Rovinpiper david.j.mee...@gmail.com 04/05/2011 22:43
So this seems to indicate that I have what I want. I have two
respiration data points at each plot on each day.
Yes; if you had only Plot+Day you'd have a completely balanced full
factorial ... for Plot and Day.
But I think I now see an
On May 5, 2011, at 2:31 PM, Asan Ramzan wrote:
Hello R-Help
How can i exctact and store the within group mean squared
difference from an
anova summary table into a varible.
In the absence of an example and code it's speculation, but something
along the lines of:
anova(fit)$`Mean Sq`
Gene,
David has given you the preferred code. I just want to
point out that the $-accessor is often not the best
thing to use. Both dat[[y]] and dat[, y] will work
just fine.
Peter Ehlers
On 2011-05-05 12:06, David Winsemius wrote:
On May 5, 2011, at 1:08 PM, Gene Leynes wrote:
This is not
Il giorno Mon, 2 May 2011 07:30:10 -0700 (PDT)
Frank Harrell f.harr...@vanderbilt.edu ha scritto:
Please elaborate.
It is simply that, generally speaking, i don't like adding numbers to a
plot. I eventually realized that riskset cardinality may be a useful
indication. However, do not
This is a very detailed question, and possibly ought to have been posted on
R-sig-geo. Please re-post there, and see if anyone steps forward. As far as
I am aware, no implementation exists. I suggest you also refer to Kostov
(2009) in Spatial Economic Analysis volume 4.
Roger
Marie-Line
On 6/05/2011 6:06 a.m., swaraj basu wrote:
Dear All,
I am trying to build a package for a set of functions. I am
able to build the package and its working fine. When I check it with
R CMD check
I get a following warning : no visible global function
On 05/05/2011 11:06 AM, swaraj basu wrote:
Dear All,
I am trying to build a package for a set of functions. I am
able to build the package and its working fine. When I check it with
R CMD check
I get a following warning : no visible global function
I have a new device that takes measurements anywhere from every second, to
every 15 minutes (depending on changes). The matrix has a date, time and Y
column (Y is the measurement). For three days it is 25,000 rows. How do I
average the measurements by every 30 minutes so my matrix is 48 rows per
I'd be tempted to do a robust fit (loess?) to the data with a
relatively small span (I'm assuming that there are errors in the
measurements and some degree of smoothing is acceptable) then
predict the fit at a regular interval (e.g., every 30 minutes).
--
Clint Bowman
I do not want smoothing as the data should have jumps (it is weight left in
feeding bunker). I was thinking of maybe using a histogram-like function and
then averaging that. Not sure if this is possible.
-
In theory, practice and theory are the same. In practice, they are not - Albert
Hello, I have a rather large set of data I need to analyze, currently I need to
work with a 20 by 20 matrix, I'm using the package bigmemory but so far
I can only allocate a 66000 by 66000 matrix, when I increase those values I get
the following error:
AdjMat -
Friends
This is an elementary question. Is there is a built in R function for
finding a sub-string in another string? Like strstr in C.
I can easily roll my own, but if there is a built in that is one less thing
I can do wrong!
cheers
Worik
[[alternative HTML version deleted]]
In your first request for help you said, How do I average the
measurements by every 30 minutes? With 25000 readings over three
days, it looks as if you are getting readings just about every
second.
Okay, why don't you use the first reading as your initial weight,
w0. Then subtract each
On 2011-05-05 14:20, Schatzi wrote:
I do not want smoothing as the data should have jumps (it is weight left in
feeding bunker). I was thinking of maybe using a histogram-like function and
then averaging that. Not sure if this is possible.
(It would be useful to include your original request -
On May 5, 2011, at 4:55 PM, Worik R wrote:
Friends
This is an elementary question. Is there is a built in R function for
finding a sub-string in another string? Like strstr in C.
I can easily roll my own, but if there is a built in that is one
less thing
I can do wrong!
I have no
Hi,
I'm trying to do some basic social network analysis with igraph in R, but
I'm new to R and haven't been able to find documentation on a couple basic
things:
I want to run igraph's community detection algorithms on a couple thousand
small graphs but don't know how to automate igraph looking
Hi Guys
I am trying to read a bunch of files in the loop but pipe function
which I use to cut few columns is somehow unable to interpolate the
file variable.
eg:
file=check.txt
data - read.table(pipe(cut -f 2,3 file), sep=\t,
col.names=c('pos','cov') )
cut: file: No such file or directory
Histograms plot data in bins - you don't get the exact value, because
each bin contains a range of values. Do you want to plot the range of
values the bin contains? Also, check ?hist to see how to set the
values of the breaks between the bins.
__
Thanks slre,
I seem to be making some progress now.
Using a colon instead of an asterisk in the code really changes things. I
had been getting residual SS and MS of zero. Which is ridiculous. Now I get
much more plausible values.
Also, When I used an asterisk instead of a colon It wouldn't
Hi!
I have 2 columns (even though the data looks like there's more columns than
just two) of data in the following format:
0,58905313R0EOL 229742002R0EOL 58905312R0EOL
1,58905317R0DBL 58905303R0DBL 58905313R0IL 58905313R0VH
58905313R0EOL 223354003R0IL 223354003R0VH 58905308R0DBL
58905308R0VM
You can ignore my question I was able to figure out the way. I guess
when I touch R after couple of weeks I am rusty.
-Abhi
On Thu, May 5, 2011 at 2:15 PM, Abhishek Pratap abhishek@gmail.com wrote:
Hi Guys
I am trying to read a bunch of files in the loop but pipe function
which I use
On May 5, 2011, at 5:15 PM, Abhishek Pratap wrote:
Hi Guys
I am trying to read a bunch of files in the loop but pipe function
which I use to cut few columns is somehow unable to interpolate the
file variable.
eg:
file=check.txt
data - read.table(pipe(cut -f 2,3 file), sep=\t,
On Thu, May 5, 2011 at 4:55 PM, Worik R wor...@gmail.com wrote:
Friends
This is an elementary question. Is there is a built in R function for
finding a sub-string in another string? Like strstr in C.
I can easily roll my own, but if there is a built in that is one less thing
I can do
On 06/05/11 07:05, P Ehlers wrote:
Gene,
David has given you the preferred code. I just want to
point out that the $-accessor is often not the best
thing to use. Both dat[[y]] and dat[, y] will work
just fine.
Admittedly one should use the preferred code, i.e. gam(y ~ s(x),data=dat)
and avoid
Hi, I'm requesting you don't berate me for asking this question:
I clearly don't have the gist of factors.
I have two dataframes, A and B.
Each of them has a column containing strings (they're labels).
I want to, one-by-one in a loop, compare the particular string in an entry from
dataframe A
On May 5, 2011, at 6:28 PM, J wrote:
Hi, I'm requesting you don't berate me for asking this question:
I clearly don't have the gist of factors.
I have two dataframes, A and B.
Each of them has a column containing strings (they're labels).
I want to, one-by-one in a loop, compare the
Hi Jonathan,
On Thu, May 5, 2011 at 6:28 PM, J jonsle...@gmail.com wrote:
Hi, I'm requesting you don't berate me for asking this question:
I clearly don't have the gist of factors.
I have two dataframes, A and B.
Each of them has a column containing strings (they're labels).
I want to,
Dave:
We are planning an intervention study for adolescent alcohol use, and I
am planning to use simulations based on a hurdle model (using the
hurdle() function in package pscl) for sample size estimation.
The simulation code and power code are below -- note that at the moment
the power
Hi all,
I run R 2.11.1 under ubuntu 10.10 and caret version 2.88.
I use the caret package to compare different models on a dataset. In
order to compare their different performances I would like to use the
same data partitions for every models. I understand that using a LGOCV
or a boot type
On Thu, May 5, 2011 at 6:28 PM, J jonsle...@gmail.com wrote:
Hi, I'm requesting you don't berate me for asking this question:
I clearly don't have the gist of factors.
I have two dataframes, A and B.
Each of them has a column containing strings (they're labels).
I want to, one-by-one in a
as.character() will enable me to do this in a loop (so thanks!).
It looks like you guys are suggesting I accomplish this in some kind of
vectorized fashion (David: yes, the number of rows are not equal - and I
need to compare ALL pairs, not just adjacent pairs, so to vectorize, I
suppose I'd have
Hello. I'd like to be able to print variable strings which contain \ as-is,
without interpreting (for example) abcde\nuvxyz as having an embedded newline
(or whatever other escaped instruction).
To do this, I've tried gsub, and here's some of my output (I've tried all kinds
of variations to
Thanks Frank
I will try rms package and give after the result.
Komine
--
View this message in context:
http://r.789695.n4.nabble.com/Draw-a-nomogram-after-glm-tp3498144p3499771.html
Sent from the R help mailing list archive at Nabble.com.
__
gsub does this because the string dsff\nfsd
does not contain a backslash - the 5th character
is a newline. The deparsed representation (used
for printing the string) of a newline is \n but
the string itself has not backslash.
You can feed the output of deparse into message (or
cat) so they show
Hi Arnau,
please send the output of sessionInfo() and the exact commands and
response that you used to install and load apTreeshape.
Cheers
Andrew
On Thu, May 05, 2011 at 04:42:58PM +0200, Arnau Mir wrote:
Hello.
I'm trying to use the functions pandit and treebase. They are in the package
Hi David,
you might have more luck with your request if you tell us what
Vermunt's LEM *does*, and provided some links to introductory reading
material ...
Cheers
Andrew
On Thu, May 05, 2011 at 03:34:02PM +, David Joubert wrote:
Hello-
Does anyone know of packages that could emulate
Even then, I think that there's a problem. If C is in the model, then
the response varies by C. The simplest way is to pick a value for C,
and then evaluate the group mean estimates of A and B (and C).
Something in my brain keeps asking whether another way to marginalize
C for the purposes of
Hi, I'm installing rgdal but I keep having failures because I have not been
able to find a good source of information for the correct configuration
settings when installing GDAL.
My error from the R install.packages(rgdal) is below.
Can someone point me to a good source to tell me how to set
Hello,
I am doing an Engle Granger test on the residuals of two I(1) processes.
I would like to get the MacKinnon (1996) critical value, say at 10%. I
have 273 observations with 5 integrated explanatory variables , so that
k=4. Could someone help me with the procedure in R?
Thank you
Hi Marco,
You're welcome.
The number at risk at given time points is a fairly standard thing to add to
survival plots. I don't think many people will confuse it with numbers of
events. Personally I find shaded confidence bands a bit more helpful but
both are useful.
Frank
Marco Barbàra-2
Hi Katrina,
the error message below is actually pretty explicit. Have you
installed the PROJ.4 library? If not, then you need to install it.
When I had to do this I think that I used macports:
sudo port install proj
Then you need to tell configure where to find it, using the protocol
On May 5, 2011, at 9:10 PM, Lee Schulz wrote:
Hello,
I am doing an Engle Granger test on the residuals of two I(1)
processes.
I would like to get the MacKinnon (1996) critical value, say at
10%. I
have 273 observations with 5 integrated explanatory variables , so
that
k=4. Could
I guess LEM is a software for latent class analysis. If so, you may
want to have a look at poLCA package.
Regards
Ronggui
On 5 May 2011 23:34, David Joubert jo...@hotmail.com wrote:
Hello-
Does anyone know of packages that could emulate what J. Vermunt's LEM does ?
What is the closest
(Regarding bootstrapping logistic regression.)
If the number of rows with Y=1 is small, it doesn't matter that n is huge.
If both number of successes and failures is huge, then ad Ripley notes
you can use asymptotic CIs. The mean difference in predicted probabilities
is a nonlinear function of
Hi all,
I had meant to make this announcement earlier but had forgotten (a
recent bug report reminded me).
The version of lattice that ships with R 2.13.0 has a fairly serious
bug in panel.abline (which would neglect to draw many negative slope
lines). If you use lattice with R 2.13.0, you
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