All values in a matrix are the same type, so if you've set up a matrix
with a character column then your numeric values will also be stored
as character. That would explain why they are being converted to
factors. It would also explain why your query isn't working.
Michael
On 11 November 2010
Dear Claudia,
you are right. Thank you very much for your explanations. So in the
non-centered case SDEV does not contain the square roots of the eigenvalues
of the covariance/correlation matrix. In in the centered case it holds
A´A=(n-1)*cov(A) (not n+1).
Have a nice day.
--
View this
Dear all,
I'm struggling with predicting expected time until death for a coxph and
survreg model.
I have two datasets. Dataset 1 includes a certain number of people for which
I know a vector of covariates (age, gender, etc.) and their event times
(i.e., I know whether they have died and when if
Thanks for the feedback. My goal is to run a simple test to show that
the data cannot be rejected as either normally or uniformally
distributed (depening on the variable), which is what a previous K-S
test run using SPSS had shown. The actual distribution I compare to my
sample only matters that
Hello,
I have a nested named list structure, like the following:
x - list(
list(
list(df1,df2)
list(df3,
list(df4,df5))
list(df6,df7)))
with df1...d7 as data frames. Every data frame is named.
Is there a way to get a specific named element
Hi
I got an barplot, and I would like to have the exact number of the bars just
above the bars anyone know how to do this?
Sorry for bad English, and I do hope that you understand what im after.
//Joel
--
View this message in context:
Hello Forum,
I have data for two groups of subjects for a ratio variable (cortisol
level) assessed three times during the day (at 0600, 0900 and 2100 hours).
So I have three means for each group. I have been asked to do a slope
analysis to compare the two groups.
According to my
Hi,
A reproducible example would have been nice, but a correct code even
more (you forgot some commas)!
So if you meant this:
x - list(
list(
list(df1,df2),
list(df3,
list(df4,df5)),
list(df6,df7)))
check str(x): (I removes the details of each df)
List of 1
$ :List of 3
That makes perfect sense.
Since I need to build up the results table sequentially as I iterate
through the data, how would you recommend it??
Thanks,
-N
On 11/11/10 12:03 AM, Michael Bedward wrote:
All values in a matrix are the same type, so if you've set up a matrix
with a character
My objective is to start having meta-data on objects that I create.
For example, consider the following function:
assign2 - function(x, ...)
{
assign(x, ...)
attr(x, creation time) - Sys.time()
x - x
}
assign2(x, 1:4)
assign2 assigns to x the vector 1:4, and it then also adds the creation
Thanks Jim,
The next issue is how i get the RP lines to cut the axes. The lines either
cut the x axis at D=35 or the y axis at H=4.5.
How do I set the axes so that the lines start and end on the axes?
http://r.789695.n4.nabble.com/file/n3037496/plot.jpeg
--
View this message in context:
If I understood you correctly, you have this matrix of indicator variables for
occurrences of terms in documents:
A - matrix(c(1,1,0,0,1,1,1,0,1,1,1,0,0,0,1), nrow=3, byrow=TRUE,
dimnames=list(paste(doc,1:3), paste(term,1:5)))
A
and want to determine co-occurrence counts for pairs of
On 11-Nov-10 04:22:55, Kerry wrote:
I'm using ks.test (mydata, dnorm) on my data.
I think your problem may lie here! If you look at the documentation
for ks.test, available with the command:
help(ks.test)
or simply:
?ks.test
you will read the following near the beginning:
Usage: ks.test(x,
Thx for the details Josh! You're right - I meant performance analytics.
I'm about 4 weeks into R now so I'm a bit of a noob. I've downloaded the source
to see what you're talking about. I'll play around with it.
The data is pretty straight forward and has something like this:
Hi Friedericksen,
This function will do it. No doubt there are more elegant ways :)
rmatch - function(x, name) {
pos - match(name, names(x))
if (!is.na(pos))
return(x[[pos]])
for (el in x) {
if (class(el) == list) {
out - getEl(el, name)
if (!is.null(out)) return(out)
You can use rbind as in your original post, but if you've got a mix of
character and numeric data start with a data.frame rather than a
matrix.
Michael
On 11 November 2010 20:30, Noah Silverman n...@smartmediacorp.com wrote:
That makes perfect sense.
Since I need to build up the results table
Dear all,
I am converting a large process to a parallel backhend using doMC and
foreach. Basically, I havea long list of input graph files and each of
them calls soem basic igraph package functions. I am parallelizing the
run, in order to save time. All works fine, and each %dopar% call ends
What you want is some sort of indexing nested lists based on names (as we
are used to for vectors, for example). As Ivan pointed out, I don't think
there's an out-of-the-box function in R that supports such indexing as it
requires some sort of mapping of the nested list's hierarchical structure.
Hi Tal,
Here's a way of doing the first bit...
assign2 - function(x, ...) {
xname - deparse(substitute(x))
assign(xname, ...)
x - get(xname)
attr(x, creation.time) - Sys.time()
assign(xname, x, pos=.GlobalEnv)
}
Michael
On 11 November 2010 20:37, Tal Galili tal.gal...@gmail.com
Hi Ivan,
thank you very much for your detailed explanations and response!
Please excuse, that I did not include a reproducible example.
Yes, what I am looking for is such a function - but I suspect, that it does not
exist already, so I have to write it on my own.
Best,
Friedericksen
On
On Thu, Nov 11, 2010 at 9:37 AM, Tal Galili tal.gal...@gmail.com wrote:
4) My real intention is to somehow change the - operator (not simply the
assign). I am unsure as to how to do that.
5) Are there any major pros/cons to the adding of such meta-data to objects?
(for example, excessive
As a hack you could do this...
assign(=, assign2)
Michael
On 11 November 2010 21:30, Barry Rowlingson
b.rowling...@lancaster.ac.uk wrote:
On Thu, Nov 11, 2010 at 9:37 AM, Tal Galili tal.gal...@gmail.com wrote:
4) My real intention is to somehow change the - operator (not simply the
On 11 November 2010 21:08, Janko Thyson janko.thy...@ku-eichstaett.de wrote:
Could it be that you forgot to supply 'getEL()'? Or do I have to use some
package to make it available?
Oops - no. The problem was me stupidly renaming the function without
modifying the code. Try this instead...
On 11/10/2010 11:27 PM, Aleksandr Andreev wrote:
OK, that loads the shape file.
But now when I do:
submap- subset(spb, as.character(spb$Name) == 'Vasilevsky Island')
the submap still has the whole city, not just the subset. Is there now
a different way of extracting a subset from a map?
I
Hello
I am trying to export and lmList by write.table(x, x.csv) and it complains.
I have tried to convert the lmList into data.frame but also complains.
Any suggestion?
Thanks
Rosario
__
R-help@r-project.org mailing list
Hi,
I just found out that Tinn-R_2.3.6.0 is now available on
http://sourceforge.net/projects/tinn-r/.
When configuring Tinn-R: R Configure Permanent (Rprofile.site) I get
error
C:\Program Files\R\R-2.12.0\bin\etc\RProfile.site
The above file was not found.
Please try to repeat the procedure!
Neat!
I think replacing
out - rmatch(el, name)
with
out - Recall(el, name)
will avoid the dependence of the code on the function name
Keith J
Michael Bedward michael.bedw...@gmail.com wrote in message
news:aanlktikv9zepxipbp7ftr8vmpgsjaptjyj_dudy-m...@mail.gmail.com...
On 11 November
Hi,
I have tried this (just to practice):
assign2 - function(x, ...){
assign(x, ..., envir=.GlobalEnv)
attr(get(x, envir=.GlobalEnv), creation.time) - Sys.time()
}
assign2(y, 1:4)
Error in attr(get(x), creation.time) - Sys.time() :
could not find function get-
Why doesn't it work?
If I
On 10/11/2010 10:53 PM, dartdog wrote:
Ok the DLL is in the C:\Program Files\R\R-2.12.0\bin\i386 directory
but as the error shows it seems that rpy2 is looking in
R.dll within C:\PROGRA~1\R\R-212~1.0
So I'm a bit lost here...
What's unclear? The version you're working with hasn't been
Please tell the authors of Tinn-R.
Uwe Ligges
On 11.11.2010 12:12, Hannu Kahra wrote:
Hi,
I just found out that Tinn-R_2.3.6.0 is now available on
http://sourceforge.net/projects/tinn-r/.
When configuring Tinn-R: R Configure Permanent (Rprofile.site) I get
error
C:\Program
Please show us the code you used to run randomForest, the output, as
well as what you get with other algorithms (on the same random subset
for comparison). I have yet to see a dataset where randomForest does
_far_ worse than other methods.
Andy
-Original Message-
From:
Hi Joel,
I got an barplot, and I would like to have the exact number of the bars just
above the bars anyone know how to do this?
Google r number above bars barplot for threads on this list how to
do it and why you might not want to.
?text
Marianne
--
Marianne Promberger PhD, King's College
If you have trouble viewing or submitting this form, you can fill it out
online:
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Survey on Statistical Computing Email Lists
This is a survey for research purposes. All results would be aggregated and
On Thu, 2010-11-11 at 01:41 -0800, dpender wrote:
Thanks Jim,
The next issue is how i get the RP lines to cut the axes. The lines either
cut the x axis at D=35 or the y axis at H=4.5.
How do I set the axes so that the lines start and end on the axes?
Hello, Is it possible to include information about phylogenetic relatedness
among species (when species are replicates for each study within a
meta-analysis) in a meta-anlaysis in the metafor package?
Alternatively, I wonder if the method f Lajeunesse 2009 American Naturalist
has been adopted in
Date: Wed, 10 Nov 2010 19:27:20 -0500
From: sa.cizm...@usherbrooke.ca
To: r-help@r-project.org
Subject: Re: [R] log-transformed linear regression
Dear List,
I would like to take another chance and see if there if someone has
anything to say to
Hello group,
Can someone please put me through how to estimate the nuisance parameters
(Saturation parameter and radius) for the Geyer saturation process?
It seems quite confusing to me how these parameters are achieved, but they
are very important in determining the interaction between points in
Hello,
Does the area-interaction process built in Spatstat actually work for data
in a polygonal windows? I try to run it but I get the error message that W
must be a rectangular object, but my window is a polygon of my study area.
Please any help?
neba
[[alternative HTML version
Hi Sachin,
please read the posting guide and include a reproducible example of what
you want to do.
For your first question you should have a look at ?axis. Supplying the
'at' argument with the positions of the desired marks and the 'labels'
with text strings like '10.000$' should do what
Hi Josh,
many thank's for your reply. I tried to read up on this more and to be frank
I got a bit confused about the exact definition of residual standardization.
It occurs to me that different people have different definitions and that it
can be done with and without the leverage of each point.
Phil
you're right Mthticker is not a table.
Mthticker is a list. I am trying to reformat all elements in the list. or
i can reformat one element and do a loop around it.
Thanks
Cameron
Mthticker
str
List of 5
$ :'zoo' series from 01/02/04 to 11/01/10
Data: chr [1:83] LHG04LHJ04
Hello, I am working on a HP Laptop with R 2.12 just installed, I tried:
R version 2.12.0 (2010-10-15)
Copyright (C) 2010 The R Foundation for Statistical Computing
ISBN 3-900051-07-0
Platform: i386-pc-mingw32/i386 (32-bit)
...
library(Hmisc)
Loading required package: survival
Loading required
On 11/11/2010 8:57 AM, R Heberto Ghezzo, Dr wrote:
Hello, I am working on a HP Laptop with R 2.12 just installed, I tried:
R version 2.12.0 (2010-10-15)
Copyright (C) 2010 The R Foundation for Statistical Computing
ISBN 3-900051-07-0
Platform: i386-pc-mingw32/i386 (32-bit)
...
library(Hmisc)
On 11.11.2010 14:57, R Heberto Ghezzo, Dr wrote:
Hello, I am working on a HP Laptop with R 2.12 just installed, I tried:
R version 2.12.0 (2010-10-15)
Copyright (C) 2010 The R Foundation for Statistical Computing
ISBN 3-900051-07-0
Platform: i386-pc-mingw32/i386 (32-bit)
...
library(Hmisc)
On Nov 11, 2010, at 6:04 AM, Rosario Garcia Gil wrote:
Hello
I am trying to export and lmList by write.table(x, x.csv) and it
complains.
I have tried to convert the lmList into data.frame but also complains.
Any suggestion?
The most important suggestion would to provide context. Why are
Hi!
I wanted to know if there could be a way to reverse lengths in
polar.plot or radial.plot.
As far as I am concerned, larger lengths imply lines or points further
from the center of the circle and I want it the other way around: so if
my length limits are from 0 to 2000 I want points with
On Nov 11, 2010, at 4:44 AM, Stefan Evert wrote:
Pasted and realigned from original posting:
term1 term2 term3 term4 term5
term1 0 2 0 1 3
term2 2 0 0 1 2
term3 0 0 0 0 0
term4 1 1 0 0 1
term5 3 2 0 1 1
Any ideas on how to do that?
If I understood you correctly, you have this matrix of
Hi all,
I have a problem while working with the uroot package, more specifically
with the ADF.test. The smallest p-value it prints is 0.0100 but I want
to know the exact p-value. R does warn you about the problem by saying the
following: In interpolpval(code = code, stat = adfreg[, 3], N =
Jüri,
How did you create the output?
An example to cluster transactions with arules can be found in:
Michael Hahsler and Kurt Hornik. Building on the arules infrastructure
for analyzing transaction data with R. In R. Decker and H.-J. Lenz,
editors, /Advances in Data Analysis, Proceedings of
Hello all! I am nabsan.
I've got error below when I try to install the Rserve0.6.3 on AIX5.3.
also I tried several version of Rserve 0.4.3 to 0.6.3.. but there were same
error message..
[ld: 0711-317 ERROR: Undefined symbol: .main]
any help please.
--
OS:AIX5.3
Thanks!
It worked fine.
--
View this message in context:
http://r.789695.n4.nabble.com/Simple-Function-tp3035585p3038017.html
Sent from the R help mailing list archive at Nabble.com.
__
R-help@r-project.org mailing list
Hi, this is a question about bootstrapping, it relates more to the concept
than to the R package boot. But I wonder below if boot can help me. I have
the below to calculate a certain point estimate:
estimate= (0.9 * 0.03 * 70 *
(((77 * (76 / 76.0)) / 83107) -
((174 * (154 / 154.0)) /
On Nov 11, 2010, at 3:44 AM, Michael Haenlein wrote:
Dear all,
I'm struggling with predicting expected time until death for a
coxph and
survreg model.
I have two datasets. Dataset 1 includes a certain number of people
for which
I know a vector of covariates (age, gender, etc.) and their
On Nov 11, 2010, at 4:16 AM, Ravi Kulkarni wrote:
Hello Forum,
I have data for two groups of subjects for a ratio variable (cortisol
level) assessed three times during the day (at 0600, 0900 and 2100
hours).
So I have three means for each group. I have been asked to do a slope
analysis to
On Thu, 2010-11-11 at 18:14 +1100, Nevil Amos wrote:
I am having a problem with NA handling in capscale function.
as I understand it from the help capscale should permit NA values when
na.action=na.mit or na.exclude,
however I am getting the error
Error in X[nas, , drop = FALSE] :
I had a few more thoughts but briefly it almost
looks like exp at low X and lin at higher X may
be something to test. Anyway, I'd be hard pressed to
see how a linear fit, that you are happy to turn to
log-log fit, could give you a number of any relevance
except maybe over a few limited
Indeed, from the predict() function of the coxph you cannot get
directly time predictions, but only linear and exponential risk
scores. This is because, in order to get the time, a baseline hazard
has to be computed and it is not straightforward since it is implicit
in the Cox model.
2010/11/11
Servet,
These data do look linear in log space. Fortunately, the model
log(y) = a + b * log(x)
does have intercept zero in linear space. To see this, consider
log(y) = a + b * log(x)
y = 10^(a + b * log(x))
y = 10^a * 10^(b * log(x))
y = 10^a * 10^(log(x^b))
y = 10^a * x^b
On 07/11/2010 9:17 PM, Rolf Turner wrote:
On 8/11/2010, at 1:24 PM, Duncan Murdoch wrote:
SNIP
... The docs do say to
limit the length of titles to 65 characters if possible; this is one
place where it matters.
SNIP
The 65 character limit had not previously impinged
Thanks very much for your answers, David and Mattia.
I understand that the baseline hazard in a Cox model is unknown and that
this makes the calculation of expected survival difficult.
Does this change when I move to a survreg model instead?
I think I'm OK with estimating a Cox model (or a
Michael,
You are looking to compute an estimated time to death -- rather than the
odds of death conditional upon time. Thus, you will want to use time to
death as your dependent variable rather than a dichotomous outcome (
0=alive, 1=death). You can accomplish this with a straight forward
I tried using panel function but wasn't sure how to get the y value for the
peak of the densityplot.
I'm thinking there should be a way to retrieve the densityplot object so
that I can get the (x,y) values of the curve directly from the object.
Anyone know how to do this?
Thank you.
Joon
On
I get an error when I call bugs(..., program=OpenBUGS,
bugs.directory=c:/Program Files/OpenBUGS/OpenBUGS312),
expecting, as suggested in help(bugs), that it would fit the
model with openbugs() via BRugs.
help(bugs)
... either winbugs/WinBUGS or openbugs/OpenBUGS, the latter
makes use of
Consider the following simulations (also fixing the pnorm instead of dnorm that
Ted pointed out and I missed):
out1 - replicate(1, {
x - rnorm(1000, 100, 3);
ks.test( x, pnorm, mean=100, sd=3 )$p.value
} )
out2 - replicate(1, {
x - rnorm(1000, 100, 3);
Thanks for the comment, James!
The problem is that my initial sample (Dataset 1) is truncated. That means I
only observe time to death for those individuals who actually died before
end of my observation period. It is my understanding that this type of
truncation creates a bias when I use a
This is a simple and sensible question that does not have a simple
answer. It's a research issue, and you should go to the literature to
see what approaches seem appropriate.
HOWEVER, one simple descriptive approach -- which, however, may have
important statistical flaws -- is to run loess on the
On Nov 11, 2010, at 12:14 PM, Michael Haenlein wrote:
Thanks for the comment, James!
The problem is that my initial sample (Dataset 1) is truncated. That
means I
only observe time to death for those individuals who actually died
before
end of my observation period. It is my understanding
Dear R developers,
I have noticed a discrepancy between the coefficients returned by R's glm()
for logistic regression and SAS's PROC LOGISTIC. I am using dist = binomial
and link = logit for both R and SAS. I believe R uses IRLS whereas SAS uses
Fisher's scoring, but the difference is
If I run the following:
windows()
bringToTop(-1)
interactive()
[1] TRUE
run - readline(prompt = Continue (Yes = 1, No = 2):)
Continue (Yes = 1, No = 2):
dummy - 1
run
[1]
it does not allow user input though the session is interactive (it jumps right
over the
Hi,
Does anybody encounter the same problem when we overlap histogram and density
that the density line seem to shift to the right a little bit?
If you do have the same problem, what should we do to correct that?
Thank you.
par(mar=c(4,4,2,1.2),oma=c(0,0,0,0))
Is the algorithm converging? Is there separation (i.e.,
perfect predictor) in the model?
Are you getting a warning about fitted probabilities of
0 or 1?, etc.
We would need much more information (preferably a reproducible
example) before we can help.
Benjamin Godlove wrote:
Dear R developers,
PLEASE follow the posting guide and give us the output of sessionInfo().
I was unable to duplicate your problem -- it worked fine for me.
sessionInfo()
R version 2.12.0 (2010-10-15)
Platform: i386-pc-mingw32/i386 (32-bit)
locale:
[1] LC_COLLATE=English_United States.1252
[2]
do you have factors (categorical variables) in the model? it could be
just a parameterization difference.
albyn
On Thu, Nov 11, 2010 at 12:41:03PM -0500, Benjamin Godlove wrote:
Dear R developers,
I have noticed a discrepancy between the coefficients returned by R's glm()
for logistic
I'm having the same problem and was wondering whether you ever found a
solution? It gives me the error Error in Survr(id, time, event) : Data
doesn't match. Every subject must have a censored time even though all my
subjects are right-censored, and to be sure, I've even used the addCenTime
Still doesn't work.
When using rbind to build the data.frame, it get a structure mostly full
of NA.
The data is correct, so something about pushing into the data.frame is
breaking.
Example code:
results - data.frame()
for(i in 1:n){
#do all the work
#a is a test label. b,c,d are
On Nov 11, 2010, at 2:09 PM, Emily wrote:
I'm having the same problem
(???: from a three year-old posting for which you didn't copy any
context.)
and was wondering whether you ever found a
solution? It gives me the error Error in Survr(id, time, event) :
Data
doesn't match. Every
Albyn Jones jones at reed.edu writes:
do you have factors (categorical variables) in the model? it could be
just a parameterization difference.
albyn
On Thu, Nov 11, 2010 at 12:41:03PM -0500, Benjamin Godlove wrote:
Dear R developers,
I have noticed a discrepancy between the
David, Mattia, James -- thanks so much for all your helpful comments!
I now have a much better understanding of how to calculate what I'm
interested in ... and what the risks are of doing so.
Thanks and all the best,
Michael
On Thu, Nov 11, 2010 at 7:33 PM, David Winsemius
Roslina Zakaria zroslina at yahoo.com writes:
Hi,
Does anybody encounter the same problem when we overlap histogram and density
that the density line seem to shift to the right a little bit?
par(mar=c(4,4,2,1.2),oma=c(0,0,0,0))
hist(datobs,prob=TRUE, main =Volume of
On Nov 11, 2010, at 2:50 PM, David Winsemius wrote:
On Nov 11, 2010, at 2:09 PM, Emily wrote:
I'm having the same problem
(???: from a three year-old posting for which you didn't copy any
context.)
and was wondering whether you ever found a
solution? It gives me the error Error in
On 11-Nov-10 18:39:34, Roslina Zakaria wrote:
Hi,
Does anybody encounter the same problem when we overlap histogram
and density that the density line seem to shift to the right a
little bit?
If you do have the same problem, what should we do to correct that?
Thank you.
[OOPS!!I accidentally reproduced my second example below
as my third example. Now corrected. See below.]
On 11-Nov-10 20:02:29, Ted Harding wrote:
On 11-Nov-10 18:39:34, Roslina Zakaria wrote:
Hi,
Does anybody encounter the same problem when we overlap histogram
and density that the
On Thu, Nov 11, 2010 at 11:33 AM, Noah Silverman
n...@smartmediacorp.com wrote:
Still doesn't work.
When using rbind to build the data.frame, it get a structure mostly full of
NA.
The data is correct, so something about pushing into the data.frame is
breaking.
Example code:
results -
You can try ctree in package party, but anyway: what is the deeper
sense in a binary split for a variable with more than 32 levels?
Regards, Sven
2010/11/10 Erik Iverson er...@ccbr.umn.edu:
Well, the error message seems relatively straightforward.
When you run str(x) (you did not provide the
Sorry for the lack of context - I found a forum (
http://r.789695.n4.nabble.com/R-help-f789696.html) that I thought was easier
to navigate and was using it for the first time. Hit the reply button, but
forgot that the mailing list recipients would not see previous message in
the thread. I've
Peter,
Your example doesn't work for me unless I
set options(stringsAsFactors=TRUE) first.
(If I do set that, then all columns of 'results'
have class character, which I doubt the user
wants.)
results - data.frame()
n = 10
for(i in 1:n){
+a = LETTERS[i];
+b = i;
+c = 3*i + 2
+
--- On Thu, 11/11/10, William Dunlap wdun...@tibco.com wrote:
From: William Dunlap wdun...@tibco.com
Subject: Re: [R] Populating then sorting a matrix and/or data.frame
To: Peter Langfelder peter.langfel...@gmail.com, r-help@r-project.org
Received: Thursday, November 11, 2010, 4:19 PM
On Thu, Nov 11, 2010 at 1:19 PM, William Dunlap wdun...@tibco.com wrote:
Peter,
Your example doesn't work for me unless I
set options(stringsAsFactors=TRUE) first.
(If I do set that, then all columns of 'results'
have class character, which I doubt the user
wants.)
You probably mean
You are right, I mistyped it.
Bill Dunlap
Spotfire, TIBCO Software
wdunlap tibco.com
-Original Message-
From: John Kane [mailto:jrkrid...@yahoo.ca]
Sent: Thursday, November 11, 2010 1:58 PM
To: Peter Langfelder; r-help@r-project.org; William Dunlap
Subject: Re: [R] Populating
On Thu, Nov 11, 2010 at 1:19 PM, William Dunlap wdun...@tibco.com wrote:
Peter,
Your example doesn't work for me unless I
set options(stringsAsFactors=TRUE) first.
Yes, you need to set options(stringsAsFactors=FALSE) (note the FALSE).
I do it always so I forgot about that, sorry.
Dear List,
I am looking to perform exploratory analyses of two (relatively) large
datasets of categorical data. The first one is a binary 80x100 matrix, in
the form:
matrix(sample(c(0,1),25,replace=TRUE), nrow = 5, ncol=5, dimnames = list(c(
group1, group2,group3, group4,group5), c(V.1, V.2,
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Hi All,
I'm trying to create labels to plot such that it doesn't show up as
scientific notation. So for example how do I get R to show 1e06 as
$1,000,000.
I was wondering if there was a single function which allows you to do that,
the same way that as.Date() allows you to show in date format on
Your errors look exactly like mine.
Changing the option flag does allow me to create the data.frame without
any errors. A quick look confirms that all the values are there and
correct.
However, R has coerced all of my numeric values to strings.
Using your sample code also turns all the
That makes perfect sense. All of my numbers are being coerced into
strings by the c() function. Subsequently, my data.frame contains all
strings.
I can't know the length of the data.frame ahead of time, so can't
predefine it like your example.
One thought would be to make it arbitrarily long
Is this what you want:
paste($, format(1e6, big.mark = ',', format = 'f'), sep='')
[1] $1,000,000
On Thu, Nov 11, 2010 at 5:51 PM,
sachinthaka.abeyward...@allianz.com.au wrote:
Hi All,
I'm trying to create labels to plot such that it doesn't show up as
scientific notation. So for example
Hi All,
I've reproduced the example from Prof. Friedrich Leisch's webpage. When I
write sweave(Example-1.Snw) OR sweave(Example-1.Rnw), (yes, I renamed
them). I get the following error:
Writing to file example-1.tex
Processing code chunks ...
1 : echo term verbatim
Error: chunk 1
Error in
On Nov 11, 2010, at 5:51 PM, sachinthaka.abeyward...@allianz.com.au
wrote:
Hi All,
I'm trying to create labels to plot such that it doesn't show up as
scientific notation. So for example how do I get R to show 1e06 as
$1,000,000.
Knowing how these things usually have even more compact
I see 4 ways to write the code:
1. make the frame very long at the start and use my code - this is
practical if you know that your data frame will not be longer than a
certain number of rows, be it a million;
2a. use something like
result1 = data.frame(a=a, b=b, c=c, d=d)
within the loop to
On Nov 11, 2010, at 6:38 PM, Noah Silverman wrote:
That makes perfect sense. All of my numbers are being coerced into
strings by the c() function. Subsequently, my data.frame contains all
strings.
I can't know the length of the data.frame ahead of time, so can't
predefine it like your
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