>From what I can see the function does not give back this information
(either in the print output or the output object structure itself).
The thing is that I have never studied the statistic used for this test, so
I am not sure how it works and if giving the SS MS is meaningful or not.
But if you
It is not clear to me how meaningful sums of squares and mean squares
are in the context of oneway.test.(). Be that as it may, the short answer
is you ***can't*** get at them from the output of oneway.test(). The
oneway.test() function simply does not return enough information.
If you ***reall
Below.
-- Bert
On Mon, Apr 9, 2012 at 7:39 PM, David Winsemius wrote:
>
> On Apr 9, 2012, at 6:24 PM, phi771 wrote:
>
>> Hello!
>> I have created a bplot-figure using this code:
>>
>> *file <- "2dcali_red.ttt"
>> ux<-as.matrix(read.table(file, dec = ","))
>>
>> mode(ux)<-'numeric'
>>
>> vel<-ux[
On Apr 9, 2012, at 10:39 PM, David Winsemius wrote:
On Apr 9, 2012, at 6:24 PM, phi771 wrote:
Hello!
I have created a bplot-figure using this code:
*file <- "2dcali_red.ttt"
ux<-as.matrix(read.table(file, dec = ","))
mode(ux)<-'numeric'
vel<-ux[,1]
ang<-ux[,2]
x<-ux[,3]
y<-ux[,4]
dat<- d
Hello everyone:
I'm a new member of this group.
I have a question about "oneway.test".
When I use "anova(lm())" to analysis the
ANOVA,
I can get the information about Sum Sq and Mean
On Apr 9, 2012, at 6:24 PM, phi771 wrote:
Hello!
I have created a bplot-figure using this code:
*file <- "2dcali_red.ttt"
ux<-as.matrix(read.table(file, dec = ","))
mode(ux)<-'numeric'
vel<-ux[,1]
ang<-ux[,2]
x<-ux[,3]
y<-ux[,4]
dat<- data.frame(ang=ang, x=x,y=y)
require(rms)
ddist2 <- da
See http://biostat.mc.vanderbilt.edu/SweaveConvert for several other
approaches. The most solid way I've found is to convert PDF to Word.
Frank
Alexander Shenkin wrote
>
> Thanks for the heads up, Paul. That's good to know. I happen to be in
> academics, as are my collaborators, but I could i
Yup. Latest version 2.15.0 for windows solved the problem alright! Thank you
Uwe.
Regards,
Fayez
From: Uwe Ligges [lig...@statistik.tu-dortmund.de]
Sent: Monday, April 09, 2012 9:56 AM
To: Aziz, Muhammad Fayez
Cc: r-help@r-project.org
Subject: Re: [R] Pan
Hello,
Jean V Adams wrote
>
> Delia,
>
> name <- data.frame(Behavior=c(1, 2, 1, 2, 1), Time=c(0, 40, 45, 55, 57))
>
> appear <- name$Time[name$Behavior==1]
> disappear <- name$Time[name$Behavior==2]
> if(length(appear) > length(disappear)) disappear <- c(disappear, 60)
> sum(disappear - appear
see below!
On Mon, Apr 9, 2012 at 7:38 PM, Gabor Grothendieck
wrote:
> We have had Linux users that successfully used Ryacas but if you can't
> get yacas to build on your particular system then you could try
> running the Windows version under wine. A Windows binary of yacas is
> available so yo
Thanks Henrique.It worked!!! I appreciate your help.
Cassie
On Mon, Apr 9, 2012 at 8:06 PM, Henrique Dallazuanna wrote:
> Try this:
>
> Sys.setlocale(category = "LC_TIME", locale = "US")
> format(as.POSIXct('0001-01-01 00:00:00') + x, "%I:%M:%S %p")
>
>
> On Mon, Apr 9, 2012 at 9:38 PM, cass
Try this:
Sys.setlocale(category = "LC_TIME", locale = "US")
format(as.POSIXct('0001-01-01 00:00:00') + x, "%I:%M:%S %p")
On Mon, Apr 9, 2012 at 9:38 PM, cassie jones wrote:
>
> Hello everyone,
>
> I am wondering if there is any routine in R which can convert time given in
> 'seconds' unit to t
On a Unix-alike system, if you have an svn client installed, it
suffices to type
svn checkout svn://svn.r-forge.r-project.org/svnroot/quantmod/
at a command prompt -- then use R CMD install on the pkg/ directory to
install the new version. No idea how to do it on Windows
Michael
On Sun, Apr
We have had Linux users that successfully used Ryacas but if you can't
get yacas to build on your particular system then you could try
running the Windows version under wine. A Windows binary of yacas is
available so you won't have to build yacas. I don't know if anyone
has tried that yet but its
Hello everyone,
I am wondering if there is any routine in R which can convert time given in
'seconds' unit to the 12 hour time format. For example, suppose the data
set looks like
x=c(36885,84000,20) #x in seconds
I want to get the output as
[1] " 11:14:45 AM"
[2] " 11:20:00 PM"
[3] "12:20:00
On Mon, Apr 9, 2012 at 12:40 PM, Jason Rodriguez
wrote:
> Hello, I have a graphics-related question:
>
> I was wondering if anyone knows of a way to create a bar chart that is
> colored with a three-part gradient that changes at fixed y-values. Each bar
> needs to fade green-to-yellow at Y=.10 a
see below!
On Mon, Apr 9, 2012 at 3:04 PM, Gabor Grothendieck
wrote:
> On Mon, Apr 9, 2012 at 4:02 PM, Gabor Grothendieck
> wrote:
>> On Mon, Apr 9, 2012 at 2:47 PM, Martin Maechler
>> wrote:
>>> Apropos:
>>>
>>> I don't have the problems, the OP had, but on my ubuntu
>>> notebook, Ryacas does
Hello!
I have created a bplot-figure using this code:
*file <- "2dcali_red.ttt"
ux<-as.matrix(read.table(file, dec = ","))
mode(ux)<-'numeric'
vel<-ux[,1]
ang<-ux[,2]
x<-ux[,3]
y<-ux[,4]
dat<- data.frame(ang=ang, x=x,y=y)
require(rms)
ddist2 <- datadist(dat)
options(datadist="ddist2")
fitn
I have time-series data looking like this:
> dataIn[sample(c(1:nrow(dataIn)), 25),]
accelerometer_y id data_block_epoch_time
782 0.8424 201300 133179733
1868 0.3432 202386 1331797384000
1828 0.3510 202346 1331797382000
1026 0
Pat Wilkins illinois.edu> writes:
>
> Greetings,
>
> I am running glm models for species counts using a poisson link function.
> Normal summary functions for this provide summary statistics in the form of
> the deviance, AIC, and p-values for individual predictors. I would like to
> obtain the
On Apr 9, 2012, at 5:33 PM, Christopher Desjardins wrote:
Hi,
I am having trouble with syntax for a for loop. Here is what I am
trying to
do.
class=c(rep(1,3),rep(2,3),rep(3,3))
out1=rnorm(length(class))
out2=rnorm(length(class))
out3=rnorm(length(class))
data=data.frame(class,out1,out2,out
class1 <- data[1==data$class,]
gets you a subset of the data into a dedicated object, but if you want to
handle arbitrarily large amounts of data or class values then the list output
of split is really much better to stay with.
Also, "data" is a predefined function, so it is not a good idea to
Hi,
I am having trouble with syntax for a for loop. Here is what I am trying to
do.
class=c(rep(1,3),rep(2,3),rep(3,3))
out1=rnorm(length(class))
out2=rnorm(length(class))
out3=rnorm(length(class))
data=data.frame(class,out1,out2,out3)
dat.split=split(data,data$class)
for(i in 1:3){
sub[i]=da
On Apr 9, 2012, at 3:17 PM, Kerapi wrote:
Hi!,
I'm really hoping someone out there will be able to help me.
I recently started my MSc dissertation on Population Projection
Matrices, which has been going well until now. I am trying to set-up
a general script that does a pairwise comparison
Delia,
name <- data.frame(Behavior=c(1, 2, 1, 2, 1), Time=c(0, 40, 45, 55, 57))
appear <- name$Time[name$Behavior==1]
disappear <- name$Time[name$Behavior==2]
if(length(appear) > length(disappear)) disappear <- c(disappear, 60)
sum(disappear - appear)
Delia Shelton wrote on 04/09/2012 12:30:23 P
Hi!,
I'm really hoping someone out there will be able to help me.
I recently started my MSc dissertation on Population Projection Matrices, which
has been going well until now. I am trying to set-up a general script that does
a pairwise comparison of all elements in my matrices.
So for exampl
It is reporting _proportion_ of the total variance. Not the variance.
The total variance is the sum of all the squared standard deviations. Only
the first 3 standard deviations are shown.
Kevin
On Mon, Apr 9, 2012 at 3:31 PM, carol white wrote:
> Hello,
> It might be a trivial question but
Hi!
I study fuzzy regression. Linear Programming(LP) methods are commonly used
for fuzzy linear regression (FLR) because they are simple and easy to
apply. I should make a simulation study with R and have come a distance.
I use "linp"
(for instence a part from my codes)
"(L<-linp(E=NULL,F=NULL,Cos
Hello,
It might be a trivial question but I just wanted to find out the relationship
between sdev and proportion of variance generated by prcomp. I got the
following result from my data set
PC1 PC2 PC3
Standard deviation 104.89454 15.40910 9.012047
Prop
On Mon, Apr 9, 2012 at 4:02 PM, Gabor Grothendieck
wrote:
> On Mon, Apr 9, 2012 at 2:47 PM, Martin Maechler
> wrote:
>> Apropos:
>>
>> I don't have the problems, the OP had, but on my ubuntu
>> notebook, Ryacas does not return expressions (just the strings),
>> and hence
>>
>> as.expression(
On Mon, Apr 9, 2012 at 2:47 PM, Martin Maechler
wrote:
> Apropos:
>
> I don't have the problems, the OP had, but on my ubuntu
> notebook, Ryacas does not return expressions (just the strings),
> and hence
>
> as.expression( )
>
> always gives NULL and e.g. the demo(Ryacas-Function)
> also f
It seems to be choking on NLST:
require("RCurl")
getURL("ftp://e4ftl01.cr.usgs.gov/MOTA/MCD15A3.005/",verbose=TRUE,ftp.use.epsv=TRUE,
dirlistonly = TRUE)
...
< 230 Guest login ok, access restrictions apply.
> PWD
< 257 "/" is current directory.
* Entry path is '/'
> CWD MOTA
< 250 CWD command su
Its kinda gross, but you can just download.file() and then parse the
result
FTP directory /MOTA/MCD15A3.005 at e4ftl01.cr.usgs.gov
FTP directory /MOTA/MCD15A3.005 at
e4ftl01.cr.usgs.gov
##
##
Apropos:
I don't have the problems, the OP had, but on my ubuntu
notebook, Ryacas does not return expressions (just the strings),
and hence
as.expression( )
always gives NULL and e.g. the demo(Ryacas-Function)
also fails:
> yacas(expression(deriv(BurrCDF(x,c,k
k*c*x^(c-1)*(x^c+1)^
Ya I hit the same error with my code that reads directories.
I'll try some other stuff . I think I hit this error before I with usgs.
On Mon, Apr 9, 2012 at 11:40 AM, Jonathan Greenberg wrote:
> Steven:
>
> Thanks -- I seem to be running into the problem with the link I sent along:
>
> > getURL
Steven:
Thanks -- I seem to be running into the problem with the link I sent along:
> getURL("ftp://e4ftl01.cr.usgs.gov/MOTA/MCD15A3.005/",verbose=TRUE,dirlistonly
> = TRUE)
Error in function (type, msg, asError = TRUE) : RETR response: 550
I'm wondering if it might be a passive ftp issue, but
Thanks to Hadley, William and Duncan for suggestions. I'm currently
implementing a
solution that is close to that of William and Duncan (and learning more about
environments
in the process). I suspect the reference classes are possibly a more reliable
long term
solution. I'll plead laziness unt
On 09/04/2012 18:52, Bert Gunter wrote:
On Mon, Apr 9, 2012 at 9:27 AM, Bazman76 wrote:
Yes I agree, there may be something pathalogical in the way at least one of
the models handles the data. That's why I was trying to get a better handle
on how the two functions spec.prgm() and spec.ar() wor
A couple of ways.
using Rcurl you can use the curlOption of dirlistonly.
otherwise you can read the page and parse. I've got some code around here
to do that.
Steve
On Mon, Apr 9, 2012 at 11:27 AM, Jonathan Greenberg wrote:
> R-helpers:
>
> I'd like to be able to store all the file informa
R-helpers:
I'd like to be able to store all the file information from an ftp site
(e.g. file and foldernames) through an R command. Any ideas how to do
this? Here's an example site to use:
ftp://e4ftl01.cr.usgs.gov/MOTA/MCD15A3.005
--j
--
Jonathan A. Greenberg, PhD
Assistant Professor
Depart
On Mon, Apr 9, 2012 at 9:27 AM, Bazman76 wrote:
> Yes I agree, there may be something pathalogical in the way at least one of
> the models handles the data. That's why I was trying to get a better handle
> on how the two functions spec.prgm() and spec.ar() work.
>
> The data has been processed by
Not really an R question. Post on a statistics site like
http://stats.stackexchange.com/
-- Bert
On Mon, Apr 9, 2012 at 9:51 AM, Pat Wilkins wrote:
> Greetings,
>
> I am running glm models for species counts using a poisson link function.
> Normal summary functions for this provide summary stati
Hi,
I am attempting to tabulate binned data. The '1' represents the appearance of
the focal mouse pup, and '2' represents the disappearance of the focal mouse
pup. The code written below is intended to calculate the total time spent
appeared out of 3600s. For Sample 1, both the hand calculat
Yes I agree, there may be something pathalogical in the way at least one of
the models handles the data. That's why I was trying to get a better handle
on how the two functions spec.prgm() and spec.ar() work.
The data has been processed by a wavelet analysis, so what you are seeing as
the "raw" d
Greetings,
I am running glm models for species counts using a poisson link function.
Normal summary functions for this provide summary statistics in the form of
the deviance, AIC, and p-values for individual predictors. I would like to
obtain the p-value for the overall model. So far, I have bee
On 05/04/2012 4:20 PM, John C Nash wrote:
In trying to streamline various optimization functions, I would like to have a
scratch pad
of working data that is shared across a number of functions. These can be
called from
different levels within some wrapper functions for maximum likelihood and ot
> Make OPCON an environment and pass it into the functions that may read it or
> alter it. There
> is no real need to pass it out, since environments are changed in-place
> (unlike lists). E.g.,
> > x <- list2env(list(one=1, two="ii", three=3))
> > x
>
> > objects(x)
> [1] "one" "three"
On Apr 9, 2012, at 12:32 PM, James Lenihan wrote:
I found this example in an Introductory R book in the chapter on
Matrices and
Arrays
The array is
m
[,1] [,2] [,3] [,4] [,5]
[1,]0 12 138 20
[2,] 120 15 28 88
[3,] 13 15069
[4,]8 286
On 09/04/2012 12:32 PM, James Lenihan wrote:
I found this example in an Introductory R book in the chapter on Matrices and
Arrays
You should probably check with the author of the book (there might be an
errata page posted somewhere), but it looks like a typo, in your code or
the original:
T
I found this example in an Introductory R book in the chapter on Matrices and
Arrays
The array is
> m
[,1] [,2] [,3] [,4] [,5]
[1,]0 12 138 20
[2,] 120 15 28 88
[3,] 13 15069
[4,]8 2860 33
[5,] 20 889 330
The code is
oops sorry
n 08.04.2012 20:39, Bazman76 wrote:
> Hi there,
>
> Can someone explain what the difference between spec.pgram and spec.ar is?
>
> I understand that they attempt to do the same thing one using an AR
> estimation of the underlying series to estimate teh sensity the other
> using
>
On Apr 9, 2012, at 16:55 , Uwe Ligges wrote:
>
>
> On 08.04.2012 20:39, Bazman76 wrote:
>> Hi there,
>>
>> Can someone explain what the difference between spec.pgram and spec.ar is?
>>
>> I understand that they attempt to do the same thing one using an AR
>> estimation of the underlying serie
On 4/5/2012 6:32 PM, mhimanshu wrote:
Hi Thomas,
Thank you so much for your suggestion.
I tried your code and it is working fine. Now when I change the values of Y
in yobs I am getting so many warnings.
say,
yobs<- data.frame(
time = 0:7,
Y = c(0.00, 3.40, 4.60 ,5.80, 5.80, 6.00, 6.00 ),
On 09.04.2012 17:01, Bazman76 wrote:
OK so I neeed to understan better what it it they are trying to measure.
I understood (incorrectly it seems) that they were simply different methods
to get the same result?
Yes. Also note this is a mailing list and you are lucky I was able to
remember th
OK so I neeed to understan better what it it they are trying to measure.
I understood (incorrectly it seems) that they were simply different methods
to get the same result?
--
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On 09.04.2012 07:16, arunkumar wrote:
HI
I want to create a dummy variable for a dataset which has various dates Eg
I've one year data. The user can choose any date range
startdate : 2012-01-01
enddate: 2012-02-01
dummy <- date >= startdate & date <= enddate
Uwe Ligges
startdate: 2
Thank you very much! This worked.
--
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Sent from the R help mailing list archive at Nabble.com.
__
R-help@r-project.org mailing list
h
On 09.04.2012 06:31, Aziz, Muhammad Fayez wrote:
Hi,
Please find the code and data following. Problem appears in lines 37 - 42 of
the code. I am using R.2.13.0 on WinXP.
Works for me with a recent version of R (i.e. 2.15.0) / grid and lattice
(i.e. 0.20-6).
Please always try with recent
On 08.04.2012 20:39, Bazman76 wrote:
Hi there,
Can someone explain what the difference between spec.pgram and spec.ar is?
I understand that they attempt to do the same thing one using an AR
estimation of the underlying series to estimate teh sensity the other using
the FFT. However when appli
On 09.04.2012 13:27, sagarnikam123 wrote:
i have attached file,
http://r.789695.n4.nabble.com/file/n4542543/1BHP_A.txt 1BHP_A.txt
i just want to validate predicting results, i give less 10 digits for
prediction& compare results it with previous input data(input file)
i<-read.table(file.choos
Thanks for the heads up, Paul. That's good to know. I happen to be in
academics, as are my collaborators, but I could imagine running into
problems down the road. The good thing seems to be that it's just the
users who want to interact with R who need the software. If
collaborators are just tou
Thanks Rich,
While it doesn't tickle me the way sweave/knitr does, SWord sounds more
or less like the thing I'm looking for. However, poking around
http://rcom.univie.ac.at/ and http://www.statconn.com doesn't reveal any
download links. It seems as though SWord has been pulled from their
lineup?
On 09/04/2012 9:58 AM, crmnaw wrote:
Hi,
Thanks for your reply. In the example I gave in the original post, your code
works. But for others, it doesn't and I'm not sure why it works for some
cases and not for others. For example:
x<-c(0.04,0.07,0.20,0.35,0.55,0.70)
order(x)
[1] 1 2 3 4 5 6
Let
On 09.04.2012 16:08, Guaramy wrote:
Tanks a lot to all of you that take the time to help me.
This is a really useful and helpful forum and i will try to help as much as
i can
Thanks for offering help to the R-help *mailing list*.
Please try to use the mailing list interface (rather than Nabb
Tanks a lot to all of you that take the time to help me.
This is a really useful and helpful forum and i will try to help as much as
i can
Best Regards Jorge
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I'm generating several scatter3d visualizations of vegetation community
distribution against a number parameters using the scatter3d function car
package,
The visualizations are fantastic and they provide great support for the
analysis. As far as I can determine, scatter3d does not save the outpu
Hi,
And regarding how to extend the nls algorithm to a larger dataset it is a
question of indicating in the nls() functions which data.frame to use.
In you example you were using a small set of x's ad y's, so if you want to
use a large set put it in a new data.frame and pass it to nls().
And if y
Hi,
Thanks for your reply. In the example I gave in the original post, your code
works. But for others, it doesn't and I'm not sure why it works for some
cases and not for others. For example:
x<-c(0.04,0.07,0.20,0.35,0.55,0.70)
order(x)
[1] 1 2 3 4 5 6
Let's say I want y to be in the order 1-2
On 09/04/2012 9:38 AM, crmnaw wrote:
Hi,
I'm trying to create a vector (or matrix row) with a specific ordering. For
example, I have the following vector:
x<-c(0.1,0.2,0.3,0.4,0.5,0.6)
that has order
order(x)
[1] 1 2 3 4 5 6
I want another vector that has the same values as x, but with a dif
Hi Berend,
I now understand what you were explaining. It works great.
Thanks a lot for your time and help. I really appreciate it.
Navin Goyal
On Mon, Apr 9, 2012 at 1:26 AM, Berend Hasselman wrote:
>
> On 09-04-2012, at 01:26, Navin Goyal wrote:
>
> > Hi,
> > I am not sure I follow the scala
You should look up what a t-test is.
Michael
On Mon, Apr 9, 2012 at 2:58 AM, ali_protocol
wrote:
> I am interested in the difference of 2 data:
>
> mat1= c(2.2, 2.3, 2.2,2.5)
> mat2= c(2.6, 2.8, 2.7,2.4)
>
> mat= mat2-mat1
>
> I perform an action on both mat1 and mat2, and I get
> mat1prime and
Hi,
I'm trying to create a vector (or matrix row) with a specific ordering. For
example, I have the following vector:
x<-c(0.1,0.2,0.3,0.4,0.5,0.6)
that has order
order(x)
[1] 1 2 3 4 5 6
I want another vector that has the same values as x, but with a different
ordering. For example, I want
On Apr 9, 2012, at 14:54 , Guaramy wrote:
> I read it believe me . The reason that a i post this is because i am making a
> thesis and a i am having this problem for over 2 weeks.
> I can´t solve it and its causing me real problems.
Well, if you have a function that is not vectorized, i.e. it wo
On 09-04-2012, at 14:54, Guaramy wrote:
> I read it believe me . The reason that a i post this is because i am making a
> thesis and a i am having this problem for over 2 weeks.
> I can´t solve it and its causing me real problems.
You didn't give us a reproducible example.
I'm assuming your ori
Your formula could be simplified to
y ~ 1 + a * exp(-b * x)
Solve this equation for x
x = ln[(y - 1)/a]/b
Use this equation to find the intersection point at a given value of y.
For example, when
y = 1.01
x = ln(0.01/a)/b
Jean
Karen Vandepoel wrote on 04/09/2012 0
Hi,
Yes, please check package "tables".
Regards,
Carlos Ortega
www.qualityexcellence.es
2012/4/9 bobo
> Could anyone please direct me on how to make a nicer table in R? THANKS FOR
> ALL THE HELP!
>
> I would like to make a table with the following in it: estimate, t value,
> significance, beta
I read it believe me . The reason that a i post this is because i am making a
thesis and a i am having this problem for over 2 weeks.
I can´t solve it and its causing me real problems.
Sorry for any inconvenience
Thanks
--
View this message in context:
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Hi,
I will try to explain what it is I need to do, how far I am in doing it yet
and where my problem is:
I have a lot of x,y values I need to fit a non linear function through.
Subsequently, I need to find the intersection point of this fitted curve
with y=1.01
The problem is I have a lot of val
On 2012-04-09 00:44, Hien Nguyen wrote:
Dear R-helpers,
I am trying to do a stepwise procedure in which I want to force some
variables in the model. I have searched around and it seems that only
leaps package allows to force the variable in the stepwise procedure. I
use the leaps package and use
If you're considering SWord, you should remember that the licence is
not the normal R licence and in commercial use will require a
commercial licence. While some academic disciplines use Word etc, the
issue may be more common outside academia.
For those of us where such requirements involve a proc
On Apr 8, 2012, at 11:48 PM, Guaramy wrote:
Sorry but i didn't understand.
The path to understanding is through study and application of the
advice in the Posting Guide.
--
David Winsemius, MD
West Hartford, CT
__
R-help@r-project.org mailing l
Sir,
I am a student in biostat and bioinformatics. I am interested to use R for my
research work related to codon usage analysis.This software was used in a
publication entitled " Online synonymous codon usage analyses with the ade4 and
seqinR packages" and a weblink
http://pbil.univ-lyon1
You might want to consider SWord, which provides similar facilities for the
Word and R
user. Word-oriented co-authors can modify the Word part of the document
without
impacting the R part of the document.
SWord is by Thomas Baier tho...@statconn.com, author of the statconnDCOM
interface
that is u
i have attached file,
http://r.789695.n4.nabble.com/file/n4542543/1BHP_A.txt 1BHP_A.txt
i just want to validate predicting results, i give less 10 digits for
prediction & compare results it with previous input data(input file)
> i<-read.table(file.choose()) #attached file taking
> i<-i$V1
> leng
Hello everybody!
I am trying to apply qcc function to time series data.
There is an example from documentation:
###
library(qcc)
data(pistonrings)
attach(pistonrings)
diameter <- qcc.groups(diameter, sample)
qcc(diameter[1:25,], type="xbar")
detach(pistonrings)
###
So, we have 5 iteams at each
Could anyone please direct me on how to make a nicer table in R? THANKS FOR
ALL THE HELP!
I would like to make a table with the following in it: estimate, t value,
significance, beta, standard errors, adjusted r squared, and residual
standard error (3 decimal points if possible, but I can do it by
Hello,
I will try to explain the problem, sorry if it will be a little long...
I'm using R to analyze results of cyclic mechanical testing, like this:
- apply quasi-sinusoidal load
- measure quasi-sinusoidal vertical and horizontal deformations
(quasi-sinusoidal load means that load "should be"
Good Afternoon,
I get it, my question was really how the data were organized, I thought
I was doing something wrong.
Thanks
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___
On 04/07/2012 05:00 AM, r-help-requ...@r-project.org wrote:
It is possible to calculate the c-index for time dependent outcomes (such as
disease) using the survivalROC package in R. My question is : is it possible to
produce a p-value for the c-index that is calculated (at a specific point in
Dear R-helpers,
I am trying to do a stepwise procedure in which I want to force some
variables in the model. I have searched around and it seems that only
leaps package allows to force the variable in the stepwise procedure. I
use the leaps package and use the regsubsets(lm1, force.in = 1, dat
PS: The following shows possibilities that are available using latticeExtra
layering:
## Best make type and attend factors
xdat = data.frame(mortality =c(5,
8,7,5,8,10,11,6,4,5,20,25,27,30,35,32,28,21,20,34,11,15,18,12,15,12,10,15,19,20),
type=
factor(c(1, 1,1,1,1,1,1,1,1,1,2,2,2,2,2,2,2,2,2,2,
On Mon, Apr 9, 2012 at 7:29 AM, Susanna Makela
wrote:
> Hello R users,
>
> I would like to generate "slanted" stacked bar graphs like those on
> the bottom of pages 1 and 2 in this document:
> http://www.wssinfo.org/fileadmin/user_upload/resources/JMP-Snapshot-SWA-HLM.pdf
> . I've also attached th
I am interested in the difference of 2 data:
mat1= c(2.2, 2.3, 2.2,2.5)
mat2= c(2.6, 2.8, 2.7,2.4)
mat= mat2-mat1
I perform an action on both mat1 and mat2, and I get
mat1prime and mat2prime:
mat1prime= c(2.5, 2.5, 2.3,2.5)
mat2prime= c(2.6, 2.8, 2.7,2.6)
matprime= mat2prime-mat1prime
I want
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