Ranney, Steven steven.ranney at montana.edu writes:
and I'm trying to fit a simple Von Bertalanffy growth curve with program:
VonB=nls(TL~Linf*(1-exp(-k*(age-t0))), data=box5.4,
start=list(Linf=1000, k=0.1, t0=0.1), trace=TRUE)
...
Everything works as it should right up until the
Does ?I() help?
See
? I()
Regards
Aval
On Wed, Sep 3, 2008 at 9:30 AM, [EMAIL PROTECTED] wrote:
R Users:
I'm wondering:
Why does my logical vector becomes a numeric vector when stuffed into a data
frame? How do I change this so that it's retained as a logical data type?
I've
thanks, but not quite what i wanted.
to be more precise: i want the whole zero-lines to be deleted, including the
attribute. in my final table only rows with non-zero rows should remain.
regarding my small toy example:
this is what i have:
test
c a b c d
a b
1 4 2 0 0 0
5 0 1 0 1
6 0
There is still no call to isoMDS in your code, and nothing we can
reproduce.
It all depends on the dissimilarity matrix you have not given us: maybe
there is no good 2D representation of it.
Looks like you need to ask a local expert about what you are doing, for
this is a statistical and
The problem is your misuse of cbind. You want
CuIEP - data.frame(Cu = CuZn$Cu, CuCen = CuZn$CuCen, StartCu, EndCu)
cbind() created a matrix, as there were numeric inputs the matrix has to
be numeric. I've absolutely no idea where you got the idea to use cbind
here, but it is helpdful to
Thank you, Professor Ripley. It is exactly what I need.
Best wishes!
Leon
Prof Brian Ripley wrote:
By reading from a file or connection, as in
foo - scan(, , n=1)
c:\tmp\foo
The parser always interprets \ in strings, so you must avoid the parser.
Deepayan,
that is exactly what I was hoping for. Thanks much!
Experimenting with it I noticed that updating two existing objects with plot
arguments seems to not work, at least not in this way:
gr1 -xyplot(rnorm(111) ~ runif(111), main = Plot A)
gr2 - xyplot(runif(111) ~ runif(111), main =
Does this help?
a - c(1,1,1,1,1,2,2,2,2,2)
b - c(4,5,6,5,4,7,8,9,8,7)
c - c(a,b,c,d,a,b,b,a,d,d)
A - cbind(a,b,c)
test - ftable(a,b,c)
test.df - data.frame(test)
test.df[test.df$Freq != 0, ]
Doesn't quite give the layout you seem to want, but effectively removes the
zeros.
Mikkel
Hello. I've being having problems to create directories in a windows
server environment . It seems that the recursive argument is not working
properly on the intranet, as it does in a local path. For example:
dir.create(server/directory1/directory2) ,
this works fine, and creates the
By the way, a different approach, with a different result, is to use
c.trellis from the latticeExtra package:
library(latticeExtra)
...
return(c(gr1, gr2))
This merges the panels into a single trellis object. Note that there
are complications with things like titles, legends and strips.
-Felix
Well, it looks like I am partly answering my own question. gnls is
clearly not going to be the right method to use to try out a
non-Gaussian error structure. The ls=Least Squares in gnls means
minimising the sum of the square of the residuals ... which is
equivalent to assuming a Gaussian error
Hello,
my test.R file contains two huge arrays (3000 entries), from which R needs to
calculate the Pearson Correlation, if I look at the file the numbers look
correct.
if I run R
R test.R --no-save
I see things like this:
0.723, 0.838, 1.002, 0.364, 0.357, 0.227, 0.982+ , 0.963, 0.535,
Hi
I am using R version 2.7.2. on a windows XP OS and have a question
concerning an analysis of covariance with count data I am trying to do,
I will give details of a scaled down version of the analysis (as I have
more covariates and need to take account of over-dispersion etc etc) but
as I am
This is what I was looking for.
Using mode instead of as.numeric is a great idea.
Many thanks.
Balázs
jholtman wrote:
Try this as a solution:
df - data.frame(a=letters[15:17], b=c(21,NA,23), c=10:12, d=15:17)
# convert to numeric
x - as.matrix(df)
mode(x) - numeric
Warning message:
Those are continuation line prompts. Most likely your file exceeds the
allowed line length for R script (1000 bytes is safe).
On Wed, 3 Sep 2008, Hanneke van Deutekom wrote:
Hello,
my test.R file contains two huge arrays (3000 entries), from which R needs to
calculate the Pearson
Hello,
I want to try oblique decision tree. I have found the mvpart package. In the
help file, there is an example of the mvpart function using the spider data
set. However, when trying the example I do not obtain an oblique decision
tree with separating rules such as p1+p25, but rather simple
Sherri;
The boxplot stats include outlier values in $out and their group number
in $group.
Let me assume that you want to put the boxes at integer positions on
the x-axis (the default is the same). The we have
y-rnorm(60)
g-gl(3,20)
labels=paste(Value, 1:length(y))
bx-boxplot(y~g)
bx.at-1:3
On Wed, 3 Sep 2008, Clusty wrote:
I want to try oblique decision tree. I have found the mvpart package. In the
What gave you the idea that mvpart was an implementation of oblique
decision trees? It does not say so anywhere.
help file, there is an example of the mvpart function using the
Hi, R users
I have a question with par(mfrow). I try to histograms and qqplots
form boot output for 5 statistics but par(mfrow=c(5,2)) or
par(mfrow=c(5,1)) does not work. R still display each figure
separately. What did I do wrong? (I check ?par)
c1.boot-boot(c1data,c1.fun,R=999)
c1.boot
When I do as.matrix I loose those columns that I specified as row headers
during cast.
Maybe its because of this:
When coercing a vector, it produces a one-column matrix, and promotes the
names (if any) of the vector to the rownames of the matrix.
but I cant figure out what does this mean or what
Hello Chibisi,
I am not shore whether I completely understand your needs: Do you want
to build a webpage which relies on a content management system (cms)? Do
you want to collect data (i.e. survey) which later on shall be analysed
using R? Or shall it be a webpage with an interactive R GUI? What
Hi Lara,
And I cant for the life of me work out why category one (semio1) is being
ignored, missing
etc.
Nothing is being ignored Lara --- but you are ignoring the fact that your
factors have been coded using the default contrasts in R, viz so-called
treatment or Dunnett contrasts. That is,
And perhaps I should also have added: fit your model without an intercept and
look at your coefficients. You should be able to work it out from there
quite easily. Anyway, you now have the main pieces.
Regards, Mark.
Mark Difford wrote:
Hi Lara,
And I cant for the life of me work out why
Dear all,
I am new in R. I want to read data which is vector from many files and
import to many object.
For example I have 10(ten) files.
file1.txt
file2.txt
:
:
file10.txt.
I want to import each file into one object as follows.
data1
data2
:
:
data10
I want to import file1.txt into
You can read in a list:
DF - lapply(dir(patt=file.*\\.txt), read.table, sep = ;, header =
TRUE)
names(DF) - paste(data, seq_along(DF), sep = )
DF[[data1]]
On Wed, Sep 3, 2008 at 7:52 AM, Sigit B Wibowo [EMAIL PROTECTED] wrote:
Dear all,
I am new in R. I want to read data which is vector
Try this:
set.seed(123)
DF - data.frame(fac = gl(5, 20), values = rnorm(100))
bp - boxplot(values ~ fac, data = DF, range = 0.5)
with(bp,
text(x = unique(group),
y = tapply(out, group, max),
labels = round(tapply(out, group, max), 2),
pos = 3))
On Tue, Sep 2,
Ok, I think I have this
it works if I use data.frame(cast(df,entityID ~ attributeID))
Ralikwen wrote:
When I do as.matrix I loose those columns that I specified as row headers
during cast.
Maybe its because of this:
When coercing a vector, it produces a one-column matrix, and
Dieter Menne [EMAIL PROTECTED] writes:
Brian Lunergan ff809 at ncf.ca writes:
I am stuck with a problem and I need help
---
R has a defined function sum()
I by chance defined a function with same name and now I m not able to
get rid of my sum() fuction. Every time
Hallo all
I realise this might be a too much to ask question, but can I improve the
Lattice plot produced by the following code? The type of figure I would like to
produce in each segment of the plot appears below the code.
library(reserving) # http://toolkit.pbwiki.com/RToolkit
haha...wrong code again, it's isoMDS not sammon in the 5th line.
Thanks for Victor Lemes Landeiro's and Brian D. Ripley's advice.
ÔÚ08-9-3£¬Prof Brian Ripley [EMAIL PROTECTED] дµÀ£º
There is still no call to isoMDS in your code, and nothing we can
reproduce.
It all depends on the
Hello,
I am trying to execute an example provided in the help files for the
playwith() function. I try to execute the first example:
library(playwith)
if (interactive()) {
options(device.ask.default = FALSE)
## Scatterplot (Lattice graphics).
## Labels are taken from rownames of data.
## Just
Hello R-User!
I still would be happy about an answer.
Further explanation:
I use distance(analogue), because I can give there a different weighting to
my used variables.
I would be interested, if there is the information in my Dissimilarity
matrix, that I used also factors and if this prevents
Dear Felix,
Thanks for the reply,
If you haven't already guessed I am new to web programming.
The sort of webpage I want to build is one that presents quantitative
information in graphs and charts, that people can interact with, e.g. select
parts of charts to zoom into, highlight values, click
different dimensions?
On Wed, Sep 3, 2008 at 8:13 AM, 陈武 [EMAIL PROTECTED] wrote:
haha...wrong code again, it's isoMDS not sammon in the 5th line.
Thanks for Victor Lemes Landeiro's and Brian D. Ripley's advice.
ÔÚ08-9-3£¬Prof Brian Ripley [EMAIL PROTECTED] дµÀ£º
There is still no call to
Hi,
Is it possible to code R to create folders or subfolders in a Windows SO?
In this case I would like to create a set of subfolders (/gis, /input,
/output, /tables, /vectorial) under a specified folder.
Paulo
__
R-help@r-project.org mailing list
Dear R community
I have a problem regarding which of the column in a matrix contains all of zero
elements. e.g.
x=c(3,3,3,3,0,0,0,0,5,5,5,5,8,8,8,8); x=matrix(x, nrow=4)
the output is
x
[,1] [,2] [,3] [,4]
[1,]3058
[2,]3058
[3,]3058
[4,]3
?dir.create
Uwe Ligges
Paulo Cardoso wrote:
Hi,
Is it possible to code R to create folders or subfolders in a Windows SO?
In this case I would like to create a set of subfolders (/gis, /input,
/output, /tables, /vectorial) under a specified folder.
Paulo
Hello.
I don't understand a particular output of portfolio.optim (tseries).
I have 4 assets and the portfolio.optim returns an asset with weight equals
to zero.
If I do a portfolio.optim with 3 assets, without the asset with weight
equals to zero,
it returns a completely different result.
try this:
x - c(3,3,3,3,0,0,0,0,5,5,5,5,8,8,8,8)
x - matrix(x, nrow=4)
which(colSums(x == 0) == nrow(x))
I hope it helps.
Best,
Dimitris
Muhammad Azam wrote:
Dear R community
I have a problem regarding which of the column in a matrix contains all of zero
elements. e.g.
Hi,
Parameters assessment in R with nls doesn't work, though it works fine with
MS Excel with the internal solver :(
I use nls in R to determine two parameters (a,b) from experimental data.
m VC0 CeQe
1 0.0911 0.0021740 3987.581 27.11637
Lara,
The first category is nor missing, nor ignored. It is used as the
reference. So the temperature effect for semio1 is only temperature. The
temperature effect for semio2 is temperature + temperature:semio2. For
semio3 it is temperature + temperature:semio3. Hence the main effect of
Hi,
I am looking for a normality test in R to see if a vector of data I have
can be assumed to be normally distributed and hence used in a linear
regression.
help.search(normality test)
suggests the Shapiro test, ?shapiro.test.
Now maybe I am interpreting things incorrectly (as is usually the
Dear list,
I'm wondering how to optimize functions/processes like the one shown
below (which simplifies something we're trying to do with very large
datasets).
In many cases I've noticed that using apply, sapply etc can help
speeding up processes, but in this case I don't see how I could do
Try squaring both sides of the formula.
On Wed, Sep 3, 2008 at 10:01 AM, Benoit Boulinguiez
[EMAIL PROTECTED] wrote:
Hi,
Parameters assessment in R with nls doesn't work, though it works fine with
MS Excel with the internal solver :(
I use nls in R to determine two parameters (a,b) from
Hello Chibisi,
you might be looking for something like Rpad
(http://www.rpad.org/Rpad/). There are some other systems like
SNetscape, Rserve, RSOAP, R.NET.Web as well. Unfortunately I have no
personal experience with these systems so far, so I can't give you any
real advice. I just know them to
I have one hundred and six independent variable that I would like to
preform a correlation analysis on. Is there anyway to only get the
values that are abolute value 0.6 or greater.
thanks
--
Stephen Sefick
Research Scientist
Southeastern Natural Sciences Academy
Let's not spend our time and
Dear Muhammad,
Try also:
x=c(3,3,3,3,0,0,0,0,5,5,5,5,8,8,8,8)
x=matrix(x, nrow=4)
which(colSums(x)==0)
[1] 2
HTH,
Jorge
On Wed, Sep 3, 2008 at 9:45 AM, Muhammad Azam [EMAIL PROTECTED] wrote:
Dear R community
I have a problem regarding which of the column in a matrix contains all of
zero
[EMAIL PROTECTED] napsal dne 03.09.2008 15:54:08:
try this:
x - c(3,3,3,3,0,0,0,0,5,5,5,5,8,8,8,8)
x - matrix(x, nrow=4)
which(colSums(x == 0) == nrow(x))
Isn't this the same?
which(colSums(x)==0)
Regards
Petr
I hope it helps.
Best,
Dimitris
Muhammad Azam wrote:
Dear R
I don't know of any implementation of the Epps-Pulley test in R, but if your
really want to do normality testing on large samples, then the best (for one
definition of best) test is the one proposed in this message:
http://finzi.psych.upenn.edu/R/Rhelp02a/archive/136160.html
--
Gregory (Greg)
Wolfgang Raffelsberger wrote:
Dear list,
I'm wondering how to optimize functions/processes like the one shown
below (which simplifies something we're trying to do with very large
datasets).
In many cases I've noticed that using apply, sapply etc can help
speeding up processes, but in this
Hello list
I have problems with little (see Debian package, r. instead R), I get an error
message relate with the symbol R_Visible, it say that it is unknown.
The same error I got from the Front-off or IDE for KDE (see debian packages, I
really don't remember the name., it is rk...
On 03/09/2008 10:33 AM, Williams, Robin wrote:
Hi,
I am looking for a normality test in R to see if a vector of data I have
can be assumed to be normally distributed and hence used in a linear
regression.
Raw data that is suitable for standard linear regression is normally
distributed, but
Hi Felix,
thanks for all the information there's plently of food for thought for me to
chew through.
Chibisi
On Wed, Sep 3, 2008 at 3:36 PM, drflxms [EMAIL PROTECTED] wrote:
Hello Chibisi,
you might be looking for something like Rpad
(http://www.rpad.org/Rpad/). There are some other
Hi
[EMAIL PROTECTED] napsal dne 03.09.2008 16:39:07:
Dear list,
I'm wondering how to optimize functions/processes like the one shown
below (which simplifies something we're trying to do with very large
datasets).
In many cases I've noticed that using apply, sapply etc can help
speeding
What is the distribution of the p-value when the null hypothesis is true?
This is an important question that unfortunately tends to get glossed over or
left out completely in many courses due to the amount of information that needs
to be packed into them.
For most appropriate tests, when the
On 9/3/2008 11:05 AM, Petr PIKAL wrote:
[EMAIL PROTECTED] napsal dne 03.09.2008 15:54:08:
try this:
x - c(3,3,3,3,0,0,0,0,5,5,5,5,8,8,8,8)
x - matrix(x, nrow=4)
which(colSums(x == 0) == nrow(x))
Isn't this the same?
which(colSums(x)==0)
No, because the column sum can be zero
(possibly an r-sig-debian question ...)
On 2 September 2008 at 16:00, Adrian Martínez Vargas wrote:
| Hello list
|
|
|
| I have problems with little (see Debian package, r. instead R), I get an
error message relate with the symbol R_Visible, it say that it is unknown.
What version or
on 09/03/2008 10:34 AM Chuck Cleland wrote:
On 9/3/2008 11:05 AM, Petr PIKAL wrote:
[EMAIL PROTECTED] napsal dne 03.09.2008 15:54:08:
try this:
x - c(3,3,3,3,0,0,0,0,5,5,5,5,8,8,8,8)
x - matrix(x, nrow=4)
which(colSums(x == 0) == nrow(x))
Isn't this the same?
which(colSums(x)==0)
Or:
which(apply(x == 0, 2, all))
On Wed, Sep 3, 2008 at 12:46 PM, Marc Schwartz [EMAIL PROTECTED]wrote:
on 09/03/2008 10:34 AM Chuck Cleland wrote:
On 9/3/2008 11:05 AM, Petr PIKAL wrote:
[EMAIL PROTECTED] napsal dne 03.09.2008 15:54:08:
try this:
x -
Hi
Excel fit is not exceptionally good. Try
fff-function(a,b) (V + b * m * a + C0 * V * b - ((C0 * V * b)^2 + 2 * C0
*
+ b * V^2 - 2 * C0 * V * m * a * b^2 + V^2 + 2 * V * m * a *
+ b + (b * m * a)^2)^(1/2))/(2 * b * m)
and with attached data frame
plot(Qe,fff(364,0.0126))
Dear John,
Yes, that's great - thanks!
Andy
John Fox wrote:
Dear Andy,
Yes, the tetrachoric correlation is a special case of the polychoric
correlation when both factors are dichotomous.
The 95-percent confidence interval that you suggest might be adequate if the
sample size is
[EMAIL PROTECTED] napsal dne 03.09.2008 17:34:24:
On 9/3/2008 11:05 AM, Petr PIKAL wrote:
[EMAIL PROTECTED] napsal dne 03.09.2008 15:54:08:
try this:
x - c(3,3,3,3,0,0,0,0,5,5,5,5,8,8,8,8)
x - matrix(x, nrow=4)
which(colSums(x == 0) == nrow(x))
Isn't this the same?
Hello,
I am trying to read in an Excel file that I saved as a .csv so I can analyze
my dissertation data! I am getting really frustrated because this is what I
keep getting:
In read.table(file = file, header = header, sep = sep, quote = quote, :
incomplete final line found by readTableHeader
Dear All,
I'm reading Frank Harrell's wonderful Regression
Modeling Strategies book and ran into a problem
following the example in Chapter 8. I'm working
on
platform: Ubuntu 8.04 (i486-pc-linux-gnu)
R version: 2.7.2 (2008-08-25)
and my command sequence was:
library(chron)
Hello Chibisi,
I am not shore whether I completely understand your needs: Do you want
to build a webpage which relies on a content management system (cms)? Do
you want to collect data (i.e. survey) which later on shall be analysed
using R? Or shall it be a webpage with an interactive R GUI? What
Hi all,
please, someone can explain me how update my R version?
thank you!
giov
--
View this message in context:
http://www.nabble.com/R-update-tp19291451p19291451.html
Sent from the R help mailing list archive at Nabble.com.
__
please delete me from this mailing list
i is not an R user and I overload my mail...
thanks
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide
Thank you very much for your time and thorough answers, Dieter Menne and
Daniel Malter.
It has to be said that I lack some of the basic statistical background and
don´t have a gut feeling if the tings I do is correct or not. Also, I use
SigmaPlot for the fitting, since I am not that familiar
First, thanks for your answer.
I find the description of mvpart package quite short:
Wrapper function for fitting and plotting rpart models
Then, I thought that mvpart was the package which seems closest to what I
would like to do (i.e. oblique decision tree also named multivariate
decision
From each R-help message's footer:
PLEASE do read the posting guide
http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
And this is really required, since we cannot guess what you did, what
the precise message is (ERROR or
try read.csv which is the same if you are using read.table with the
sep=, . is the final line complete? You are welcome to send me the
data and I can try and figur it out, but this is not a lot to go on.
On Wed, Sep 3, 2008 at 11:00 AM, catherine workman [EMAIL PROTECTED] wrote:
Hello,
I am
On 03/09/2008 11:00 AM, catherine workman wrote:
Hello,
I am trying to read in an Excel file that I saved as a .csv so I can analyze
my dissertation data! I am getting really frustrated because this is what I
keep getting:
In read.table(file = file, header = header, sep = sep, quote = quote, :
look here:
http://www.r-project.org/mail.html#instructions
On Wed, Sep 3, 2008 at 12:23 PM, Giovanni Tarquinio
[EMAIL PROTECTED] wrote:
please delete me from this mailing list
i is not an R user and I overload my mail...
thanks
__
depends on the OS, but just download the binary from cran.
On Wed, Sep 3, 2008 at 11:16 AM, giov [EMAIL PROTECTED] wrote:
Hi all,
please, someone can explain me how update my R version?
thank you!
giov
--
View this message in context:
On Wed, 3 Sep 2008, Clusty wrote:
First, thanks for your answer.
I find the description of mvpart package quite short:
Wrapper function for fitting and plotting rpart models
Then, I thought that mvpart was the package which seems closest to what I
would like to do (i.e. oblique decision
Petr Pikal wrote:
Hi
[EMAIL PROTECTED] napsal dne 03.09.2008 16:39:07:
In many cases I've noticed that using apply, sapply etc can help
speeding up processes, but in this case I don't see how I could do so.
a - runif(1000,0.5,1.6)
C - 2
M - 1000
system.time( for (i
Dear Stephen,
Perhaps
thishttp://www.nabble.com/Re:-applying-cor.test-to-a-(m,-n)-matrix---SUMMARY-to17150239.html#a17150239post
could helps. In general:
# Function
correl.stats=function(X, method = pearson, use = complete , conf.level =
0.95){
require(forward)
combs=t(fwd.combn(colnames(X),
Hi Paul,
Take a look at gam() from package mgcv (gam = generalized additive models),
maybe this will help you. GAMs can work with other distributions as well.
Generalized additive models consist of a random component, an additive
component, and a link function relating these two components.
Dear Uwe, Petr and Berend,
Thank's a lot !!
I just checked and it's
c(cumprod(a)[M-1],cumprod(a[-1])[M-1]))
that gives an identical result to the my initial loop (at 10x speed of
my initial loop ... )
Wolfgang
Berend Hasselman a écrit :
Petr Pikal wrote:
Hi
[EMAIL PROTECTED] napsal
On Wed, 3 Sep 2008, Clusty wrote:
First, thanks for your answer.
I find the description of mvpart package quite short:
Wrapper function for fitting and plotting rpart models
I agree that this could be improved.
Then, I thought that mvpart was the package which seems closest to what I
Note:
n - 100;
x - rnorm(n);
t1 - system.time({
xp - prod(x);
xpA2 - xp/x[n];
xpB2 - xp/x[1];
});
is calculating the product only once and is *constant in memory*.
With the suggest approach you are not only calculating the same thing
twice but you are also allocating a lot of memory,
You might also consider the additive model fitting with rqss() in the
package quantreg.
url:www.econ.uiuc.edu/~rogerRoger Koenker
email[EMAIL PROTECTED]Department of Economics
vox: 217-333-4558University of Illinois
fax: 217-244-6678
reg.line - function(y, x, title)
{plot(y~x, main=title, xlab=TSS, ylab=Aluminum)
line - lm(y~x)
d - summary(line)
legend(topleft, legend=paste(expression(r^2), = ,d$r.squared,
sep= ), bty=n)
abline(line)}
reg.line(1:10, 10:1, line)
how do I get the legend to print the plotmath symbol
Hi list
Is there a possibility to filter certain values out of an R output?
In my case: I want to create a vector of p-values in a for loop that
invokes for every increment cor.test() on two vectors.
I haven't found a way yet to tell cor.test() to only return the p-
values instead of the
Hello,
I am a researcher in sleep and circadian rhythms who is having much trouble
deciding on proper statistical analyses. Before I state my question, I
provide a brief synopsis of the looming problem; I am interested in activity
bout distributions across a 24 hr day. In addition to looking at
Dear Tobias,
From ?cor.test:
# Data set
x - c(44.4, 45.9, 41.9, 53.3, 44.7, 44.1, 50.7, 45.2, 60.1)
y - c( 2.6, 3.1, 2.5, 5.0, 3.6, 4.0, 5.2, 2.8, 3.8)
# Including text
res=cor.test(x, y, method = pearson)
res
# Just the p.value
res$p.value
Also, you might be interested in
I have a data frame containing sequences and I am interested in changing a few
sequences in a window and the swapping the original sequence back after I have
completed my analysis.
My temporary data frame that I am creating seq.in.window does not like the way
I am making me assignment. The
On 9/3/2008 12:20 PM, Tobias Binz wrote:
Hi list
Is there a possibility to filter certain values out of an R output?
In my case: I want to create a vector of p-values in a for loop that
invokes for every increment cor.test() on two vectors.
I haven't found a way yet to tell cor.test()
Hello, I'm trying to plot data that has gaps in the timeline because my data
only has the business day in it. When I do a line plot I get the data and
then a blank area where a line goes the tail of the last data point to the
head of the next data point. Is there a way I can do a line plot where
I have had luck using the circular and the CircStats packages to look at sleep
related data. I have fit mixtures of vonmeises distributions to time of
occurance responses (have not looked at duration with time of occurance).
There are also tools for doing regression models with circular
Hi,
I am getting accumulated data from PostgreSQL, ie for every day in
which a condition is true I get the number (count) of cases. Starting
date is 2008-01-01 and end day the last day for which the condition
is true (which is not necessarily today).
I obviously do not get records (dates) with
and I'm trying to fit a simple Von Bertalanffy growth curve with program:
VonB=nls(TL~Linf*(1-exp(-k*(age-t0))), data=box5.4,
start=list(Linf=1000, k=0.1, t0=0.1), trace=TRUE)
...
Everything works as it should right up until the confint(VonB) statement.
When I ask
for the confidence
I am using the tree package. One option is the cv.tree, which is
supposed to run cross-validations on your tree models. Is there
somewhere I can find some documentation on this function? I have the
help file that comes with the library, but I need more, especially on
what the output is.
The zoo package can represent and plot irregular time series.
There are three vignettes that describe it plus the help pages.
See ?zoo
?plot.zoo
?xyplot.zoo
e.g.
library(zoo)
z - zoo(1:3, Sys.Date() + c(1, 2, 5))
plot(z)
library(lattice)
xyplot(z)
On Wed, Sep 3, 2008 at 3:00 PM, Dr Eberhard W
On 3 September 2008 at 20:00, Dr Eberhard W Lisse wrote:
| Hi,
|
| I am getting accumulated data from PostgreSQL, ie for every day in
| which a condition is true I get the number (count) of cases. Starting
| date is 2008-01-01 and end day the last day for which the condition
| is true (which is
On Wed, 03-Sep-2008 at 02:18PM -0400, stephen sefick wrote:
| reg.line - function(y, x, title)
| {plot(y~x, main=title, xlab=TSS, ylab=Aluminum)
| line - lm(y~x)
| d - summary(line)
| legend(topleft, legend=paste(expression(r^2), = ,d$r.squared,
| sep= ), bty=n)
| abline(line)}
|
|
Hi Chibisi
I'm sort of jumping into this thread but from your description of charts /
plotting below I thought you might like to take a look at flot:
http://code.google.com/p/flot/
From the page
Flot is a pure Javascript plotting library for jQuery. It produces graphical
plots of arbitrary
Hello.
I'm having trouble installing rgl. I have a theory as to the problem.
First, the error message and session info.
install.packages(rgl)
trying URL 'http://probability.ca/cran/src/contrib/rgl_0.81.tar.gz'
Content type 'application/x-gzip' length 1636939 bytes (1.6 Mb)
opened URL
Hi,
I succeeded to read-in a transaction-table tr_dat with single items per
line via RODBC of the form:
Transact-ID ItemID
1item1
1item2
1item3
2item2
2item3
...
how do I create a transaction object of the arules package from such a
table?
I tried
Is there a way to convert a character (such as m) to a value between 0
and 255?
Thanks in advance,
Dave
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