On 03/07/2010 05:23 AM, david hilton shanabrook wrote:
I am using color2D.matplot to plot a matrix about 400 by 200.
The values in the matrix are 0:5 and NA. The resulting plot is not color, but
shaded b/w. I tried to figure out how to add colors, I would like something
like c(blue, green, r
Jim Lemon wrote:
I think Rolf's guess about the provenance of the request was based
upon the carefully copied format, which Blind Freddie could see was
homework (and the requester actually admitted it!). Personally, I
never answer requests that have the due date for the assignment at the
botto
Peter Dalgaard wrote:
(Notice, BTW, that I too have taken to sending mailing lists via Gmail.
Nothing to do with hiding my identity, but I'm now working behind an
Exchange server, and those beasts won't do server-side filtering, so
webmail would be a pain if regular mail got mixed with r-help
On 03/06/2010 05:51 AM, Sharpie wrote:
TheSavageSam wrote:
I am wishing to write my own random distribution simulation function using
C programmin language(for speed) via R. I am familiar with R programming
but somewhat new to C programming. I was trying to understand "Writing R
extensions" -gu
Hi all,
Just letting you know about another challenge that might interest
someone on the list.
http://click.email.innocentive.com/?ju=fe5e1d71756c07787412&ls=fe0417717764067b711d7075&m=fef91270706102&l=fe8815797c62037d72&s=fe1e16787c640d797c1371&jb=ffcf14&t=
Predictive Data Analysis
(#923157
Dear R-users,
i have the following exmple for which i want to use ecm.mix from the
mix-package.
with da.mix after using em.mix i get the error "improper
posterior--empty cells", which is not uncommen because of 17 * 5 * 3 =
255 cells.
so the next attempt is to use the ecm.mix for the restric
Hi all,
Let's say I have a data.frame and wants to turn each of it's columns into a
factor.
My instinct would be to use as.factor with apply. But this won't work, and
result with a data.frame of characters.
I found another solution for how to achieve this, but I would also like to
understand - *WH
Hi,
I was wondering if there is a way to control
the strip size in xyplot for example using the
strip.default function. Or do I need to redraw
the strips by myself?
I would like to make
the strip slimmer and I have reduced the font
size using cex. However, I don't know how
to change the size of t
On Thu, Mar 4, 2010 at 11:47 PM, Jim Lemon wrote:
> I had a working hypothesis about this, but decided to check the data before
> replying. Looking at the ten most recent obvious pseudonyms, all were from
> free email accounts like gmail or yahoo. A few of these included all or part
> of the name
Dear friends:
I have just read an article entitled " Monitoring of nosocomial invasive
aspergillosis and early evidence of an outbreak using cumulative sum tests
(CUSUM)", which is published in "Clinical Microbiology and Infection". We have
great need to estimate the fluctuation of incidence
The basic reason because apply works with matrices - it first turns
the input into a matrix, processes each column and then returns a
matrix. See colwise in the plyr package for a function that works
column wise on a data frame, returning a data frame.
Hadley
On Sun, Mar 7, 2010 at 11:07 AM, Tal
Hello
On 3/7/10, sdzhangping wrote:
> package. Can you give me some materials about Cusum test? An example is more
> appreciated.
>
There seem to be several packages dealing with 'cusum'. Try this:
library(sos)
findFn('cusum')
Also, perform a search on Rseek. Regards
Liviu
___
hello list,
the topic is covered extensively but from none of the postings i could
conclude the correct statement for my design: a 2-level within and a 2-level
between subjects factor, both fixed, subjects as random factor.
i want to test wheter there is a within effect and if it is different fo
Thanks for the explanation (and the function) Hadley.
Tal
Contact
Details:---
Contact me: tal.gal...@gmail.com | 972-52-7275845
Read me: www.talgalili.com (Hebrew) | www.biostatistics.co.il (Hebrew) |
www.r-statistics.com (Eng
On 03/06/2010 09:35 AM, j verzani wrote:
...
Sorry about that one. On the errata page I have:
page 16, exercise 1.12 #5
This one is most easily done using c() and the sequence operator :. (Please
ignore request to use just seq and rep.)
And all over the world, perhaps even on other planets
On Sun, 7 Mar 2010, sdzhangping wrote:
Dear friends:
I have just read an article entitled " Monitoring of nosocomial invasive aspergillosis and early
evidence of an outbreak using cumulative sum tests (CUSUM)", which is published in "Clinical
Microbiology and Infection". We have great need t
What happened to the category k-means package for performing k-means
clustering on categorical variables? I expected it to become more
prominent after the Netflix Challenge recommendation engine contest
concluded, but instead it seems to have dropped from view. Where was
it and where is it now, a
Nevermind, I found it:
http://cran.at.r-project.org/web/packages/knncat/index.html
On Sat, Mar 6, 2010 at 10:43 PM, James Salsman wrote:
> What happened to the category k-means package for performing k-means
> clustering on categorical variables? I expected it to become more
> prominent after th
On Sun, Mar 7, 2010 at 4:28 AM, Jim Lemon wrote:
> On 03/06/2010 09:35 AM, j verzani wrote:
>>
>> ...
>> Sorry about that one. On the errata page I have:
>>
>> page 16, exercise 1.12 #5
>> This one is most easily done using c() and the sequence operator :.
>> (Please
>> ignore request to use j
In absence of workable code, I assume you are trying to find the mean of column
by referring to its name (which is read from another file).
try...v<-mean(Dataset[, paste(varlist[i])])
Kulwinder
> From: dwinsem...@comcast.net
> To: jw...@klaru-baycrest.on.ca
> Date: Wed, 24 Feb 2010 20:56:26 -0
Well, it takes a bit of hacking with concatenation functions followed by get,
eval, and parse,
but I've written functions/scripts which build functions like the f1,f2,...
below out of various strings.
Heck, I even (don't ask :-) ) wrote a tool to invert an array along a specified
dimension.
SVGAnnotation works fine for me under Windows. I guess Cleber was
using R < 2.10.
Regards,
Yihui
--
Yihui Xie
Phone: 515-294-6609 Web: http://yihui.name
Department of Statistics, Iowa State University
3211 Snedecor Hall, Ames, IA
2010/3/6 Uwe Ligges :
>
>
> On 06.03.2010 18:35, Cleber Borges w
http://www.harding.edu/fmccown/R/#autosdatafile
http://www.harding.edu/fmccown/R/#autosdatafile
I am tring to get a barchat by factors,
following the example in that link above.
===
x=c(145,40,40,120,180,
140,155,90,160,95,
195,150,205,110,160,
45
On 06.03.2010 21:49, rkevinbur...@charter.net wrote:
In browsing the source I see the following construct:
res<- switch(type, working = , response = r, deviance = ,
pearson = if (is.null(object$weights))
r
else r * sqrt(object$weights), partial = r)
I under
On 07.03.2010 17:42, casperyc wrote:
http://www.harding.edu/fmccown/R/#autosdatafile
http://www.harding.edu/fmccown/R/#autosdatafile
I am tring to get a barchat by factors,
following the example in that link above.
===
x=c(145,40,40,120,180,
140,155,90,160,95,
Probably a bug in the RdbiPgSQL package you are using. Please ciontact
its maintainer and send a reproducible example.
Best,
Uwe Ligges
On 06.03.2010 04:08, Peter wrote:
hi,
I have been attempting to run this script and am getting some strange
results. The script connects to a database and r
On 2010-03-07 3:12, Hadassa Brunschwig wrote:
Hi,
I was wondering if there is a way to control
the strip size in xyplot for example using the
strip.default function. Or do I need to redraw
the strips by myself?
I would like to make
the strip slimmer and I have reduced the font
size using cex. H
Try the RpgSQL package and see if you get the same problem.
On Fri, Mar 5, 2010 at 10:08 PM, Peter wrote:
> hi,
>
> I have been attempting to run this script and am getting some strange
> results. The script connects to a database and retrieves a series of tables,
> using sequential sql statement
I am trying to follow this example for multinomial logistic regression
http://www.ats.ucla.edu/stat/r/dae/mlogit.htm
However, I cannot get it to work properly.
This is the output I get, and I get an error when I try to use the mlogit
function. Any ideas as to why this happens?
> mydata <- r
Dear R-users,
I have two regular hourly time series data which were recorded in time
zone GMT+1, and now I would like to merge them together for further
analyses. Here I used zoo and merge.zoo for my purposes and everything
worked fine except the timestamp shifted 2 hours after merging which
On Mar 7, 2010, at 12:06 PM, cmc wrote:
I am trying to follow this example for multinomial logistic regression
http://www.ats.ucla.edu/stat/r/dae/mlogit.htm
However, I cannot get it to work properly.
This is the output I get, and I get an error when I try to use the
mlogit
function. Any
Without reproducible code (that means we can copy your code from your
post, paste it into our session and see the same problem that you see)
there is not much that can be said that addresses your specific
situation but in terms of general advice:
- the inappropriate use of time zones is a frequent
http://n4.nabble.com/file/n1583733/100307070476876317b486a941.jpg
I want to get a histogram by factors.
Thanks.
--
View this message in context:
http://n4.nabble.com/barplot-with-factors-problem-tp1583671p1583733.html
Sent from the R help mailing list archive at Nabble.com.
_
On 2010-03-07 10:30, David Winsemius wrote:
On Mar 7, 2010, at 12:06 PM, cmc wrote:
I am trying to follow this example for multinomial logistic regression
http://www.ats.ucla.edu/stat/r/dae/mlogit.htm
However, I cannot get it to work properly.
This is the output I get, and I get an error
Is it possible to vectorize anova over the output of a vectorized lm? I
have a gene expression matrix with each row being a gene and columns for
samples. There are several factors with interactions. I can get p values by
looping over the matrix with lm and anova, but I would like to make this as
c
I have tried doing it whithout attaching the mydata file, and i still get the
same output.
When i create mldata using mlogit.data i do not generate the chid and alt
columns, is this where the problem is, if so how do i fix this problem.
Cheers for any help
--
View this message in context: http:
Looks like a problem that the maintainer should be copied with:
> maintainer("mlogit")
[1] "Yves Croissant "
Not sure why it should affect your systems and not mine, but here is
my sessionInfo() if it helps:
> sessionInfo()
R version 2.10.1 RC (2009-12-09 r50695)
x86_64-apple-darwin9.8.0
lo
On 05.03.2010 15:24, kloyt...@mappi.helsinki.fi wrote:
Hi,
I have a list p with different size dataframes and length of over 8000.
I'm trying to
calculate correlations between the rows of dataframes of this list and
columns of another
dataset (type data.frame also) so that first column is corre
Simple example:
# Classification Tree with rpart
library(rpart)
# grow tree
fit <- rpart(Kyphosis ~ Age + Number + Start,
method="class", data=kyphosis)
Now I would like to know how can I measure the "importance" of each of my
three explanatory variables (Age, Number, Start) in the model
Mark Kimpel wrote:
>
> Is it possible to vectorize anova over the output of a vectorized lm?
>
library(nlme)
fm1 <- lmList(distance ~ age | Subject, Orthodont)
lapply(fm1,anova)
Dieter
--
View this message in context:
http://n4.nabble.com/vectorizing-ANOVA-over-a-vectorized-linear-model
Ah, this clarifies what the reshape documentation referred to as the
"timevar" argument. I'd gone wrong because I'd been confusing
reshape's functionality with that of SPSS's cassestovars command,
which is where I'm coming from. Your recommendation seems to lead to
exactly that I was lookin
Hi Mark,
If efficiency is a concern you might want to read "Computing Thousands
of Test Statistics Simultaneously in R" by Holger Schwender and Tina
Müller, http://stat-computing.org/newsletter/issues/scgn-18-1.pdf.
If you just want to do it, see the examples in
http://had.co.nz/plyr/plyr-intro-0
David,
Here's my sessionInfo:
> sessionInfo()
R version 2.11.0 Under development (unstable) (2010-03-02 r51195)
i386-pc-mingw32
locale:
[1] LC_COLLATE=English_Canada.1252 LC_CTYPE=English_Canada.1252
[3] LC_MONETARY=English_Canada.1252 LC_NUMERIC=C
[5] LC_TIME=English_Canada.1252
attached bas
There is now a document called "Some hints
for the R beginner" whose purpose is to get
people up and running with R as quickly as
possible.
Direct access to it is:
http://www.burns-stat.com/pages/Tutor/hints_R_begin.html
JRR Tolkien wrote a story (sans hobbits) called
'Leaf by Niggle' that has a
It worth adding that, for evaluating the quality of a model,
randomization is fairly useless.
Basically, as long as your model is slightly better than noise, it can
show a significant difference from the average randomized model. In
the qsar studies that we do, the samples sizes can be in the hund
hello
can you show me how to create a data.frame from two factors x and y. column 1
should be equal to x and column 2 is 1 if it is common to y and 0 if it is not.
x=factor(c("A","B","C","D","E","F","G"))
y=factor(c("B","C","G"))
the output should look like this:
A0
B1
C1
D0
E
Hi,
try this,
data.frame(x,as.numeric(x %in% y))
HTH,
baptiste
On 7 March 2010 21:06, joseph wrote:
> hello
>
> can you show me how to create a data.frame from two factors x and y. column 1
> should be equal to x and column 2 is 1 if it is common to y and 0 if it is
> not.
>
> x=factor(c("A
Hi,
data.frame(x=x,y=as.numeric(x%in%y))
HTH,
Stephan
joseph schrieb:
hello
can you show me how to create a data.frame from two factors x and y. column 1
should be equal to x and column 2 is 1 if it is common to y and 0 if it is not.
x=factor(c("A","B","C","D","E","F","G"))
y=factor(c("B",
Thanks Gabor,
You're right. The problem comes from the environment variable TZ. I just
tried the Sys.getenv("TZ") and it's nothing there. After I have set the
environment variable TZ as the same as the data, let's say
Sys.setenv(TZ="GMT+1"), the problem is gone.
In order to complete the prob
And just a small followup. To find out what class each column is, you wanted
lapply(a,class)
$x1
[1] "numeric"
$x2
[1] "factor"
$x3
[1] "factor"
With regard to your solution, and why it works, it is my
understanding that data frames are in some sense actually lists, each
column correspond
On my Vista system I get an error message saying that there is no such
timezone as GMT+1.
You may be better off not using time zones and just adjusting the
times yourself. You could just use chron and avoid the entire time
zone problem in the first place.
In the first of the two approaches below
Thanks Gabor,
Just one comment on the error you received on the Vista machine. I also
noticed this problem on Windows as well, and found there is no
specification of "GMT+1" on Windows XP. The closest setting for "GMT+1"
on windows is "Africa/Lagos", however there is no this problem on *unix
On 2010-03-07 10:41, cmc wrote:
I have tried doing it whithout attaching the mydata file, and i still get the
same output.
When i create mldata using mlogit.data i do not generate the chid and alt
columns, is this where the problem is, if so how do i fix this problem.
I've suggested not nami
On Mar 7, 2010, at 3:20 PM, Don MacQueen wrote:
And just a small followup. To find out what class each column is,
you wanted
lapply(a,class)
$x1
[1] "numeric"
$x2
[1] "factor"
$x3
[1] "factor"
With regard to your solution, and why it works, it is my
understanding that data frames are
Thatnk you.
The documentation indicates as you indicated that if there is not an exact
match then the next element is chosen. But it does not indicate the case that
contains an exact match but there is not value to be returned (=, case). From
what you indicate this is treated as if it was not a
Dear R People:
The aggregate function works very well on regular time series.
Is there a version for zoo or its that would take daily data and
convert it to monthly, please?
Thanks in advance,
Sincerely,
Erin
--
Erin Hodgess
Associate Professor
Department of Computer and Mathematical Sciences
See ?aggregate.zoo, e.g.
library(zoo)
z <- zoo(1:1000, as.Date("2000-01-01") + 0:999)
aggregate(z, as.yearmon, mean)
or replace mean with whatever summarization you want.
On Sun, Mar 7, 2010 at 5:29 PM, Erin Hodgess wrote:
> Dear R People:
>
> The aggregate function works very well on regular t
Deepayan:
Thank you so much for your time and hints on my question. The examples do
help me move further. I will let you know once I solve the whole question.
Have a good day!
Doug
On Mar 6 2010, Deepayan Sarkar [via R] wrote:
>
>
>
>On Wed, Mar 3, 2010 at 12:35 PM, DougNiu wrote:
>>
>> I
Dear Dhruv,
You could create interaction variables manually (assuming A is your
dependent variable). Just multiply the variables together.
cd.int<-C*D
ce.int<-C*E
cde.int<-C*D*E # what about D*E, or interactions with B?
Include those in your model, such as A~B+C+D+E+cd.int+cd.int+ce.int+cde.int.
On 07/03/2010 5:26 PM, rkevinbur...@charter.net wrote:
Thatnk you.
The documentation indicates as you indicated that if there is not an exact
match then the next element is chosen. But it does not indicate the case that
contains an exact match but there is not value to be returned (=, case). F
Hadley,
Thanks for pointing me to some good articles. Unfortunately, I have already
read Holger's and my main concern is computational efficiency. The buzzword
on this list regarding efficient code is "vectorization". I am, frankly,
surprised that there is a way to vectorize analysis of complex mo
x <- c(0,0,1,2,3,0,0,4,5,6)
How to identify the regions of non-zeros and average c(1,2,3) and c(4,5,6) to
get 2 and 5.
Thanks
_
Hotmail: Trusted email with Microsoft¡¯s powerful
I have split my original dataframe to generate a list of dataframes each of
which has 3 columns of factors and a 4th column of numeric data.
I would like to use lapply to apply the fitdistr() function to only the 4th
column (x$isi) of the dataframes in the list.
Is there a way to do this or am
I have split my original dataframe to generate a list of dataframes each of
which has 3 columns of factors and a 4th column of numeric data.
I would like to use lapply to apply the fitdistr() function to only the 4th
column (x$isi) of the dataframes in the list.
Is there a way to do this or am I
This topic refer to independent variables reduction, as we know ,a lot of
method can do with it,however, for pre-processing independent varibles, a
method like the sentence below can reduce many variable, How can I
understand it?
what is significant correlation at 5% level, what is the criterion
Try this:
> x <- c(0,0,1,2,3,0,0,4,5,6)
> # partition the data
> x.p <- split(x, cumsum(x == 0))
> # now only process groups > 1
> x.mean <- lapply(x.p, function(a){
+ if (length(a) == 1) return(NULL)
+ return(list(grp=tail(a, -1), mean=mean(tail(a, -1
+ })
> # now only return the real
It would have been nice if you had at least posted what the structure of
myList is. Assuming that this is the list of your data frames, then the
following might work:
lapply(myList, function(x) fitdistr(x$isi,
densfun='gamma',start=list(scale=1, shape=2)))
On Sun, Mar 7, 2010 at 7:30 PM, Dgnn
In the code of the mlogit function, the call is saved under the name
'mldata' that you use to store the result of the mlogit.data function.
When calling mlogit with 'mldata' as the data argument, there is a
conflict between these two objects with the same name, one of class
'call' and one of class
Thanks that solved my problem.
Hadassa
On Sun, Mar 7, 2010 at 7:04 PM, Peter Ehlers wrote:
> On 2010-03-07 3:12, Hadassa Brunschwig wrote:
>
>> Hi,
>>
>> I was wondering if there is a way to control
>> the strip size in xyplot for example using the
>> strip.default function. Or do I need to redr
Hi Mark,
Unless you are fitting millions of very very very simple models, I
doubt that extracting p-values is going to be a limiting factor in the
speed of your analysis.
Hadley
On Mon, Mar 8, 2010 at 3:47 AM, Mark Kimpel wrote:
> Hadley,
>
> Thanks for pointing me to some good articles. Unfort
try this:
x <- c(0,0,1,2,3,0,0,4,5,6)
rl <- rle(x == 0)
grp <- rep(seq_along(rl$lengths), rl$lengths)
res <- tapply(x, grp, mean)
res[res > 0]
I hope it helps.
Best,
Dimitris
On 3/8/2010 3:48 AM, Daren Tan wrote:
x<- c(0,0,1,2,3,0,0,4,5,6)
How to identify the regions of non-zeros and a
library(MASS)
dat <- data.frame(
col1=as.factor(sample(1:4, 100, T)),
col2=as.factor(sample(1:4, 100, T)),
col3=as.factor(sample(1:4, 100, T)),
isi=rnorm(100)
)
dat <- split(dat, as.factor(sample(1:3, 100, T)))
lapply(dat, function(x, densfun) fitdistr(x$isi, densfun), 'normal')
On Mon, 8 Mar 2010, yves croissant wrote:
In the code of the mlogit function, the call is saved under the name
'mldata' that you use to store the result of the mlogit.data function.
When calling mlogit with 'mldata' as the data argument, there is a
conflict between these two objects with the sam
Hi
r-help-boun...@r-project.org napsal dne 05.03.2010 13:40:54:
>
> I have created a large dataframe (d) by getting data from file using
> read.table
>
> I now have 79 columns and 3 million rows. How can I plot the 6th column?
I
> tried plot(d[,6]) but it doesn't look right. When I try to do j
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