary axis, which is currently entitled as c(-100,200).
Hello,
I need help to build an array within an array, i. e., I have this:
tt[,,c(1,2)]
, , metadados.class_7.R
CS WRC LRA
Inicial 1.000 1.000 1.000
Final 0.5974482 0.6095162 0.5866560
Indep 0.4335460 0.4799575 0.4169591
Inicial 0.9925572 0.9925572
On Mon, 2008-04-21 at 15:43 +0200, Birgit Lemcke wrote:
Hello R users!
I got again an error message.
Something here is causing compiled code to segfault (crash). I don't
know what the problem is here exactly --- I'll let those much more
acquainted with R look into that --- but you seem to be
If you just want the title, look at ?mtext.
Charles Annis, P.E.
[EMAIL PROTECTED]
phone: 561-352-9699
eFax: 614-455-3265
http://www.StatisticalEngineering.com
-Original Message-
From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] On
Behalf Of [EMAIL PROTECTED]
Sent: Monday, April 21,
Erich:
Past posts on this list have pointed out various pros and cons of different
methods of data transer to R from Excel, in particular, loss of precision,
formatting problems, etc. Do you have any comments about to what degree any
of these alternatives may be susceptible or immune from these
Hello Gavin,
thanks for you answer.
If I use it without with I get back the same error.
The with thing was only to try out for functions that do not
contain a data-argument. I still try to learn and therefor I
sometimes just try.
It is understood that I am on the way to simplify the model
Hello,
Im trying to create a bar chart. I have a file with two different columns. I
have created bar charts before where I am only reading from one column but
now i wish to read from two columns. Here is the code i use when creating
bar charts when reading from one column:
satisfaction
Dear R-users,
I've been working with three different data sets (X, Y and Z) with the same
dimension (i.e, n \times k). What I needed to do was to conform a 4th data
set, i.e. FINAL, which first row was the X's first row, its second row was
the Y's first row, and its third row was the Z's first
Dear Statisticians,
I would like to analyse my data with a GLM with binomial error distribution
and logit link function. The point is that I want a model fitted without
intercept, i.e. the fitted curve should start at y=0.5 for x=0.
I tried it with the following code:
glm(value~0+ppm,
Dear all
I am working on a Java class which calls R functions. To do that, I use JRI.
Everything works fine up to now. The only problem I have is the following:
when I call the command plot(
.) from java, a blank R Graphics window
appears and if I click on it the program does not answer any more.
On Mon, 2008-04-21 at 17:21 +0200, Birgit Lemcke wrote:
Hello Gavin,
thanks for you answer.
If I use it without with I get back the same error.
The with thing was only to try out for functions that do not
contain a data-argument. I still try to learn and therefor I
sometimes just try.
A summary, for those interested and posterity...
Thanks to Christos Hatzis who is correct, the package 'AlgDesign' (which I'd
overlooked) has gen.mixture which Creates a candidate list of mixture
variables.
gen.mixture(4,c(egg, flour, butter))
Thanks also to a private e-mailer who suggested
On 21 Apr 2008, at 12:33, Prof Brian Ripley wrote:
Is it possible to download a compiled snapshot of 2.7.0 for Windows
XP?
Yes, http://cran.r-project.org/bin/windows/base/rtest.html
And it is due for release tomorrow.
Many thanks! I can see the progress :)
But please forgive my
On 4/21/08, K. Elo [EMAIL PROTECTED] wrote:
Dear all,
I use the barchart-function (lattice) for plotting stacked barcharts.
The data is a summary table (data frame) of likert-scale-evaluations
(strongly agree, agree...strongly disagree) to different issues
constructed as follows
This is not a dump question. This is a serious problem and it depends on
what you know or assume about the relastionship between x1 and x4. If
you assume linear interaction, you might want to introduce some
interaction term to the model for example.
Uwe Ligges
Thiemo Fetzer wrote:
Hello
Will this do it for you:
# Seed and data frames X, Y and Z
set.seed(123)
X=matrix(rnorm(300),ncol=5)
Y=matrix(rpois(300,10),ncol=5)
Z=matrix(rexp(300,1),ncol=5)
index - seq(1, by=3, length=nrow(X))
FINAL - matrix(ncol=5, nrow=3*nrow(X))
FINAL[index,] - X
FINAL[index + 1,] - Y
FINAL[index + 2,] -
There is a package in the R souces called windlgs in
.../src/gnuwin32/windlgs which has examples, but you may want to go some
way that is cross platform and avoid Windows specific programming at all.
Best wishes,
Uwe Ligges
Ingrida B wrote:
Dear, List members,
My student are creating some
On Mon, 21 Apr 2008, Hans-Joerg Bibiko wrote:
On 21 Apr 2008, at 12:33, Prof Brian Ripley wrote:
Is it possible to download a compiled snapshot of 2.7.0 for Windows XP?
Yes, http://cran.r-project.org/bin/windows/base/rtest.html
And it is due for release tomorrow.
Many thanks! I can see
boxplot(x[,c(2,15,28,41,54,67,80,93,106)], ylab=mg/s, names=c(RM215,
RM202, RM198, RM190, RM185, RM179, RM148, RM119, RM61))
this is the code I am using to make a standard box plot. Is there a way to
get the number of NA observations plotted onto the graph easily. I can
always go in and
Dear R experts,
I am trying to optimize my script, because right
now it requires a lot of memory. The goal is to
generate four plots in one page. Every plot
corresponds to the means and sem's calculated for a
given variable at different days. In order to obtain
the means and sem's I apply the
fp0 = C:\\Documents and Settings\\myname\\My Documents\\Research\\ #upper
file path
fp1 = PIname\\PIproject\\Working00\\
#middle file path
fp2 = DateData\\Dates.xls
#lower file path
dbase =
Thank you so much to Jim and Mark for their advices. Now I solved the
problem I had using a new approach.
Best,
Jorge
On Mon, Apr 21, 2008 at 12:53 PM, jim holtman [EMAIL PROTECTED] wrote:
Will this do it for you:
# Seed and data frames X, Y and Z
set.seed(123)
Hi everyone,
I'm studying the manual name: Analysis of Epidemiological Data Using
R and Epicalc, maked by: Virasakdi Chongsuvivatwong and Edward McNeil.
And I can't find the data base that they use in some examples, this are
the names:
Chapter7.Rdata,Chapter8.Rdata,Chapter9.Rdata
Somebody can
Hi there,
Try this:
# Function to read data in R from Excel
FromExcel=function(yourfile,spreadsheet){
require(RODBC)
channel=odbcConnectExcel(yourfile)
sqlTables(channel)
mydata=sqlFetch(channel, spreadsheet)
attach(mydata)
mydata
}
mydata=FromExcel(C:/mydata/2008/yourfile.xls,yourspreadsheet)
Hi again,
Deepayan Sarkar wrote (21.4.2008):
Write your own panel function (which may or may not be a simple
exercise depending on your level of expertise in R). You could use
panel.barchart as a starting point. Basically, you need to insert
some calls to panel.text() (or something
Dear R community, I am printing a jpeg file (using plot) and my y-axis
label becomes partly cut (at the left) by a very close margin of document.
See example: http://www.igm.jhmi.edu/~gehret/progr_collect_data/beta.jpg
Can you please help me fix this? I tried din and fig in the parameter
and
Have a look at the addtable2plot function in the
plotrix package.
It should do what you want.
--- stephen sefick [EMAIL PROTECTED] wrote:
boxplot(x[,c(2,15,28,41,54,67,80,93,106)],
ylab=mg/s, names=c(RM215,
RM202, RM198, RM190, RM185, RM179,
RM148, RM119, RM61))
this is the code I am
HI, pretty basic question: is that possible to change the code of the
function within library? If so what should I do? I work on R linux (ubuntu),
thanks a lot
--
View this message in context:
http://www.nabble.com/Change-the-core-code-tp16808285p16808285.html
Sent from the R help mailing
Hi,
I need to analyze the influences of several factors on a variable that is a
measure of fecundity, consisting of 73 observations ranging from 0 to 5. The
variable is continuous and highly positive skewed, none of the typical
transformations was able to normalize the data. Thus, I was thinking
José Ignacio Bustos Melo wrote:
Hi everyone,
I'm studying the manual name: Analysis of Epidemiological Data Using
R and Epicalc, maked by: Virasakdi Chongsuvivatwong and Edward McNeil.
And I can't find the data base that they use in some examples, this are
the names:
Hello :)
I am happy to hear that I am not necessarily asking stupid questions.
The thing is, that I have data on x1 and x4 for the whole sample. However,
theoretically, it is clear that the informational content of x1 is not as
high as of x4. x4 provides more accurate information to the subjects
Hello!
I was thinking again about the possible interaction between x1 and x4.
Theoretically it makes sense, that the influence of x4 on y is the stronger,
the less informative is x1. It can be argued that the higher x1, the less
informative it is x1.
How could I incorporate this relationship in
andrea previtali wrote:
Hi,
I need to analyze the influences of several factors on a variable that is a
measure of fecundity, consisting of 73 observations ranging from 0 to 5. The
variable is continuous and highly positive skewed, none of the typical
transformations was able to normalize
Duncan Murdoch [EMAIL PROTECTED] wrote:
One thing I'd like to do, but didn't have time to implement before
2.7.0, is to have history set to some finite size, e.g. a default might
be the last 3 or 10 plots. The problem with record=TRUE is that it
keeps a record of all the plots, so memory
I often just download the source, find the appropriate function, create an
file with an alternate version of it (i.e. plotMeans becomes plot.Means) and
modify it to suit. I load all custom functions like that from my .rprofile.
I guess you _could_ recompile the library, but, that might break
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