Hi
>
> On 17/11/11 17:33, arunkumar wrote:
> > Hi
> >
> > I need to add an year to and date field in the dataframe.
> >
> > Please help me
> >
> > X Date
> > 1 2008-01-01
> > 2 2008-02-01
> > 3 2003-03-01
>
> I can't find anything built in. This is probably because "year" is an
> ill-def
This mean
First, I am no expert but I am analyzing some marketing data.
I have information on two versions of the same site, and I have data
on the number of times people filled out a form on each version
of the site.
Sample data:
Site 1 Site 2
Fill
Please don't ask non-R programming questions on R-help.
This is best asked on R-sig-mac, including details of how you built
your R. But the short answer is not to mix macports with the CRAN
build of R. And there are matching libs for CRAN's build at
http://r.research.att.com/libs/
On Thu, 1
Thanks a lot for the guidance. I will take a look at these options.
Ravi
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View this message in context:
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Sent from the R help mailing list archive at Nabble.com.
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R-hel
hello mr martin
i need the program of lower incomplet Gamma function, and i dont know any thing
about mathematica, my gamma function is Y(1/2,r^2/4*teta), r is variable.
pleasr help me, its emmidiate for me.
all the best
somaye node
[[alternative HTML version deleted]]
_
Basically all I want to do is predict a scalar response using some
curves. I've got as far as doing a regression (using fRegress from the
fda package) but have no idea how to apply the results to a NEW set of
curves (for prediction).
I have N=536 curves, and 536 scalar responses. Here's what I'
Dear Researches,
I am using RF (in regression way) for analize several metrics extract from
image. I am tuning RF setting a loop using different range of mtry, tree
and nodesize using the lower value of MSE-OOB
mtry from 1 to 5
nodesize from1 to 10
tree from 1 to 500
using this paper as refery
Hi everyone, I tried to solve this problem but I could not find the
solution. I have about 105 matrices of equal size in the memory of**R, I
need to do is extract from these matrices some known positions and
create a new matrix with these columns. Show you an example with only
three matrices (b
Hello, I am wondering if someone can help me. I have the following function
that I derived using nls() SSlogis. I would like to find its derivative. I
thought I had done this using deriv(), but for some reason this isn't
working out for me.
Here is the function:
asym <- 84.951
xmid <- 66.90742
sca
Ok guys, as requested, I will add more info so that you understand why a
simple vector operation is not possible. It's not easy to explain in few
words but let's see. I have a huge amount of points over a 2D space.
I divide my space in a grid with a given resolution,say, 100m. The main loop
that I
All,
the MacOSX binary build of the biOps package is broken on cran, so I am trying
to compile from source. I am very close; the trick is apparently that this
package depends on fftw3, libjpeg and libtiff. My fftw3 is in /usr/local/, but
my libjpeg and libtiff are in /opt/local/ since i got th
Hi,
I need to make a subset of my species abundance matrix with only species
(columns) that have a total abundance(column sum) greater than 0.5 to do
ordination in vegan package. I used following code but it is not working. Can
you please give me a solution.
gl1<- subset(grassland[,5:44], colS
Dear Dennis,
Many thanks.I was wondering if there was a way to edit the variable and put
\n's in it. Is there ?
Thank you,
Ashim
On Thu, Nov 17, 2011 at 4:07 PM, Dennis Murphy wrote:
> Hi:
>
> This worked for me - I needed to modify some of the strip labels to
> improve the appearance a bit an
Hi All,
I need to a calculation W%*%d. However I know that this matrix is symmetric
(since W=t(d)%*%w). My question is considering that I only need to
calculate the lower/ upper triangle (n(n+1)/2 elements) rather than the n^2
elements of the entire matrix. Is there a way to do this efficiently.
Hi Joshua!
2011/11/17 Joshua Wiley :
> One possibility (though it does not concatenate per se):
>
> combined <- list(corpus.1, corpus.2)
Thanks I will look into it.
> *if* (there are only attributes in corpus.1 OR corpus.2) OR (the
> attribute names in corpus.1 and corpus.2 are unique), then yo
Dear all,
I want to draw ticks on the 3rd and 4th row of a lattice. How do I do this
? In my search of the help, I discovered a parameter alternating,which kind
of says where the ticks will be but does not suffice for me.
I am running this command : -
barchart(X03/1000~time|Company,
dat
Hi the list,
I define a class 'C' that inherit from two classes 'A' and 'B'. 'A' and 'B'
have no slot with similar names.
setClass(
Class="C",
contains=c("A","B")
)
To define the get operator '[' for class "C", I simply use the get of "A" or
"B" (the constante 'SLOT_OF_A
Peter Minting hotmail.com> writes:
>
>
> Dear R help,
> I am trying to work out if I am justified in
> log-transforming data and specifying Gamma in the same glm.
> Does it have to be one or the other?
No, but I've never seen it done.
> I have attached an R script and the datafile to show w
Looks like the function I was looking for was source(), but thanks Joshua I
certainly do need to make a package once I finish this set of re-coding
into R from Matlab. Fingers crossed the effort is worth it.
Thanks,
Sachin
On Fri, Nov 18, 2011 at 1:34 PM, Sarah Goslee wrote:
> ?source
>
> source
Here's a function Josh Wiley provided in another thread:
spec.cor <- function(dat, r, ...) {
x <- cor(dat, ...)
x[upper.tri(x, TRUE)] <- NA
i <- which(abs(x) >= r, arr.ind = TRUE)
data.frame(matrix(colnames(x)[as.vector(i)], ncol = 2), value = x[i])
}
Michael
On Thu, Nov 17, 2011
?source
source("/path/to/foo.R") will load it into R.
Sarah
On Thu, Nov 17, 2011 at 8:26 PM, Sachinthaka Abeywardana
wrote:
> Hi All,
>
> I have written a function (say) called foo, saved in a file called
> foo.R. Just going by Matlab syntax I usually just change my folder path and
> therefore
Hi Josh,
You're absolutely right. I suppose one could set up some sort of S3
thing for Henri's problem:
c <- function(..., recursive = FALSE) UseMethod("c")
c.default <- base::c
c.corpus <- function(..., recursive = FALSE) {ans = c.default(...);
attributes(ans) <- c(do.call(attributes, ...))}
Bu
Hi Sachin,
Nope, R does not work that way. You do have several options, though.
For a function or two, consider creating/editing a workspace .Rprofile
file.
https://www.google.com/?q=Rprofile
should bring up a fair number of pages describing this, you might look at a few.
If you find yourself
Hi Michael,
require(sos)
findFn("{meta}", sortby = "Function")
## see that only two functions have the exact name, 'meta'
## one is titled, "Meta Data Management" in the package 'tm'
## seems a pretty likely choice
Also, the fact that it is a truly terrible idea does not mean it is not easy:
mvi
Hi All,
I have written a function (say) called foo, saved in a file called
foo.R. Just going by Matlab syntax I usually just change my folder path and
therefore can call it at will.
When it comes to R, how is the usual way of calling/loading it? because R
doesnt seem to automatically find the fun
What package is all this from()?
You might check if there is a special rbind/cbind method provided. I don't
think you can easily change the behavior of c()
Michael
On Nov 17, 2011, at 4:43 PM, Henri-Paul Indiogine wrote:
> Greetings!
>
> I loose all my metadata after concatenating corpora.
Hi Henri-Paul,
This can be rather tricky. It would really help if you could give us
a reproducible example. In this case, because you are dealing with
non standard data structures (or at least added attributes), the data
exactly as R "sees" it. This means either A) code to create some data
that
Excellent; thanks Josh.
Joshua Wiley-2 wrote:
>
> Hi Brad,
>
> You do not really need to reshape the correlation matrix. This seems
> to do what you want:
>
> spec.cor <- function(dat, r, ...) {
> x <- cor(dat, ...)
> x[upper.tri(x, TRUE)] <- NA
> i <- which(abs(x) >= r, arr.ind = TRUE)
On Nov 17, 2011, at 10:37 AM, Nasrin Pak wrote:
Hi all;
It seemed to be easy at first, but I didn't manage to find the answer
through the google search. I have a set of data for every second of
the
experiment, but I don't need such a high resolution for my analysis.
I want
to replace every
Hi Brad,
You do not really need to reshape the correlation matrix. This seems
to do what you want:
spec.cor <- function(dat, r, ...) {
x <- cor(dat, ...)
x[upper.tri(x, TRUE)] <- NA
i <- which(abs(x) >= r, arr.ind = TRUE)
data.frame(matrix(colnames(x)[as.vector(i)], ncol = 2), value = x[
This is probably not the prettiest or most efficient function ever, but this
seems to do what I wanted.
spec.cor <- function(dat, r, ...){
require("reshape")
d1 <- data.frame(cor(dat))
d2 <- melt(d1)
d2[,3] <- rep(rownames(d1), nrow(d2)/length(uni
I'm starting to get a clearer idea of what you mean: there are two
(possibly three) routes you can go:
1) If your matrices are sparse (mostly zero) there's some specialized
work on multiplying them quickly
2) You can look at the RcppArmadillo package which interfaces to a
very high quality linear
Hi Michael,
Thanks for that. The X1 and X2 are vectors are typically 1000 by 3
matrices, and hoping to scale up to much larger dimensions (say 20,000 by
3).
I do appreciate your help and seems like this is the best way to do this, I
was just wondering if I could squeeze out just a bit more perfor
These might get you started
Analysing spatial point patterns in R by Adrian Baddeley
CSIRO and University of Western Australia
http://www.csiro.au/files/files/p10ib.pdf
Spatial Regression Analysis in R: A Workbook, by Luc Anselin
Spatial Analysis Laboratory
http://geodacenter.asu.edu/system/files
I fail to see why you would need another idea: you asked how to
multiply matrices efficiently, I told you how to multiply matrices
efficiently.
if you want to calculate X1-X2 times W times X1-X2, then simply do so:
X1 <- matrix(1:6, 3)
X2 <- matrix(7:12, 3)
W = matrix(runif(9), 3)
t(X1-X2) %*% W
Greetings!
I loose all my metadata after concatenating corpora. This is an
example of what happens:
> meta(corpus.1)
MetaID cid fid selfirst selend fname
1 0 1 11 2169 2518WCPD-2001-01-29-Pg217.scrb
2 0 1 14 9189 9702 WCPD-2003-01-1
I'm not quite sure of what you mean by not worry if it's 1d R matrices. X1
and X2 are both n by d matrices and W is d by d.
Thanks for the help though. Any other ideas?
Thanks
Sachin
On Friday, November 18, 2011, R. Michael Weylandt <
michael.weyla...@gmail.com> wrote:
> The fastest is probably
Well, for
" What is the command for combining certain number of data into their
average value?"
one way would be (calling the data vector, x)"
colMeans(matrix ( x[ seq_len(30 * floor(length(x)/30))], nrow=30))
Note that this will leave out the mean of any values with indices
beyond the largest
There is no single command to do all of what you want.
Read the posting guide for advice on how to ask questions that are more
likely to receive helpful answers.
The mean() function is a command for "combining certain number of data
into their average value".
The write.csv() function will creat
Please post this to r-sig-mixed-models instead. Or, better yet,
consult your local statistician, as your question indicates a profound
lack of understanding that may require more back and forth discussion
than can occur on an internet help site.
-- Bert
On Thu, Nov 17, 2011 at 5:36 AM, arunkumar1
It might not be as general as you have in mind, but this works:
X = array(1:24, c(2,3,4))
rowSums(X, dims = 2)
Combined with aperm() it's pretty powerful.
Michael
On Thu, Nov 17, 2011 at 11:24 AM, Simone Salvadei
wrote:
> I'm looking for a function that allows to sum the elements of an array
I can't see how it's stored like that and the email servers garble it
up. Use dput() to create a plain text representation and paste that
back in.
Thanks,
Michael
On Thu, Nov 17, 2011 at 9:37 AM, muzz56 wrote:
> Hi Michael,
> Here is a sample of the data.
>
> Gene Array1 Array2 Array3 Array4 Ar
Sorry -- that came off as very muddled.
What I meant to say:
To make it (almost) certain you will get different results on each
machine, you can reset the PRNG seed on each machine in some way
unique to that machine. What immediately came to mind was IP address,
which you can access with somethin
http://search.dilbert.com/comic/Random%20Number%20Generator
In all seriousness, you could set the seed differently on each machine
after putting jobs through Torque (i.e., as part of the batch script,
maybe using some piece of hardware id you can get through system()
somehow or other: possibly net
On Nov 17, 2011, at 10:13 AM, Miguel Lacerda wrote:
> Hi,
>
> I am using the persp function to plot 3D surfaces, but the plots have
> "little white lines" when I print them to a pdf file (visible in
> Acrobat, Foxit, Evince, Xpdf and Gimp). This does not happen when I
> create png or tiff images.
Hello Michael,
Thanks again for your reply. Actually, I am working with wind data.
I have some sample data for actual load.
scan("/home/sam/Desktop/tt.dat") ->tt ## This is the input for the actual
output of the generation
t = ts(tt, start=8, end=24, frequency=1,)
I have another random sequen
Hello:
I have some trouble making a prediction from an AR(p) model. After I have the
AR(p) model fitted , I want to use a new data set to make predictions.
But I get the error: Error in newdata - object$x.mean : non-numeric argument
to binary operator.
A small version of my original data lo
Hi. I am new to R and actually have several questions related to this topic.
A row in my data looks like the following:
418 12 6/21/2010 9:37:12 40.7219593 -73.9962579
1.3406345525960568
0.019682641058810173
In order, the columns are id, week, date, time, latitude, long
You can print out the nodes and their corresponding clusters into a file by
this:
> write.table (hc,file="hc_40clusters.cvs", quote=FALSE, sep=" ")
--
View this message in context:
http://r.789695.n4.nabble.com/hierarchical-clustering-within-a-size-limit-tp3515354p4080551.html
Sent from the R
I'm looking for a function that allows to sum the elements of an array
along a dimension that can be different from the classical ones (rows or
columns).
Let's suppose for example that:
- A is an array with dimensions 2 x 3 x 4
- I want to compute B, a 2 x 3 matrix with elements equal to the sum
I am trying to run a lme model and some contrast for a matrix .
lnY
[1] 10.911628 11.198557 11.316971 11.464869 11.575233 11.612101 11.755903
11.722035 11.757705 11.863744 11.846515 11.852721 11.866936 11.838452
11.946680 11.885509
[17] 11.583309 11.750082 11.756005 11.630797 11.705536 11.566722 1
Hi all;
It seemed to be easy at first, but I didn't manage to find the answer
through the google search. I have a set of data for every second of the
experiment, but I don't need such a high resolution for my analysis. I want
to replace every 30 row of my data with their average value. And then sa
Hi Michael,
Here is a sample of the data.
Gene Array1 Array2 Array3 Array4 Array5 Array6 Array7 Array8 Array9 Array10
Array11 Fth1 26016.01 23134.66 17445.71 39856.04 27245.45 23622.98 37887.75
49857.46 25864.73 21852.51 29198.4 B2m 7573.64 7768.52 6608.24 8571.65
6380.78 6242.76 6903.92 7330.6
Dear R help,
I am trying to work out if I am justified in log-transforming data and
specifying Gamma in the same glm.
Does it have to be one or the other?
I have attached an R script and the datafile to show what I mean.
Also, I cannot find a mixed-model that allows Gamma errors (so I cannot find
Hi
I'm running the same R script (throuth linux shell) of several cpu's. This
R program uses random numbers and the result should be different every time.
But if put jobs (through Torque) for several cpu's I get the same result. As
a resealt my program saves numbers in file with randomly generate
Hi
How to include a factor or class variable to a fixed effect of lmer
function. when i included it throws an error. Please help
My code
data <- read.delim("C:/TestData/data.txt")
Mon=as.factor(data$Month)
lmerform= "Y~ X2 +X3 + Month:Mon + (1|State)+ (1+ X5|State)"
lmerfit=lmer(formula=l
Hi,
Thanks for your reply. Based on your suggestions, I managed to simplify the
code, but only a little. I don't see how I could do without a loop, given the
nestedness of the hierachy. See the code below, which is working, but I'd like
to simplify it.
# sample data
theCodes <- c('STAT.01',
Tyler Rinker hotmail.com> writes:
>
>
> try: library(pscl)
>
> There's a zeroinfl for zero inflated neg. binom.
>
> Tyler
> >
> > Dear All,
> > I am trying to fit some data both as a negative binomial and a zero
> > inflated binomial.
> > For the first case, I have no particular problems,
This kind of error seems to surprise R users. It surprises me that it doesn't
happen much
more frequently. The "BFGS" method of optim() from the 1990 Pascal version of
my book was
called the Variable Metric method as per Fletcher's 1970 paper it was drawn
from. It
really works much better with
Hstrangeif possible, this might be solvable by simply
updating to the release version R 2.14. If it's at all possible, I'd
start there.
Can you find the object it's unhappy about? On my machine, I do the following
1) Open Finder
2) Macintosh HD -> Library -> Frameworks -> R.framework
My responses are in brackets below, plus a final note after the main text.
- Original Message -
From: Uwe Ligges
To: Scott Raynaud
Cc: "r-help@r-project.org"
Sent: Thursday, November 17, 2011 9:16 AM
Subject: Re: [R] modelling and R misconceptions; was: package installtion
This is hope
See my responses in brackets below.
- Original Message -
From: Rolf Turner
To: Scott Raynaud
Cc: "r-help@r-project.org"
Sent: Wednesday, November 16, 2011 6:04 PM
Subject: Re: [R] package installtion
On 17/11/11 05:37, Scott Raynaud wrote:
> That might be an option if it weren't my m
On Thu, Nov 17, 2011 at 7:33 AM, R. Michael Weylandt
wrote:
> Hmmm...sorry -- the only thing I can suggest is maybe striking some
> sort of deal that you change when it gets however many months out of
> date: if you look here
> (http://cran.r-project.org/src/contrib/Archive/MASS/), you can see th
Hi,
On Thu, Nov 17, 2011 at 11:49 AM, Alaios wrote:
> Dear all I have a txt file with the following contents
> 1 50.790643000 6.063498
> 2 50.790738000 6.063471
> 3 50.791081000 6.063380
> 4 50.791189000 6.063552
>
>
> I am usind read.table('myfile.
I'm having trouble replicating/understanding why that would happen since I do
it all the time. The only thing that raises a hint of suspicion is using the
blank space separator , but I'm pretty sure that's fine What does str() give?
Possibly factors?
If you are sure that's happening as describ
Just looking at the ambiguity in "adding a year"
> dates <- as.Date(c('2007-03-01','2008-02-29'))
> tmp <- as.POSIXlt(dates)
> tmp$year <- tmp$year+1
> dates2 <- as.Date(tmp)
> dates2
[1] "2008-03-01" "2009-03-01"
> dates2 - dates
Time differences in days
[1] 366 366
KJ
"MacQueen, Don" wrote in
Dear all I have a txt file with the following contents
1 50.790643000 6.063498
2 50.790738000 6.063471
3 50.791081000 6.063380
4 50.791189000 6.063552
I am usind read.table('myfile.txt',sep=" ")
which unfortunately returns only integers and not dou
Here is an example that could probably be described as "adding a year":
dates <- c('2008-01-01','2009-03-02')
tmp <- as.POSIXlt(dates)tmp$year <- tmp$year+1
dates2 <- format(tmp)
> dates
[1] "2008-01-01" "2009-03-02"
> dates2
[1] "2009-01-01" "2010-03-02"
## to begin to understand how it works, g
One more thing: trying to defend R's honor, I've run optimx instead of
optim (after dividing the IV by its max - same as for optim). I did
not use L-BFGS-B with lower bounds anymore. Instead, I've used
Nelder-Mead (no bounds).
First, it was faster: for a loop across 10 different IVs BFGS took
6.14
Hi,
Please copy your replies to r-help so others may participate in the
discussion.
2011/11/17 Jie TANG :
> yes ,I have tried readLines
> by
> config<- readLines(configfile,ok=TRUE,n=-1)
> #but when strsplit is used as below
> food<-unlist(strsplit(config[2].":"))
> #here "food " is a vector
> bu
This is hopeless, since you never seem to listen to our advice,
therefore this will be my very last try:
So you actually need local advice, both for statistical concepts and R
related. No statistics software can estimate effects of variables that
you observed to be constant (e.g. 0) all the ti
See Sarah's reply here:
http://www.mail-archive.com/r-help@r-project.org/msg152883.html
Michael
On Thu, Nov 17, 2011 at 7:54 AM, haohao Tsing wrote:
> hi,R users:
> I have such a text
>
> num = 3
> testco = 12
> testno = 1;12;3
> infp = test1;test2;test3
> How can I read this text by readLines?
Why are you trying to take the matrix trace of a regression model?
(That's the only hit for mtrace on my system at least)
Perhaps you mean to use traceback() or, even more useful,
options(error = recover)
Michael
On Thu, Nov 17, 2011 at 9:49 AM, Scott Raynaud wrote:
> I believe the problem is a
I don't know if this is faster, but ...
out <- do.call(rbind,
lapply(s, function(x)data.frame(x$category,x$name,as.vector(x$series
## You can then name the columns of out via names()
Note: No fancy additional packages are required.
-- Bert
On Wed, Nov 16, 2011 at 6:39 PM, Kevin Burton
I believe the problem is a column of zeroes in my x matrix. I have tried the
suggestions in the documentation,
so now to try to confirm the probelm I'd like to run debug. Here's where I
think the problem is:
###~~ Fitting the model using lmer funtion ~~###
(fitmodel <-
Hi,
On Thu, Nov 17, 2011 at 9:37 AM, Jie TANG wrote:
> hi everyone .
> Here I have a text where there are some integer and string variables.But I
> can not read them by readLines and scan
I've seen this question several times this morning. If that's you,
please do not post multiple times. If yo
hi everyone .
Here I have a text where there are some integer and string variables.But I
can not read them by readLines and scan
the text is :
weight ;30;130
food:2;1;12
color:white;black
the first column is the names of the variables and others are the value of
them.
the column in different l
I think something like this should do it, but I can't test without data:
rownames(mydata) <- mydata[,1] # Put the elements in the first column
as rownames
mydata <- mydata[,-1] # drop the things that are now rownames
Michael
On Thu, Nov 17, 2011 at 9:23 AM, Musa Hassan wrote:
> Hi Michael,
> Th
On 11/17/2011 06:28 AM, vioravis wrote:
> I am looking for online courses to learn Spatial Statistics using R.
> Statistics.com is offering an online course in December on the same topic
> but that schedule doesn't suit mine. Are there any other similar modes for
> learning spatial statistics using
Hi Ravi,
You would probably get more answers to this if you posted to the list
r-sig-geo. The following course was advertised a week ago and might
match your needs:
http://www.itc.nl/personal/rossiter/teach/degeostats.html
You might also find the videos from this years' GEOSTAT course in Landa
vioravis gmail.com> writes:
>
> Thanks, Raphael. Just checked their website. It appears that they currently
> do not have any online courses planned.
>
You may find that this site:
http://geostat-course.org/
has a wider range of possible courses.
Dear All,
The following equation is a linear mixed-effects model with linear trend and
AR(1) error structure,
y = B0 + B1x + bo + b1x + e; e~AR(1)
where y is a response, x is the predictor, B0 and B1 are fixed effects and b0
and b1 are random effects.
Coud someone please advice me a fun
On Wed, Nov 16, 2011 at 11:22 PM, muzz56 wrote:
> Thanks to everyone who replied to my post, I finally got it to work. I am
> however not sure how well it worked since it run so quickly, but seems like
> I have a 2000 x 2000 data set.
Behold the great and mighty power that is R! Don't worry -- on
Thanks Gabor and Jan.
The batch files solution seems the way to go.
Will implement it!
Rubén
-Mensaje original-
De: Gabor Grothendieck [mailto:ggrothendi...@gmail.com]
Enviado el: jueves, 17 de noviembre de 2011 13:58
Para: Rubén Roa
CC: r-help@r-project.org
Asunto: Re: [R] RV: Reporting
Or, for this specific application
rowSums(XXX, na.rm = TRUE)
Michael
On Thu, Nov 17, 2011 at 5:51 AM, Jim Holtman wrote:
> change to
>
>> row.sums.m <- apply(dummy.curr.res.m,1,sum, na.rm = TRUE)
>
> Sent from my iPad
>
> On Nov 17, 2011, at 5:18, Vikram Bahure wrote:
>
>> Dear R users,
>>
>>
The fastest is probably to just implement the matrix calculation
directly in R with the %*% operator.
(X1-X2) %*% W %*% (X1-X2)
You don't need to worry about the transposing if you are passing R
vectors X1,X2. If they are 1-d matrices, you might need to.
Michael
On Thu, Nov 17, 2011 at 1:30 AM,
Thanks, Raphael. Just checked their website. It appears that they currently
do not have any online courses planned.
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On Thu, Nov 17, 2011 at 3:54 AM, Rubén Roa wrote:
>
> I've just found that there is a conflict between tools used to build R
> packages (Rtools) and ADMB due to the need to put Rtools compiler's location
> in the PATH environmental variable to make Rtools work.
>
> On a Windows 7 64bit with Rto
hi,R users:
I have such a text
num = 3
testco = 12
testno = 1;12;3
infp = test1;test2;test3
How can I read this text by readLines?
[[alternative HTML version deleted]]
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Hi there,
I have read the help page of nls(), there is lower or upper for defining
the bounds of parameters. For example,
nls(y ~ 1-a*exp(-k1*x)-(1-a)*exp(-k2*x), data=data.1, start=list(a=0.02,
k1=0.01, k2=0.0004), upper=c(1,1,1), lower=c(0,0,0))
I hope to define k1 > k2, but I don't find
Well, if your problem is that a workspace is being loaded automatically
and you don't want that workspace, you have several options:
1. Use a different directory for each project so that the file loaded
by default is the correct one.
2. Don't save your workspace, but regenerate it each time.
3.
Something like this?
library(plyr)
ddply(df, .(group), function(x){
x[sample(nrow(x), 1), ]
})
Best regards,
Thierry
ir. Thierry Onkelinx
Instituut voor natuur- en bosonderzoek
team Biometrie & Kwaliteitszorg
Gaverst
Hello,
I have got a dataframe which looks like:
y <- c(1,5,6,2,5,10) # response
x <- c(2,12,8,1,16,17) # predictor
group <- factor(c(1,2,2,3,4,4)) # group
df <- data.frame(y,x,group)
Now I'd like to resample that dataset.
I want to get dataset (row)
per group. So per total sample I get 4 rows
i
change to
> row.sums.m <- apply(dummy.curr.res.m,1,sum, na.rm = TRUE)
Sent from my iPad
On Nov 17, 2011, at 5:18, Vikram Bahure wrote:
> Dear R users,
>
> I am new to R and have some query.
>
> I am having a dataset with binary output 0's and ones. But along with it it
> has NA's too. I want
Hello,
I currently run aov in the following way:
> throughput.aov <-
> aov(log(Throughput)~No_databases+Partitioning+No_middlewares+Queue_size,data=throughput)
> summary(throughput.aov)
Df Sum Sq Mean Sq F valuePr(>F)
No_databases 1 184.68 184.675 136.6945 < 2.2e-16
Hi:
This worked for me - I needed to modify some of the strip labels to
improve the appearance a bit and also reduced the strip font size a
bit to accommodate the lengths of the strings. The main thing was to
change \\n to \n.
Firstly, I created a new variable called Indic as a character variable
Dear R users,
I am new to R and have some query.
I am having a dataset with binary output 0's and ones. But along with it it
has NA's too. I want to sum all the rows and get the sum total for each
column.
But whenever there is a NA in an row the sum of the row is returned as NA
so I am not able
Take a look on http://geodacenter.asu.edu/ , Training section.
On Thu, Nov 17, 2011 at 4:28 AM, vioravis wrote:
> I am looking for online courses to learn Spatial Statistics using R.
> Statistics.com is offering an online course in December on the same topic
> but that schedule doesn't suit mine
On 17.11.2011 10:31, Dennis Murphy wrote:
Hi:
Here's one way:
apply(matrix(var.names, ncol = 2, byrow = TRUE), 1, function(x)
paste(x[1], x[2], sep = ','))
[1] "a,b" "c,d" "e,f"
Or for short and slightly faster for huge data use column-wise
operations as in:
apply(matrix(var.names, nrow=
Le mercredi 16 novembre 2011 à 20:02 -0800, Quercus a écrit :
> Hey everyone,
>
> I am new to R, and I'm making a scatter plot graph where i have a bunch of
> plots/points that fall into 9 unique categories. I want each category to
> have a unique colour, however, with the coding I have (below),
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