Thank you very much for the help. I decided to go with the 'plyr' package but
it is nice to have options.
Have a lovely day.
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Dears Dejian and David:
Thank you for your help.
Maybe dataframes2xls package have the same problem.
But "xlsx" take too much time to write it down.
Kenneth.
El lun, 06-09-2010 a las 20:56 -0400, David Winsemius escribió:
> On Sep 6, 2010, at 8:09 PM, Dejian Zhao wrote:
>
> > The maximum numb
By the way, ordinal regression would require huge datasets because my dependent
variable has around 20 different responses... but again, one might say that
with so many ordinal responses, it is as if we have a linear/interval
variable, right? I just hoped that there would be a two-way kruskal-w
On Tue, Sep 7, 2010 at 1:41 AM, Stephen Liu wrote:
> Hi Johnathan,
>
> Tried,
> Ctrl-x R Enter
>
> The same, no response.
No, not ctrl-x r. The previous poster told you to do "Alt-x R" (where
Alt might be "option" on your keyboard). In emacs-speak, this modifier
key (alt/opt) is written as "M" ..
thanks for the response
we are talking about 7 cities. If I run a two-way anova, I find the residuals
skewed and non-normal. I'll try the rlm method and see what happens. Thanks to
all of you for the support.
Dr. Iasonas Lamprianou
Assistant Professor (Educational Research and Evaluation)
Hi,
On Tue, Sep 7, 2010 at 12:39 AM, Stephen Liu wrote:
> Hi Steve,
>
> Thanks for your advice.
>
>> Actually, for that to work, the path for your R executable needs to be
>> in emacs' exec-path:
> http://www.emacswiki.org/emacs/ExecPath
>
> appending the /sw/bin directory to the exec-path and PA
Hi Johnathan,
Tried,
Ctrl-x R Enter
The same, no response.
B.R.
Stephen L
Sent from my iPhone
On Sep 7, 2010, at 12:53 PM, Jonathan Christensen wrote:
Hi Stephen,
Just to check: when you say you type "M-x R", are you typing the letter "M"?
M-x in Emacs-speech means Meta-x, i.e., Alt-x.
Jo
Scott, it seems to have bug in your code:
> A=matrix(c(1,-2,3,9,-1,3,0,-4,2,-5,5,17),ncol=4,byrow=T)
> ref(A)
[,1] [,2] [,3] [,4]
[1,]1 -2.5 2.17 7.83
[2,]0 1.0 3.00 5.00
[3,]0 0.0 1.00 2.00
the row echelon is apparently incorrect.
--
View this
sorry: start=rep(1,6) since there are 6 parameters in the model.
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_
Since the default initial value is not good enough. You should choose one
based on your experience or luck. I choose start=rep(1,5) since there are
parameters in the model.
> polr(Species~Sepal.Length+Sepal.Width+Petal.Length+Petal.Width,iris,
> start=rep(1,6), method = "logistic")
Call:
polr(fo
Hi Stephen,
Just to check: when you say you type "M-x R", are you typing the letter "M"?
M-x in Emacs-speech means Meta-x, i.e., Alt-x.
Jonathan
On Mon, Sep 6, 2010 at 7:01 PM, Stephen Liu wrote:
> Hi Dirk,
>
> Thanks for your advice.
>
>
> Emacs and ESS already installed.
>
> $ apt-cache pol
Hi Steve,
Thanks for your advice.
> Actually, for that to work, the path for your R executable needs to be
> in emacs' exec-path:
http://www.emacswiki.org/emacs/ExecPath
appending the /sw/bin directory to the exec-path and PATH variables (useful for
Mac OS X users running LaTeX):
(setenv "PATH"
Hi,
On Mon, Sep 6, 2010 at 9:01 PM, Stephen Liu wrote:
> $ apt-cache policy ess
> ess:
> Installed: 5.3.8~svn3917-1
> Candidate: 5.3.8~svn3917-1
> Version table:
> *** 5.3.8~svn3917-1 0
> 500 http://ftp.hk.debian.org lenny/main Packages
> 100 /var/lib/dpkg/status
I believe th
On Sep 6, 2010, at 7:54 PM, Brant Inman wrote:
R-helpers,
I am using the package "drc" to fit a 4 parameter logistic model.
When I
use the predict function to get prediction on a new dataset, I am not
getting the requested confidence or prediction intervals. Any idea
what
is going on?
On Sep 6, 2010, at 9:22 PM, tooblue wrote:
I simply put,
NEVER=subset(infants$bwt,ISNO1)
UNTILPREGNANT=subset(infants$bwt, ISNO2)
ONCENOTNOW=subset(infants$bwt, ISNO3)
and I wanna combine those three.
I do it like
ISNO=NEVER&UNTILPREGNANT&ONCENOTNOW
The "&" operator does not do concatenat
I simply put,
> NEVER=subset(infants$bwt,ISNO1)
> UNTILPREGNANT=subset(infants$bwt, ISNO2)
> ONCENOTNOW=subset(infants$bwt, ISNO3)
and I wanna combine those three.
I do it like
ISNO=NEVER&UNTILPREGNANT&ONCENOTNOW
and R tells me
1: In NEVER & UNTILPREGNANT :
longer object length is not a mu
Hi:
See inline.
On Mon, Sep 6, 2010 at 9:29 AM, Viki S wrote:
>
> Hi Jim,
> Thanks,
>
> Thatæ right. But the problem is that it
> introduces unnecessary quotes, perhaps due to the format of first
> column data in this case :
>
> x<-cbind(c("row:1", "row:2", "row:3"), c("4889", "9987", "494"))
Maybe Friedman test
On Mon, Sep 6, 2010 at 4:47 PM, David Winsemius wrote:
> The usual least-squares methods are fairly robust to departures from
> normality. Furthermore, it is the residuals that are assumed to be normally
> distributed (not the marginal distributions that you are probably looki
Hello, I am brand new to R and need a little help understanding what I'm
doing wrong here! Please excuse me, I might have the terminology incorrect.
What I've got is: A dataset ami07.data with a variable ami_fy07. There are
2929 values in this dataset, each representing a value for a unique
hospi
The quotes are not unnecessary since 'row:1' is not a valid name and
therefore you will need quotes to refer to the data:
> x<-cbind(c("row:1", "row:2", "row:3"), c("4889", "9987", "494"))
> x1<-as.list(x[,2])
> names(x1)<-x[,1]
> x1
$`row:1`
[1] "4889"
$`row:2`
[1] "9987"
$`row:3`
[1] "494"
>
Hi Dirk,
Thanks for your advice.
Emacs and ESS already installed.
$ apt-cache policy emacs
emacs:
Installed: 22.2+2-5
Candidate: 22.2+2-5
Version table:
23.1+1-4~bpo50+1 0
1 http://backports.org lenny-backports/main Packages
*** 22.2+2-5 0
500 http://ftp.hk.debian.
On Sep 6, 2010, at 8:09 PM, Dejian Zhao wrote:
The maximum number of rows in excel 2003 or below is 65535, less
than your number of rows, so if you export your data into "xls"
files, probably you cannot see all your data in excel. Exel 2007 can
hold as many as 1048575 lines, thus "xlsx" fi
The maximum number of rows in excel 2003 or below is 65535, less than
your number of rows, so if you export your data into "xls" files,
probably you cannot see all your data in excel. Exel 2007 can hold as
many as 1048575 lines, thus "xlsx" file is a better choice.
On 2010-9-7 0:03, Kenneth
R-helpers,
I am using the package "drc" to fit a 4 parameter logistic model. When I
use the predict function to get prediction on a new dataset, I am not
getting the requested confidence or prediction intervals. Any idea what
is going on? Here is code to reproduce the problem:
---
library(
Hi Folks:
I found the following paper to be an interesting example of how even
well designed and conducted studies -- randomized trials, even -- can
be rendered problematic by systematic effects beyond the control of
the investigators:
http://www.nature.com/nrurol/journal/v7/n9/full/nrurol.2010.1
The design is a repeated measures with 3 instances. There are 3 groups:
Controls, Heavy Cocaine Users, Light Cocaine Users.
I reshaped the data so that there was one variable for the 3 instances
called AvgTrials. Time is the indicator of each instance.
Here is the model call:
mod5 <- lme(AvgT
Try this (after making sure that Col_1 in data2 matches your column
names in data1
> data1 <- read.table(textConnection("Taxon stage1 stage2 stage3 stage4
+ T1 0 0 1 1
+ T2 0 1 1 0
+ T3 0 0 0
We would need to at least see 10 lines either side of the error to
understand its context. Also take a look with you editor (hex editor
would be handy) to see if there is some unprintable character around
it. Can you isolate just that portion of the code? Try putting it
inside a function to see
On Sat, 4 Sep 2010, raje...@cse.iitm.ac.in wrote:
Hi,
I have the following piece of code,
repeat{
ss<-read.socket(sockfd);
if(ss=="") break
output<-paste(output,ss)
}
but somehow, output is not receiving all the data that is coming through the
socket.My suspicion is on the if statement. wh
On 09/06/2010 09:41 PM, raje...@cse.iitm.ac.in wrote:
Hi,
I have a list which looks like this...
str(y)
List of 10
$ : chr [1:4] "ABCD" "5" "0" "1"
$ : chr [1:4] "DEF" "15" "1" "16"
$ : chr [1:4] "AAA" "2" "17" "8"
$ : chr [1:4] "SSS" "15" "25" "1"
$ : chr [1:4] "III" "15" "26" "4"
I am using the standard phonetic font "Doulos SIL" in a graph
(http://scripts.sil.org/cms/scripts/page.php?site_id=nrsi&id=DoulosSILfont)
This is an example:
windowsFonts(IPA="TT Doulos SIL")
barplot(c(1,2,3,4,5),names=c("\u{0251}","\u{0252}","\u{0253}","\u{0254}","\u{0255}"),family="IPA")
Howe
The usual least-squares methods are fairly robust to departures from
normality. Furthermore, it is the residuals that are assumed to be
normally distributed (not the marginal distributions that you are
probably looking at) , so it does not sound as though you have yet
examined the data prop
Hi
I have got a long script which will not run for me as i keep getting errors :
> source("clusterfixV1_4.r")
Error in source("clusterfixV1_4.r") :
clusterfixV1_4.r: unexpected symbol at
158: eck[k,2] <- as.numeric(1)
159: #ClusterInfo[k,2] <- "Clustered
I have sorted all the
Dear friends, two questions
(1) does anyone know if there are any non-parametric equivalents of the two-way
ANOVA in R? I have an ordinal non-normally distributed dependent variable and
two factors (gender and city of birth). Normally, one would try a two-way
anova, but if R has any non-parame
Dear community,
I am currently trying to fit an ordinal logistic regression model with the
polr function. I often get the same error message :
"attempt to find suitable starting values failed", for example with :
require(MASS)
data(iris)
polr(Species~Sepal.Length+Sepal.Width+Petal.Length+Peta
On Sep 6, 2010, at 2:07 PM, tooblue wrote:
I simply put, plot(density(), main="", + xlab = "XXX"), it
says that
I have an unexpected "=" in it.
It may be a case of a confused parser. You have an extraneous "+" in
there:
> = rnorm(100)
> plot(density(), main="",
On Mon, Sep 6, 2010 at 11:07 AM, tooblue wrote:
>
> I simply put, plot(density(), main="", + xlab = "XXX"), it says that
> I have an unexpected "=" in it.
You just have an extra ' + ' before the xlab argument:
plot(density(rnorm(100)), main = "", xlab = "XXX")
ought to do it.
Chee
On Mon, Sep 6, 2010 at 12:15 PM, Dimitri Shvorob
wrote:
>
> I have a (very big - 1.5 rows) dataframe with a (POSIXt" "POSIXlt") column h
> (hour). Surprisingly, I cannot calculate a simple aggregate over the
> dataframe.
>
>> n.h1 = sqldf("select distinct h, count(*) from x group by h")
> Error i
I simply put, plot(density(), main="", + xlab = "XXX"), it says that
I have an unexpected "=" in it.
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__
Hi
This question is far less simple than the title suggests, please read
carefully, thanks.
I have 2 sets of data, both read into R
>data1<-read.table ("1.txt", header=T, sep="\t")
>data2<-read.table ("2.txt", header=T, sep="\t")
>data1
Taxon stage1 stage2 stage3 stage4
T1 0
On Mon, Sep 6, 2010 at 11:56 AM, trb1 wrote:
>
> Thank you very much for your post.
> Your answer has been very helpful.
> Is it possible to merge >2 time series?
>
zz is my posted code was formed by merging two univariate and one
multivariate series.
--
Statistics & Software Consulting
GKX Gro
Hi,
Looking at the picture, I think you are just talking about plotting
two datasets. Here is an example I made up, that looks sort of like
your picture:
# make a barplot
barplot(-50:50)
# add points into the existing plot at the coordinates set by x and y
# and use a line to connect them
points
On Sep 6, 2010, at 12:15 PM, Dimitri Shvorob wrote:
I have a (very big - 1.5 rows) dataframe with a (POSIXt" "POSIXlt")
column h
(hour). Surprisingly, I cannot calculate a simple aggregate over the
dataframe.
n.h1 = sqldf("select distinct h, count(*) from x group by h")
Error in sqliteE
Dear useRs,
the deadline for submission of abstracts is approaching for ERCIM'10.
Please upload your abstract until 2010-09-08 if you would like to give
a presentation at our track on "Statistical Algorithms and Software" at the
3rd International Conference of the ERCIM WG on
COMPUTING & STA
Successful!
Thank you!
2010/9/6 David Winsemius
>
> On Sep 6, 2010, at 1:13 PM, tamas barjak wrote:
>
> Hello!
>>
>> I need some help.
>> How I know it to draw the formula of the poisson distribution?
>>
>> expr<-expression(P(xi == k) == frac(lambda^k, factorial(k))*e^-lambda)
>> --->
>> not
On Sep 6, 2010, at 1:13 PM, tamas barjak wrote:
Hello!
I need some help.
How I know it to draw the formula of the poisson distribution?
expr<-expression(P(xi == k) == frac(lambda^k, factorial(k))*e^-
lambda) --->
not good
?plotmath
(Do not see factorial as a plotmath "function"
Try:
ex
Le 06/09/10 19:17, rajesh j a écrit :
> Is there a c++ equivalent switch statement in R?
yes, with the same name. See the "R Language definition" :
http://cran.r-project.org/doc/manuals/R-lang.html#switch
If this is not what you want, maybe you could at least share some C++ code
snippet of what
Hi,
I am not sure if this make sense at all. I'd like to sample a matrix, which
follows a wishart / inverted wishart distribution. However, the (1,1)
element of this matrix should always be equal to 1. How can I handle it in
R? Any suggestion is greatly appreciated. Thanks a lot.
Sonia
[
Hi Everyone,
I have two different data set in 2 different scale.
I want to plot these two data in the same plot
in their respective scale. So the plot will have 2 different scale.
I have added an image below to show how it should look.
does any bode has any idea how this can be done.
2 differen
Is there a c++ equivalent switch statement in R?
--
Rajesh.J
[[alternative HTML version deleted]]
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/posti
Hi Jim,
Thanks,
That´s right. But the problem is that it
introduces unnecessary quotes, perhaps due to the format of first
column data in this case :
x<-cbind(c("row:1", "row:2", "row:3"), c("4889", "9987", "494"))
x1<-as.list(x[,2])
names(x1)<-x[,1]
> x1
$`row:1`
[1] "4889"
$`row:2`
[1] "9987"
Hi,
I am not sure if this make sense at all. I'd like to sample a matrix, which
follows a wishart / inverted wishart distribution. However, the (1,1)
element of this matrix should always be equal to 1. How can I handle it in
R? Any suggestion is greatly appreciated. Thanks a lot.
Sonia
[
I have a (very big - 1.5 rows) dataframe with a (POSIXt" "POSIXlt") column h
(hour). Surprisingly, I cannot calculate a simple aggregate over the
dataframe.
> n.h1 = sqldf("select distinct h, count(*) from x group by h")
Error in sqliteExecStatement(con, statement, bind.data) :
RS-DBI driver:
Hello!
I need some help.
How I know it to draw the formula of the poisson distribution?
expr<-expression(P(xi == k) == frac(lambda^k, factorial(k))*e^-lambda) --->
not good
on the screen the " k! " not the Poisson Formula, but "factorial(k)"
Thanx!
[[alternative HTML version deleted]]
[forwarding back to r-help for archiving/further discussion]
On 10-09-05 08:48 PM, Sally Luo wrote:
> Prof. Bolker,
>
> Thanks for your reply and the helpful info.
>
> I still have a few questions.
>
> 1. I also tried to use different methods other than "BFGS" without
> changing their def
I use the following sintaxis for the packages:
For WriteXLS I use:
writeXLS(todo2009,"todo2009.xls")
And for dataframes2xls I use:
dataframe2xls::write.xls(todo2009,"todo2009.xls")
El lun, 06-09-2010 a las 12:34 -0400, David Winsemius escribió:
> On Sep 6, 2010, at 12:25 PM, Kenneth Roy Cabre
On Sep 6, 2010, at 12:25 PM, Kenneth Roy Cabrera Torres wrote:
Thank you Ivan for you answer:
El lun, 06-09-2010 a las 18:11 +0200, Ivan Calandra escribió:
Hi,
Are you sure you used the correct syntax and object names? It might
just
be because of that...(reading the error messages)
Im sur
On 6 September 2010 at 09:18, Stephen Liu wrote:
| Hi folks,
|
| Debian 504 64-bit
Good. All you need is
sudo apt-get install ess
| I found following document;
| http://www.biostat.wisc.edu/~kbroman/Rintro/
|
| Whether it is the right document for installing Emacs+ESS and R so that R can
Thank you Ivan for you answer:
El lun, 06-09-2010 a las 18:11 +0200, Ivan Calandra escribió:
> Hi,
>
> Are you sure you used the correct syntax and object names? It might just
> be because of that...(reading the error messages)
Im sure, because it works with write.csv or write.table.
> There is a
Hi folks,
Debian 504 64-bit
I found following document;
http://www.biostat.wisc.edu/~kbroman/Rintro/
Whether it is the right document for installing Emacs+ESS and R so that R can
run on Emacs?
TIA
B.R.
Stephen L
__
R-help@r-project.org mailing l
Hi Dave,
You can look at the function ?bxp it might work for you. Alternately,
create a meaningless boxplot object, and then just edit that data, in
which case I know it will work with bxp().
# Create a boxplot, the data does not matter
x <- boxplot(1:10)
x # view the data for the boxplot
x$sta
There are lots of options for path analysis in R.
If you go to http://www.rseek.org and type path analysis into the search box,
you will get lots of information on functions/packages, and more general
info as well.
Beyond that, we'd need more specifics about your task.
Sarah
On Mon, Sep 6, 2010
Hi,
Are you sure you used the correct syntax and object names? It might just
be because of that...(reading the error messages)
There is another function, xlsReadWrite::write.xls(), that I like a lot:
it is really easy to use and does not require Perl or Python.
HTH,
Ivan
Le 9/6/2010 18:03, K
Hi R users:
I don't know if you have had the following problem trying to
export to an "xls" format file in a non windows platform.
I try to use the following packages:
1. dataframes2xls (version 0.4.4) (with phyton 2.7 and 3.1)
2. WriteXLS (version 1.9.0) (with perl and testPerl working)
Even "x
Also, when I create the data.frame with matrix and try to rbind, I get
warnings..
Warning messages:
1: In `[<-.factor`(`*tmp*`, ri, value = "4") :
invalid factor level, NAs generated
2: In `[<-.factor`(`*tmp*`, ri, value = "5") :
invalid factor level, NAs generated
3: In `[<-.factor`(`*tmp*`, r
Thank you very much for your post.
Your answer has been very helpful.
Is it possible to merge >2 time series?
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But If I do that how will I resize later?
On Mon, Sep 6, 2010 at 8:54 PM, ONKELINX, Thierry
wrote:
> This will give a matrix with 0 rows.
>
> data.frame(matrix(nrow = 0, ncol = 22, dimnames = list(NULL,
> LETTERS[1:22])))
>
> But you should avoid growing dataframes is the final dataframe is goin
Dear list,
I am using a external program that outputs Q1, Q3, median, upper and lower
whisker values for various datasets simultaneously in a tab delimited format.
After importing this text file into R, I would like to plot a boxplot using
these given values and not the original series of data
Hi.
which package i need to install to be able to run "Path analysis" using r?
many thanks, Guy
--
Guy Rotem
Department of Life Sciences
The Spatial Ecology Lab
Ben Gurion University of the Negev
P.O.B. 653 Beer-Sheva 84105
ISRAEL
+972-52-3354485 (mobile)
+972-8-6461350 (lab)
[[alte
that was my first idea as well, but as the result shows, the minimized function
value of the wider interval is greater.
In
addidtion, the problem also exists, if the minimized parameter in the
case of the larger interval also already lies within the smaller
interval:
f<-function(delta,P,U){
This will give a matrix with 0 rows.
data.frame(matrix(nrow = 0, ncol = 22, dimnames = list(NULL,
LETTERS[1:22])))
But you should avoid growing dataframes is the final dataframe is going
to be large. You are very likely to get memory problems. It is much to
better to create a large enough datafr
On Sep 6, 2010, at 10:47 AM, Dimitris Rizopoulos wrote:
one way is the following:
M <- cbind(c(1,1,1,1,2,2,3,3,3,3), c(2,2,2,3,4,4,4,5,5,6),
c(1,2,3,4,5,6,7,8,9,10))
ind <- do.call(paste, c(as.data.frame(M[, 1:2], sep = "\r")))
M[, 3] <- ave(M[, 3], ind, FUN = "sum")
unique(M)
I had been
Hi again!
I'm trying to follow your general goal from your questions today but
it's not easy.
First, declaring a data.frame of 0 rows is a bad idea. It is much faster
to define the length and number of rows from the beginning and to fill
it then.
Second, I don't know how to do it! What I kn
On Mon, Sep 6, 2010 at 10:24 AM, trb1 wrote:
>
> Hi
>
> How would I analyse time series with
> - different lengths (i.e. one has 9 entries and the other has 14 entries)
> - different frequency (i.e. dates are random - no repeated length)
> - multiple values for the same time entry (e.g. 2009-10-23
Hi,
first off, I wanna ask how do I declare a data.frame of 0 rows and n
columns?
Coming to my problem,
I have a data.frame of 22 columns by dynamic rows which I insert using
rbind. The total number of rows could go upto 2,00,000. The problem is that
after about 800 or 900 get inserted rbind sta
On 06/09/2010 9:19 AM, Juliet Ndukum wrote:
I am building a package say mypkg. Five months ago, when I built the package I
got the mypkg manual in pdf format.
Today, after making updates, I build the same package, same name, and steps;
unfortunately I do not get the manual in pdf format.
Rath
Hello everyone.
I would kindly request your help concerning how R converts data between
different structrures.
In the following example please keep attention on the following two
1)
I create
f <- GaussRF(x=x, y=y, model=model, grid=TRUE,param=c(mean, variance, nugget,
scale, alpha))
with
ima
On Mon, Sep 6, 2010 at 3:29 PM, Kennedy wrote:
> I want to reduce the matrix according to the following: If the values of the
> two first columns are the same in two or more rows the values in the third
> column of the corresponding rows should be added and only one of the rows
> should be keept.
one way is the following:
M <- cbind(c(1,1,1,1,2,2,3,3,3,3), c(2,2,2,3,4,4,4,5,5,6),
c(1,2,3,4,5,6,7,8,9,10))
ind <- do.call(paste, c(as.data.frame(M[, 1:2], sep = "\r")))
M[, 3] <- ave(M[, 3], ind, FUN = "sum")
unique(M)
I hope it helps.
Best,
Dimitris
On 9/6/2010 4:29 PM, Kennedy wrot
Hi everyone. I have these data:
probClass<-seq(0,0.9,0.1)
prob1<-c(0.0070,0.0911,0.1973,0.2949,0.3936,0.5030,0.5985,0.6869,0.7820,0.8822)
prob2<-c(0.0066,0.0791,0.2358,0.3478,0.3714,0.3860,0.6667,0.6400,0.7000,1.)
# which I'm plotting as follows:
plot(probClass,prob1,xlim=c(0,1),ylim=c(0,1),
Hi,
I have a matrix that looks like this
a <- c(1,1,1,1,2,2,3,3,3,3)
b <- c(2,2,2,3,4,4,4,5,5,6)
c <- c(1,2,3,4,5,6,7,8,9,10)
M <- matrix(nr=10,nc=3)
M[,1] <- a
M[,2] <- b
M[,3] <- c
> M
[,1] [,2] [,3]
[1,]121
[2,]122
[3,]123
[4,]
On Sep 5, 2010, at 10:32 AM, Daniele Sluijters wrote:
Hello,
I'm sorry to just pop-up on the mailing list like this and ask a
relatively non-R related question but I had no idea whom else to
contact on this matter.
I'm working on a completely different port of an application to OS X
whic
Gabor, David, thank you.
David, your last suggestion is what I need.
Regards,
Sergey
On Mon, Sep 6, 2010 at 16:12, David Winsemius wrote:
>
> On Sep 6, 2010, at 9:56 AM, Sergey Goriatchev wrote:
>
>> Hello everyone.
>>
>> Say we have the following:
>>
>> a <- matrix(c(-75, 3, 5, 9, 2, 3, 5), nr
Hi
How would I analyse time series with
- different lengths (i.e. one has 9 entries and the other has 14 entries)
- different frequency (i.e. dates are random - no repeated length)
- multiple values for the same time entry (e.g. 2009-10-23 below)
i.e. my data takes the form:
1st time series
2
On Sep 6, 2010, at 9:56 AM, Sergey Goriatchev wrote:
Hello everyone.
Say we have the following:
a <- matrix(c(-75, 3, 5, 9, 2, 3, 5), nrow=1, dim=list("06092010",
c("ES", "PT", "Z ", "CF", "GX", "ST", "EO")))
b <- matrix(c(-5, 2, 4, 12, 5), nrow=1, dim=list("06092010", c("PT",
"CF", "AT", "EM
On Mon, Sep 6, 2010 at 9:56 AM, Sergey Goriatchev wrote:
> Hello everyone.
>
> Say we have the following:
>
> a <- matrix(c(-75, 3, 5, 9, 2, 3, 5), nrow=1, dim=list("06092010",
> c("ES", "PT", "Z ", "CF", "GX", "ST", "EO")))
> b <- matrix(c(-5, 2, 4, 12, 5), nrow=1, dim=list("06092010", c("PT",
>
Dear all,
Thanks very much for the replies and for the help.
This whole data set consists of about 7000 individuals who have had
multiple blood pressure measures taken over time so I just used one
individual as an example. I'm sorry if it looked like homework...it isn't.
Jim your solution wo
Hello everyone.
Say we have the following:
a <- matrix(c(-75, 3, 5, 9, 2, 3, 5), nrow=1, dim=list("06092010",
c("ES", "PT", "Z ", "CF", "GX", "ST", "EO")))
b <- matrix(c(-5, 2, 4, 12, 5), nrow=1, dim=list("06092010", c("PT",
"CF", "AT", "EM", "ST")))
d <- cbind(a, b)
I want to calculate sums of
Dear Edward,
You have no degrees of freedom left to estimate those p-values. Your
design does not allows for the model your implemented. We need a brief
summary of your design in order to help you further.
HTH,
Thierry
---
Hi Paul, Ivan,
Hartstikke bedankt and thanks alot for sharing these thoughts. I can see
'listing up' multiple symmetrical data sets makes a lot of sense. As
does using lapply() on them which i understand to be more
efficient/faster than for().
Goodo- with your concensus (and helpful examples
Hi Adrian,
dat=data.frame(matrix(0,3,3))
write.xlsx(dat,"z:/dat.xlsx",sheetName="sheet1",append=F)
write.xlsx(dat,"z:/dat.xlsx",sheetName="sheet2",append=F)
The above code works and creates new worksheets. But if I want to append
to an existing worksheet I seem to get an error.
wri
Everyone -
What do the NaN's mean here? Is this analysis a problem?
Linear mixed-effects model fit by maximum likelihood
Data: tmp.dat
AIC BIClogLik
1611.251 1638.363 -797.6253
Random effects:
Formula: ~1 | group_id
(Intercept) Residual
StdDev: 0.0003077668 9.23671
Thank you, David:
I obviously didn't look hard enough. This is exactly what I need.
Charles Annis, P.E.
charles.an...@statisticalengineering.com
561-352-9699
http://www.StatisticalEngineering.com
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org
On Sep 6, 2010, at 4:03 AM, B W wrote:
->Hello,I am trying to take the information from the summary of my
best fit logisticregression model for the occurrence of a high
elevation plant spp. and create the appropriate equation that will
calculate probability of occurrence, given the data.
Hi again,
see ?rbind
Ivan
Le 9/6/2010 14:11, raje...@cse.iitm.ac.in a écrit :
> Hi,
>
> is it possible to insert a vector as a row in a data.frame?
> [[alternative HTML version deleted]]
>
> __
> R-help@r-project.org mailing list
> https://stat
It is easy to store a list of that size:
> x <- list(1:1e6, 1:1e6, 1:1e6)
> object.size(x)
12000112 bytes
> str(x)
List of 3
$ : int [1:100] 1 2 3 4 5 6 7 8 9 10 ...
$ : int [1:100] 1 2 3 4 5 6 7 8 9 10 ...
$ : int [1:100] 1 2 3 4 5 6 7 8 9 10 ...
Now it really depends on how you a
Hi!
I'm sure there's an easier way, but that works for me:
test_list <- list(c("ABC","5","0"), c("DEF","10","1")) ##just a part of
your example, think about using dput() to create a copy/pastable example
test_df <- t(as.data.frame(test_list)[-1,])
rownames(test_df) <- t(as.data.frame(test_list
try this:
> x <- "|1,ab,2.34|2,cd,3.44|"
> # split by the "|" and remove vectors of zero characters
> x.sp <- strsplit(x, '|', fixed = TRUE)[[1]]
> x.sp <- x.sp[nchar(x.sp) > 0]
> # now split by comma
> x.comma <- strsplit(x.sp, ',')
> # you can now access you data
> x.comma
[[1]]
[1] "1""ab"
I am building a package say mypkg. Five months ago, when I built the package I
got the mypkg manual in pdf format.
Today, after making updates, I build the same package, same name, and steps;
unfortunately I do not get the manual in pdf format.
Rather I get the following message:
cd: can't cd
Hi, this is more related to understanding some statistics while using R; I've
see such output in a paper:
out <- glm(response~Var1+Var2+Var3..,family=binomial,data=mydata)
summary(out)
stepAIC(out)
anova(out, test='Chisq')
I understand that stepAIC is used to select the model with the lowest AIC
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