Brian Quinif bquinif at gmail.com writes:
What I need to do is make tables with the std. errors beneath the
estimates in parentheses (standard econ style). How can I make a
dataframe with that format?
Best prepare the output in R as strings; there are alternatives in LaTeX, but
it can be
Linda Lei wrote:
Hi All,
Could you please help with the error I get from the following codes?
Library(cluster)
data(iris)
bc1 - bclust(iris[,2:5], 3, base.method=clara, base.centers=5)
Then I got:
Committee Member:
CG Pettersson cg.pettersson at evp.slu.se writes:
Fitting and reducing a first model for grain yield went smoothly. When I
wanted to look at the fixed effects with estimable() I got an error
message claiming that I was using the wrong variable names, estimable
wanted the variable names in
I am ashamed to be asking this question, but I couldn't find the
solution anywhere. Searching for if and R is not very
productive...
I cannot get a simple if statement to work.
I have data on college students. I want to make a string variable
that has the names of the years. That is, when the
Dear R users,
When I run the following code, R crashes:
require(cclust)
x - matrix(c(0,0,0,1.5,1,-1), ncol=2, byrow=TRUE)
cclust(x, centers=x[2:3,], dist=manhattan, method=kmeans)
While this works:
cclust(x, centers=x[2:3,], dist=euclidean, method=kmeans)
I'm posting this here because I am
How about the following, if you really want characters or just leave as
factor
i - round(runif(10, 1, 4))
years - as.character(factor(i, labels = c(Freshman, Sophomore,
Junior, Senior)))
HTH,
ken
__
R-help@stat.math.ethz.ch mailing list
Hi everyone
What you are trying to do does work, I do it all the time, but
there are
a few details to work out. Because of the above differences I
actually
generate the DESCRIPTION* files from my Makefile, differently
depending
on the target.
I do too. Having a DESCRIPTION.in is
Hi Ken,
Thanks for the help, but I should have mentioned that I want to do
this within a loop. Perhaps it would be better for me to explain my
exact situation.
I am running two loops so that I can calculate estimates for years
1,2,3,4 and a categorical GPA variable that takes on three values
On Fri, 7 Apr 2006, Brian Quinif wrote:
I am ashamed to be asking this question, but I couldn't find the
solution anywhere. Searching for if and R is not very
productive...
I cannot get a simple if statement to work.
I have data on college students. I want to make a string variable
that
On Fri, 07 Apr 2006 09:02:46 +0200,
Timo Becker (TB) wrote:
Dear R users,
When I run the following code, R crashes:
require(cclust)
x - matrix(c(0,0,0,1.5,1,-1), ncol=2, byrow=TRUE)
cclust(x, centers=x[2:3,], dist=manhattan, method=kmeans)
While this works:
cclust(x,
Timo Becker wrote:
Dear R users,
When I run the following code, R crashes:
require(cclust)
x - matrix(c(0,0,0,1.5,1,-1), ncol=2, byrow=TRUE)
cclust(x, centers=x[2:3,], dist=manhattan, method=kmeans)
While this works:
cclust(x, centers=x[2:3,], dist=euclidean, method=kmeans)
I'm
On Wed, 5 Apr 2006 12:58:00 -0700,
Linda Lei (LL) wrote:
Hi All,
For the function bclust(e1071), the argument base.method is
explained as must be the name of a partitioning cluster function
returning a list with the same components as the return value of
'kmeans'.
In my
Vassily Shvets wrote:
Just a note to say if you use msm you may find that
statetable returns an incorrect statetable: you can
then build an incorrect model and not even know that
statetable was the villain.
shfets
__
R-help@stat.math.ethz.ch
On Fri, Apr 07, 2006 at 02:58:00AM -0400, Brian Quinif wrote:
I have data on college students. I want to make a string variable
that has the names of the years. That is, when the year variable i is
equal to 1, I want to have a variable called years equal to
Freshmen.
I tried this
years
On Fri, Apr 07, 2006 at 09:55:47AM +0200, Philipp Pagel wrote:
On Fri, Apr 07, 2006 at 02:58:00AM -0400, Brian Quinif wrote:
I tried this
years - Freshmen if i==1
years - Sophomores if i==2
What you are looking for is not an if clause but logical indexing:
years[years==Freshmen] - 1
Leon wrote:
Hi,
I am using rpart to do leave one out cross validation, but met some problem,
Data is a data frame, the first column is the subject id, the second column
is the group id, and the rest columns are numerical variables,
Data[1:5,1:10]
sub.id group.id X3262.345 X3277.402
Uwe Ligges schrieb:
Timo Becker wrote:
Dear R users,
When I run the following code, R crashes:
require(cclust)
x - matrix(c(0,0,0,1.5,1,-1), ncol=2, byrow=TRUE)
cclust(x, centers=x[2:3,], dist=manhattan, method=kmeans)
While this works:
cclust(x, centers=x[2:3,], dist=euclidean,
Why do some functions build chm help and others do not? Is there something
in the .Rd file that specifies this?
Thanks,
Steve
-Original Message-
From: Duncan Murdoch [mailto:[EMAIL PROTECTED]
Sent: Tuesday, April 04, 2006 2:22 PM
To: Steven Lacey
Cc: 'Thomas Lumley';
On 4/7/2006 6:03 AM, Steven Lacey wrote:
Why do some functions build chm help and others do not? Is there something
in the .Rd file that specifies this?
They all build it (if you have the appropriate variable defined in the
makefile). The difference is that other formats are redone with every
Hello,
I want to make a fuzzy classification from a dissimilarity matrix
(calculated with daisy from package 'cluster'). I have tried to use
fanny (package cluster) but I have the same problems than described in a
previous message
(http://tolstoy.newcastle.edu.au/R/help/05/05/4546.html) i.e. it
On 4/7/06, Brian Quinif [EMAIL PROTECTED] wrote:
I tried that, and the problem seems to be that the results of the
'paste'ing are enclosed in quotes...that is the paste yields
'101.tex'. Of course, that is what I said I wanted above, and it
should work for that particular instance, but I also
Consider this:
y - matrix(1:8, ncol=2)
is.matrix(y[-c(1,2),])
[1] TRUE
is.matrix(y[-c(1,2,3),])
[1] FALSE
is.matrix(y[-c(1,2,3,4),])
[1] TRUE
It seems like an inconsistent behaviour:
- with 2 or more rows we have a matrix
- with 1 row we do not have a matrix and
- with 0 rows we have a
I have a basic question on the mechanics of fdim. I am calculating
the information dimension of tree distributions. What I have are
positions of trees within a plot (x data in one column, y data in
another). My question is this: how does fdim combine the data from
both columns to
To enhance my understanding, and that of my students, I have a question
about cox.zph in the survival package.
If I have correctly gleaned the high-level point from the 1994
Biometrika paper of Grambsch and Therneau, it looks to me like
cox.zph provides a mechanism to test for a simple trend in
this is documented: check ?[. You need to specify drop = FALSE,
i.e.,
is.matrix(y[-c(1,2,3), , drop = FALSE])
Best,
Dimitris
Dimitris Rizopoulos
Ph.D. Student
Biostatistical Centre
School of Public Health
Catholic University of Leuven
Address: Kapucijnenvoer 35, Leuven, Belgium
Tel:
[EMAIL PROTECTED]
y - matrix(1:8, ncol=2)
is.matrix(y[-c(1,2),])
[1] TRUE
is.matrix(y[-c(1,2,3),])
[1] FALSE
is.matrix(y[-c(1,2,3,4),])
[1] TRUE
It seems like an inconsistent behaviour:
- with 2 or more rows we have a matrix
- with 1 row we do not have a matrix and
- with 0 rows we have a
Use drop = FALSE in your subscripting calls. That will retain matrixness.
For example:
y - matrix(1:8, ncol = 2)
is.matrix(y[-c(1,2,3),,drop = FALSE]
More info is on the help page for [. You can type: ?[ to get it from
the command line.
Hope this helps,
Matt Wiener
-Original
I'm interested in any thoughts that people have about this idea -
what errors do you commonly see, and how can they be dealt with?
Another (selfish) idea: a guide for programmers coming from other languages
that highlights the differences between languages. For example, a perl to R
guide or C
On Fri, 7 Apr 2006, Dimitris Rizopoulos wrote:
this is documented: check ?[. You need to specify drop = FALSE,
i.e.,
is.matrix(y[-c(1,2,3), , drop = FALSE])
It's not only documented, it's a FAQ.
-thomas
Best,
Dimitris
Dimitris Rizopoulos
Ph.D. Student
Biostatistical
Dear R-users
Can anyone please tell me how to format a label of a plot?
I found that R uses a floating point number (e.g. 2.0) sometimes and an
integer number (e.g. 2) other times in lables of a plot even though no
floating number is needed to show the lables.
How can I get R to use only integer
Hi,
I want to generate a heatmap for my data (in a matrix). However, the
data has some missing values (represented as blank).
I get the following errors (with the blanks and with blanks replaced by
NA and including the option rm.na = TURE):
filename = input_heatmap.txt
g -
On Fri, 7 Apr 2006, Kevin E. Thorpe wrote:
To enhance my understanding, and that of my students, I have a question
about cox.zph in the survival package.
If I have correctly gleaned the high-level point from the 1994
Biometrika paper of Grambsch and Therneau, it looks to me like
cox.zph
Hi there,
I have a statistics question on a classification problem:
Suppose I have 1000 binary variables and one binary dependent variable. I
want to find a way similar to PCA, in which I can find a couple of
combinations of those variables to discriminate best according to the
dependent
Do you mean the axis tick annotations, the xlab, the ylab or what? (It is
the xlab and ylab that are quite explicitly called 'axis label' in the
documentation.)
If you mean the axis tick annotations, please show us an example where R
uses an unnecessary (in your view) '.0'. The internal code
I'm using a very long irregular time-series of air temperature and
relative humidity of this kind (this is an extract only)
its.format(%
Y%d%m %X)
base
T H
20020601
12.00.00 27.1 47
20020601 15.00.00 29.1 39
20020601 18.00.00 27.4 39
20020601 21.00.00 24.0 40
On 4/7/06 11:54 AM, Himanshu Ardawatia [EMAIL PROTECTED]
wrote:
Sean Davis wrote:
On 4/7/06 11:01 AM, Himanshu Ardawatia [EMAIL PROTECTED]
wrote:
Hi,
I want to generate a heatmap for my data (in a matrix). However, the
data has some missing values (represented as blank).
I
Hi,
'g' must be a numeric matrix and obviously it's not (suspect it's character?).
If 'g' is numeric in your example depends on how you read data into
'filedata' and this step is missing in your code (maybe you just forgot to
use read.table?). See ?read.table for that, and use argument
aggregate(base,by=list(as.factor(format(dates(base),%Y%m%d))),mean,na.
rm=T)
-Original Message-
From: [EMAIL PROTECTED]
[mailto:[EMAIL PROTECTED] On Behalf Of Vittorio
Sent: Friday, April 07, 2006 11:38 AM
To: r-help@stat.math.ethz.ch
Subject: [R] Aggregating an its series
I'm using a
It sounds as though 'logic regression' might help. See
Ruczinski I, Kooperberg C, LeBlanc ML (2003). Logic Regression,
Journal of Computational and Graphical Statistics, 12, 475-511.
and the LogicReg package.
Peter Ehlers
Weiwei Shi wrote:
Hi there,
I have a statistics question on a
Weiwei,
You could also look into using one of several methods of classification
that calculate the weight of individual predictors in producing a correct
result based on some version of cross-validation. One that I use relatively
often is randomForest (in the randomForest package).
Sean
On
Jeanne == Jeanne Vallet [EMAIL PROTECTED]
on Fri, 7 Apr 2006 12:43:59 +0200 writes:
Jeanne Hello, I want to make a fuzzy classification from a
Jeanne dissimilarity matrix (calculated with daisy from
Jeanne package 'cluster'). I have tried to use fanny
Jeanne (package
Hello. Is there any documentation on the lmer function in the lme4 package
beyond what was published in the May 2005 R News (vol.5/1)? As well, has
the nonlinear version of lmer appeared yet?
Bill Shipley
North American Editor, Annals of Botany
Editor, Population and Community Biology
Hello. Besides the short paper in the May 2005 edition of R News, I have
not found any documentation concerning the lmer function in the lme4
package. Does anyone know of anything more substatial? Also, does anyone
know if the nonlinear equivalent of lmer exists yet?
Thanks.
Bill Shipley
The goal of logic regression is to find predictors that are Boolean
(logical) combinations of the original predictors, which might be more like
what I need.
The rf's variable importance evaluations and oob might help feature
selection but not feature construction, which is more of my concern.
Suppose I have a list where I want to extract only the elements that occur
in every component. For instance in the list foo I want to know that the
numbers 2 and 3 occur in every component. The solution I have seems
unnecessarily clunky. TIA, Andy
foo - list(x = 1:10, y=2:11, z=1:3)
bar
Jerry Friedman's RuleFit finds important rules, which I guess can be
thought of as features...
Andy
From: Weiwei Shi
The goal of logic regression is to find predictors that are Boolean
(logical) combinations of the original predictors, which
might be more like what I need.
The rf's
The vignette in the mlmRev package might be of some help.
Andy
From: Bill Shipley
Hello. Besides the short paper in the May 2005 edition of R
News, I have not found any documentation concerning the lmer
function in the lme4 package. Does anyone know of anything
more substatial? Also,
Hello,
I'm trying to figure out how to run the stepAIC function starting with the
NULL model. I can call the null model (e.g., lm(y ~ NULL)), but using
this object in stepAIC doesn't seem to work.
The objective is to calculate AICc. This can be done if stepAIC can be
run starting with the NULL
Joshua B Robins wrote:
Hello,
I'm trying to figure out how to run the stepAIC function starting with the
NULL model. I can call the null model (e.g., lm(y ~ NULL)), but using
this object in stepAIC doesn't seem to work.
I don't know what doesn't seem to work means here, but your null
Here is one solution:
all(unlist(lapply(foo, function(x) c(2,3) %in% x)))
[1] TRUE
This doesn't have the restriction of assuming that the components
of the list have unique elements, as the original solution does.
Patrick Burns
[EMAIL PROTECTED]
+44 (0)20 8525 0696
http://www.burns-stat.com
Thanks in advance,
I have searched and read the archives extensively and have been unable
to find a question (solution) exactly mimicking my problem. I have
created a scatterplot simulation of data through which I have drawn a
linear regression in this manner:
points(y~x, pch=*, col=black)
Thanks to the help of many on this list, I am now an R user and have
been able to write some functioning code to do matching estimation.
I have two for loops (i in 1:3, and j in 0:2). Within the loops, I
had been creating matrices of relevant estimation coefficents in order
to make lots of LaTeX
Hello,
I have a script that must be executed by R at the CLI completely
... without halting execution on encountering an error (object not found).
Rcmdr works fine, but source behaves strangely in R-2.1...
Thanks,
--
A. Mani
Member, Cal. Math. Soc
[[alternative HTML version
use a 'list';
x - list()
for (i in 1:3)
for (j in 0:2) {
.your calculations.
x[[as.character(i)]][[as.character(j)]] - yourResults
}
On 4/7/06, Brian Quinif [EMAIL PROTECTED] wrote:
Thanks to the help of many on this list, I am now an R user and have
been able to write
I kind of understand this recommendation, but I can't figure out how
to work with x once I have created it. Here is some fully
reproducible code. It creates a 4x1 matrices within the loops. Once
done with the looping, I want to cbind these matrices together. I
know it's a simple question, but
I don't have a direct answer to your question, but in case you are
interested in a general introduction to time series capabilties in R, I
will suggest the following:
1. Ch. 14 in Venables and Ripley (2002) Modern Applied Statistics
with S, 4th ed. (Springer)
2.
If the matrices you're storing inside the loop are all the same dimension
(as in your example), it's probably better to store them in an array; e.g.:
i1 - 2:3
i2 - 11:12
x - array(0, c(4, 1, length(i1), length(i2)))
for (i in seq(along=i1)) {
for (j in seq(along=i2)) {
x[, , i, j] -
Thanks for the suggestion, but I do not know how to get it to work for
the actual matrix I want to create. This is not reproducible, but
here is the form of the matrix I want:
#above I compute satt.yr.hrs and satt.full.load using Match()
Estimates - matrix(c(satt.yr.hrs$est,
Another possibility is to create a data.frame. I've just put
one statistic, sum, in it but you could put in as many as
you like:
g - expand.grid(i = 1:2, j = 1:2, k = 1:3)
f - function(x) with(x, c(i = i, j = j, k = k, sum = i+j+k))
do.call(rbind, lapply(split(g, rownames(g)), f))
i j k
Maybe clt.examp() in TeachingDemos is what you want.
2006/4/8, A Mani [EMAIL PROTECTED]:
Hello,
I have a script that must be executed by R at the CLI completely
... without halting execution on encountering an error (object not found).
Rcmdr works fine, but source behaves strangely
Hi Andy,
I apologize for the flat structure of my code. I do not know much
about programming conventions. I think your solution will work quite
well, but could you clarify a couple of things?
In the line
x - array('', c(4, 1, 6, 2, 3))
where do the numbers 4, 1, 6, 2, 3 come from. Here are
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