On Thu, Jul 12, 2012 at 9:39 PM, Martin Ivanov wrote:
> Dear R users,
>
> I have a lot of experience with traditional R graphics, but I decided to turn
> to trellis as
> it was recommended for spatial graphs by the sp package. In traditional R
> graphics
> I always first set the size of the dev
On Jul 13, 2012, at 08:18 , darnold wrote:
> Hi,
>
> I'm looking for some ideas on how to reproduce the attached image in R.
> There are three samples, each of size n = 10. The first is drawn from a
> normal distribution with mean 60 and standard deviation 3. The second is
> drawn from a normal
Dear R users,
I am struggling with the colorkey on a levelplot lattice graphic.
I want that no ticks are printed on the colorkey. That is, I want their size
tck=0.
Merely setting tck=0 t in the colorkey parameter does not work. Setting it in
the lattice.par.set()
removes the ticks from the leve
Hi, Arnold,
looking at the example section of
?stripchart
may help you.
Hth -- Gerrit
On Thu, 12 Jul 2012, darnold wrote:
Hi,
I'm looking for some ideas on how to reproduce the attached image in R.
There are three samples, each of size n = 10. The first is drawn from a
normal distributi
Hi,
I'm looking for some ideas on how to reproduce the attached image in R.
There are three samples, each of size n = 10. The first is drawn from a
normal distribution with mean 60 and standard deviation 3. The second is
drawn from a normal distribution with mean 65 and standard deviation 3. The
t
plot(1:5, main = paste0("Figure 1: x=", x))
Michael
On Jul 12, 2012, at 3:30 PM, JeffND wrote:
> Hi folks,
>
> To simplify my question, consider an example:
>
> x=runif(1,0,1)
> plot(1:5,1:5)
>
> Now I want to add a title to the above plot showing the vaue of x, so if the
> generated x is 0.
Dear All,
I am using the function vuong from pscl package to compare 2 non nested models
NB1
(negative binomial I ) and Zero-inflated model.
NB1 <- glm(, , family = quasipoisson), it is an
object of class: "glm" "lm"
zinb <-
zeroinfl( dist = "negbin") is an object of class: "zeroinfl"
when a
Thanks. It works very good.
Chi
On Thu, Jul 12, 2012 at 7:27 PM, arun wrote:
> Hello,
>
> I saw your reply in nabble. Sorry about that. I thought the dataset had
> only few columns.
>
> #You can read first line of a file using:
> readLines("foo.txt",n=1)[1]
>
>
> #The more generic colname subs
Hello,
I saw your reply in nabble. Sorry about that. I thought the dataset had only
few columns.
#You can read first line of a file using:
readLines("foo.txt",n=1)[1]
#The more generic colname substitution
dat1<-read.table(text="
2.5 3.6 7.1 7.9
100 3 4 2 3
200 3.1 4
Hi, for my dataset is actually retrieve from a access file not a textfile.thanks
Date: Thu, 12 Jul 2012 11:58:30 -0700
From: ml-node+s789695n4636343...@n4.nabble.com
To: jubil...@live.com.sg
Subject: Re: plot graph by first letter
Hi,
Try this:
dat1<-read.table(text="
Name
*Dear **Madam**/**Sir,*
*What would happen if we irrigate the degraded land using the desalinated
saline water in Minqin Oasis, Gansu Province, China? Can we prevent the
disappearing of Minqin Qasis?*
*
*
*Background:*
*1. **The Minqin Oasis is surrounded by two dese
Dear All:
Could anybody help me figure out why I get the Error message below while I
running the example code of bugs() function in R2OpenBUGS packages? I have
tried the code both in Win 7 and Ubuntu 12.04, but they show the same
message. My R version is 2.15.1 in win7 and 2.15.0 in Ubuntu. Many
Thanks Yasir, this helps a lot.
BTW, is there an R command to read just the first line of the file?
Yasir Kaheil wrote
>
> just do this:
> colnames(r)<-substr(colnames(r),2,nchar(colnames(r)))
>
> This will remove the X.
> Later when you want to use the headed to plot something, cast it as
> n
Thanks, but I don't want to specify the column names by hand, since I have a
lot of similar files.
arun kirshna wrote
>
> Hi,
>
> Try this:
>
> dat1<-read.table(text="
> 2.5 3.6 7.1 7.9
> 100 3 4 2 3
> 200 3.1 4 3 3
> 300 2.2 3.3 2 4
> ",sep="",header=TR
I'm sorry but I see to be getting an error.Â
> data.tmp <- aggregate(MyTo ~ Date + Hour, data = tUnitsort[, cols], max)
> Error in `[.default`(xj, i) : invalid subscript type 'builtin'
From: Rui Barradas [via R]
To: jcrosbie
Sent: Thursday, July 12, 2012
My first guess would be there's not enough memory on the machine to support two
of the tasks you're running, so swapping ensues.
--Todd
--
Why is it so hard to remember the spelling of "mnemonic"?
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.or
This is additional to the post above (which I forgot to write):
What I mean with "How to interpret the coefficients as they are very large
numbers" is that it seems to be a bit ridiculous to think that survival is
e.g. 8.815e+34 (hazard ration taken from the interaction LT:Food:Temp2)
times lower
Hi folks,
To simplify my question, consider an example:
x=runif(1,0,1)
plot(1:5,1:5)
Now I want to add a title to the above plot showing the vaue of x, so if the
generated x is 0.3, graphically the tile should be like "Figure 1: x=0.3".
How do I achieve this in R?
Thanks.
Jeff
--
View this me
I was just trying to say that Ole might know the technical difference
between "NULL" and "ZERO", but it seemed to be rather a linguistic
confusion...
BTW. the subject contains "null", but the body of the mail correctly states
"zero" :-)
2012/7/12 Patrick Connolly
> On Tue, 10-Jul-2012 at 11:19P
Hi,
Here is the summary-output of the Coxph-model I used (the output is based on
the best final model i.e. all significant explanatory variables and their
interactions are included):
coxph(formula = Y ~ LT + Food + Temp2 + LT:Food + LT:Temp2 +
Food:Temp2 + LT:Food:Temp2)
n= 555
Tena koe Pragya
It is advisable to use the extractor functions where they are available; i.e.,
coef(j) and coef(j)[1] or coef(j)['(Intercept)'] for the intercept.
However, j$coefficients["(Intercept)"] does give you what I suspect you want.
Check its structure:
str(j$coefficients["(Intercept)"
Hi everyone,
I am fitting a simple linear regression model in R. My
command is j=lm( Y ~ Sex + begsal + time + int)
Call:
lm(formula = Y ~ Sex + begsal + time + int)
Coefficients:
(Intercept) Sex begsal time int
191.916 -241.8053.96
what would be nice is to understand the amount of cpu time vs. the elapsed time
consumed. you might be paging if you don't have sufficient real memory. I
would venture a guess that the cpu time is very similar and the differnce is
elapsed time due to some contention like memory.
Sent from my
And I guess I would also either leave out the vertical axis line
or add box(bty="l") after the axis(1, ) call.
Peter Ehlers
On 2012-07-12 19:48, Peter Ehlers wrote:
On 2012-07-12 18:39, David L Carlson wrote:
Sorry about that. I got rid of the box another way and then switched to using
pa
On 2012-07-12 18:39, David L Carlson wrote:
Sorry about that. I got rid of the box another way and then switched to using
pars=
without making sure it worked. This works:
flies$group <- factor(flies$group, 5:1) # 1
levels(flies$group) <- paste0("Group ", 5:1) # 2
oldpar <- par(bty="n")
boxplot
Sorry about that. I got rid of the box another way and then switched to using
pars=
without making sure it worked. This works:
flies$group <- factor(flies$group, 5:1) # 1
levels(flies$group) <- paste0("Group ", 5:1) # 2
oldpar <- par(bty="n")
boxplot(long ~ group,
data = flies,
Added your code:
flies <- read.table("example12_1.dat",header=TRUE,sep="\t")
flies$group <- factor(flies$group,5:1)
levels(flies$group) <- paste0("Group ",5:1)
boxplot(long ~ group,
data = flies,
pars = list(las=1, ylim=c(10,110), xaxt="n", bty="n"),
horizontal = TRUE,
If you read the posting guide you will see that only a few types of attachments
will be allowed through. Also it is much preferred to paste data into your
email using dput:
> dput(flies)
structure(list(long = c(40L, 37L, 44L, 47L, 47L, 47L, 68L, 47L,
54L, 61L, 71L, 75L, 89L, 58L, 59L, 62L, 79L
Looks like data was not attached. Here is is.
longgroup
40 1
37 1
44 1
47 1
47 1
47 1
68 1
47 1
54 1
61 1
71 1
75 1
89 1
58 1
59 1
62 1
79 1
96 1
58 1
62 1
70 1
72 1
74 1
96
The example you gave had only one split. If your real situation has three
splits, you'll have to take a look at testtree$csplit matrix and decide
how you want to define the new grouping variable. Here's one way to do it
...
Jean
library(rpart)
library(rpart.plot)
test_set <- data.frame(
Hi,
I managed to use the attached data set and figure out the following:
flies <- read.table("example12_1.dat",header=TRUE,sep="\t")
boxplot(long ~ group,
data = flies,
horizontal = TRUE,
col = "red")
I'm very new to R and would like some help with the following:
1. Chan
Hello,
I've not been following this thread but this seems ndependent from
previous posts. Try the following.
url <- "http://r.789695.n4.nabble.com/file/n4636337/BR3_2011_New.csv";
tUnitsort <- read.csv(url, header=TRUE)
cols <- sapply(c("Date", "Hour", "BlockNumber", "MyTo"), function(x)
On Thu, 12 Jul 2012, Lee, Laura wrote:
Hi all!
I have built a model to predict interactions with turtles and the model
includes an offset for effort:
ZIP<-zeroinfl(Sturgeon~fMesh+fSeason+offset(LogEffort),dist="poisson",link="logit",data=data)
Note that this includes the offset both in the
I wrote a little function called first() to help with situations like
this. It returns a 1 every time an element of a vector is different from
the previous element, and a 0 otherwise.
first <- function(x) {
L <- length(x)
c(1, 1-(x[-1]==x[-L]))
}
sd <- 1
residuals <- c(1, 2.1, 3, 4
I have independent event sequences for example as follows :
Independent event sequence 1 : A , B , C , D
Independent event sequence 2 : A, C , B
Independent event sequence 3 :D, A, B, X,Y, Z
Independent event sequence 4 :C,A,A,B
Independent event sequence 5 :B,A,D
I want to able to find
Thank you all for your help.
Felipe D. Carrillo
Supervisory Fishery Biologist
Department of the Interior
US Fish & Wildlife Service
California, USA
http://www.fws.gov/redbluff/rbdd_jsmp.aspx
>
>From: Gabor Grothendieck
>To: Felipe Carrillo
>Cc: "r-help@r-proj
If you don't understand the methodology, why are you using it??
Use the glmnet package instead and read the supporting documents. If it's
too technical for you, consult a statistician or do something else.
I do NOT do brain surgery.
-- Bert
On Thu, Jul 12, 2012 at 1:04 PM, SHIR SHIRY wrote:
>
Dear all
I am using lars package to do lasso in R. I dont undesrtand what max.steps
do?and how I can understand from the outputs to obtain the last steps in this
packagethanks for your helpbest
[[alternative HTML version deleted]]
__
R-help@r-
P.S.: I should have mentioned: The operating system is Windows XP.
--
View this message in context:
http://r.789695.n4.nabble.com/Two-R-sessions-on-multicore-computer-seem-to-inhibit-each-other-tp4636336p4636355.html
Sent from the R help mailing list archive at Nabble.com.
__
just do this:
colnames(r)<-substr(colnames(r),2,nchar(colnames(r)))
This will remove the X.
Later when you want to use the headed to plot something, cast it as numeric:
plot(colMeans(r)~as.numeric(colnames(r)))
-
Yasir Kaheil
--
View this message in context:
http://r.789695.n4.nabble.com/re
To add to Gabor's remarks:
This has nothing to do per se with R -- it is your insufficient
understanding of the underlying mathematical issues.
It is pretty trivial to do the plinear algorithm by hand. Fit linear
regressions
weight ~ lm(weight ~z)
where z is exp(gamma*week) for a suitable sequen
Another new question:
I want to be able to subset the data based on whether or not that data point
was recorded on a holiday. The is.holiday() function from the chron package
would be perfect for this. However, when I try it, the following happens
(I'm also using the timeDate package):
> holida
Hi,
Try this:
dat1<-read.table(text="
2.5 3.6 7.1 7.9
100 3 4 2 3
200 3.1 4 3 3
300 2.2 3.3 2 4
",sep="",header=TRUE)
#Either
colnames(dat1)<-c("2.5","3.6","7.1","7.9")
#or
colnames(dat1)<-c(2.5,3.6,7.1,7.9)
#produce character column names
is.character
Hi all!
I have built a model to predict interactions with turtles and the model
includes an offset for effort:
ZIP<-zeroinfl(Sturgeon~fMesh+fSeason+offset(LogEffort),dist="poisson",link="logit",data=data)
I wasn't clear about one aspect of the response to a similar question I
recently posted..
On Tue, 10-Jul-2012 at 11:19PM +0200, Erdal Karaca wrote:
|> german "Null" == english "zero" :-)
German "Gift" == English "poison" :-(
|>
|> 2012/7/10 Rolf Turner
|>
|> >
|> >
|> > In addition to taking cognisance of Richard Heiberger's reply you
|> > should also learn to distinguish b
On Thu, Jul 12, 2012 at 3:40 PM, Felipe Carrillo
wrote:
> I get a different error now:
>> nls(weight ~ cbind(1, exp(gamma*week)), weightData, start = list(gamma=
>> 0.2), alg = "plinear")
> Error in nls(weight ~ cbind(1, exp(gamma * week)), weightData, start =
> list(gamma = 0.2), :
> step fac
I get a different error now:
> nls(weight ~ cbind(1, exp(gamma*week)), weightData, start = list(gamma=
>0.2), alg = "plinear")
Error in nls(weight ~ cbind(1, exp(gamma * week)), weightData, start =
list(gamma = 0.2), :
step factor 0.000488281 reduced below 'minFactor' of 0.000976562
The help
On Thu, Jul 12, 2012 at 3:02 PM, Felipe Carrillo
wrote:
> Thanks Bert, I increased the number of iterations:
>
> M_model <- nls(weight ~ alpha +
> beta*exp(gamma*week),control=nls.control(maxiter=200), weightData,
> start = c(alpha = 0.0, beta = 1, gamma = 0.2), trace = TRUE)
>
> Bu
I can reproduce the errors. I'll take a look.
Thanks,
Max
On Thu, Jul 12, 2012 at 5:24 AM, Dominik Bruhn wrote:
> I want to use the caret package and found out about the timingSamps
> obtion to obtain the time which is needed to predict results. But, as
> soon as I set a value for this option,
Thanks Bert, I increased the number of iterations:
M_model <- nls(weight ~ alpha +
beta*exp(gamma*week),control=nls.control(maxiter=200), weightData,
start = c(alpha = 0.0, beta = 1, gamma = 0.2), trace = TRUE)
But now the 'start' argument seems to be the problem.
Looking at the
HI,
Much more simplified code that my previous one.
a<-list("abc","def","ghi", "jkl", "mno", "pqr")
paste(a[],sep=" ")
[1] "abc" "def" "ghi" "jkl" "mno" "pqr"
A.K.
- Original Message -
From: purushothaman
To: r-help@r-project.org
Cc:
Sent: Thursday, July 12, 2012 8:22 AM
Subject:
I have a text file like this
2.5 3.6 7.1 7.9
100 3 4 2 3
200 3.1 4 3 3
300 2.2 3.3 2 4
I used "r <- read.table("a.txt", header=T)"
The row names becomes X2.5, X3.6... What I need is the row names are
numeric, so I can use the row names as number
Thank you,
I am sorry but I am still trying to figure out how to make the function
work.
I have a column called tUnitsort$BlockNumber which can range from 0 to 6.
I have another two columns with the date and the hour ending for the given
day.
Example
DateHour BlockNumber MyTo New
I have a spatial salinity field s and a model g(s) ~ Xb where the X comes from
slightly modified GAM basis functions.
I am trying to deal with the following set of requirements:
1. The underlying physics are linear, and plain salinity (the identity link) is
the correct response to my covariates.
Hi Gabor,
I updated my RStudio installation to 0.96.316 (which did not modify my R build
of 2.14.0) and that did it. Thanks for the help!
David Marx
Please note our office has moved:
David Marx | Data Analyst Specialist, Claims Dept | SoundExchange, Inc.
733 10th Street, NW | 10th Floor | Was
Hi William,
Glad to know that.
So, I guess "difftime()" did the trick if both befong to the class "Date".
A.K.
From: William Mabe
To: arun
Sent: Thursday, July 12, 2012 1:12 PM
Subject: Re: [R] unable to subtract dates in R
Thanks Arun! That worked.
On
I am trying to write a loop to forecast realized volatility over successive
days for the purpose of VaR prediction using the HAR-RV-CJ model which is as
follows:
log(RV_t+1) = β_0 + β_CD log(CV_t) + β_CW log(CV_t-5) + β_CM
log(CV_t-22) + β_JD log(J_t + 1) + β_JW log(J_t-5 + 1) + β_JM log(J_t-2
I have a graph of residuals and I am attempting to get a list of the indexes
of each time the residual is greater than 2 standard deviations or less than
-2 standard deviations, but only the first point of the section. And then
I'd also need the first point where the point returns to the range betw
I am trying to write a loop to forecast realized volatility over successive
days for the purpose of VaR prediction using the HAR-RV-CJ model which is as
follows:
log(RV_t+1) = β_0 + β_CD log(CV_t) + β_CW log(CV_t-5) + β_CM
log(CV_t-22) + β_JD log(J_t) + β_JW J_t-5 + β_JM J_t-22 + e_t
where RV
Hi,
Try this:
dat1<-read.table(text="
Name Age
Angel 20
Amelia 20
Bernard 19
Stephanie 20
Vanessa 22
Angeline 23
Camel 21
",sep="",header=T
Thanks Simon! I can use the functional approach to bridge the gap between
this and more traditional forms of data assimilation.
At the moment, though, what I am doing is related to your work with soap
films, in which your response comes from linearly combine solutions of a PDE
evaluated at the po
Hi!
Would be grateful if somebody helped me understand this error message after
trying to run a panel data:
PanelData <- read.xls("/Users/Bahman/Desktop/Taylor
rule/Data/PanelData2.xls")
attach(PanelData)
# Panel Data regresson for all countrys.
> Answer <- plm(NOM.INT.RATE ~ INFL + TARGET.INF
Hello!
I'm doing a svar and when I make the estimation the next error message
appears:
In SVAR(x, Amat = amat, Bmat = bmat, start = NULL, max.iter = 1000, :
The AB-model is just identified. No test possible.
Could you help me to interpret it please.
Also I have the identification assumption
Dear R-helpers,
I am puzzled by the following observation:
On my home dual core Windows desktop computer, I am used to running two R
sessions in parallel. These do very well in using the full CPU of the
computer (half each) and don't seem to slow each other down.
Today I have started some large
Hi,
I'm trying to run a permutation test on paired samples.
First I tried the package "exactRankTests":
require("exactRankTests")
x <- c(1.83,0.50,1.62,2.48,1.68,1.88,1.55,3.06,1.30)
y <- c(0.878,0.647,0.598,2.05,1.06,1.29,1.06,3.14,1.29)
wilcox.test(x,y,paired = TRUE,alternative = "gre
try this:
> readcsv<-function(filepath)
+ {
+ output<-read.csv(filepath)
+ }
> readcsv
function(filepath)
{
output<-read.csv(filepath)
}
> x <- capture.output(readcsv)
> x
[1] "function(filepath)" "{"
"output<-read.csv(filepath)"
[4] "}"
>
On Thu, Jul 12, 2012 at 5:00 AM, purushothaman
Read the Help file!
?nls ## Note the "control" argument
?nls.control
-- Bert
On Thu, Jul 12, 2012 at 10:47 AM, Felipe Carrillo
wrote:
> Hi:
> Using nls how can I increase the numbers of iterations to go beyond 50.
> I just want to be able to predict for the last two weeks of the year.
> Thi
See ?nls.control (referenced from ?nls).
On 12/07/2012 18:47, Felipe Carrillo wrote:
Hi:
Using nls how can I increase the numbers of iterations to go beyond 50.
I just want to be able to predict for the last two weeks of the year.
This is what I have:
weight_random <- runif(50,1,24)
Hi:
Using nls how can I increase the numbers of iterations to go beyond 50.
I just want to be able to predict for the last two weeks of the year.
This is what I have:
weight_random <- runif(50,1,24)
weight <- sort(weight_random);weight
weightData <- data.frame(weight,week=1:50)
If you are going to be doing a lot of this then you might want to
consider using logspline density estimates (logspline package) instead
of kernel density estimates.
On Wed, Jul 11, 2012 at 8:33 AM, firdaus.janoos wrote:
> Hello,
>
> I wanted to know if there is a simple way of getting the invers
If you want something other than an arrow (or an arrow that looks
different from those produced by the arrows function) then look at the
my.symbols function in the TeachingDemos package.
On Mon, Jul 9, 2012 at 9:37 PM, Manish Gupta wrote:
> Hi,
> I am working on stacked bar plot and want to add m
You can use the locator function to retrieve the user coordinates of a
point that you click on in the plot, then use those points with
rasterImage to add the image. So replace the last 2 lines of
Michael's answer with something like:
> barplot(VADeaths, border = "dark blue")
> tmp <- locator(1)
>
It is not clear to me what you want in the lattice context, but perhaps
?print.trellis
may be relevant.
Note that "inner" and "outer" margins are not directly translatable in
lattice, as there are multiple levels of plots involved, all determined by
viewport contexts. Ergo my confusion about wh
Dear R users,
I have a lot of experience with traditional R graphics, but I decided to turn
to trellis as
it was recommended for spatial graphs by the sp package. In traditional R
graphics
I always first set the size of the device region absolute units (e.g. mm) and
then I
firmly fix the inne
Hi David,
please make sure that you have the 0.6 version of igraph installed.
You might need to upgrade your R installation to install the 0.6
version of igraph.
Best,
Gabor
On Wed, Jul 11, 2012 at 10:36 AM, David Marx wrote:
> Hi,
>
> I've installed the igraph package and have been otherwise u
Hello,
There's a package, lavaan, that implements FIML as an option of function
sem(). I have never used it, though, so I can't say much about it.
Hope this helps,
Rui Barradas
Em 12-07-2012 16:20, John Fox escreveu:
Dear Joshua,
If I understand correctly what you want to do, the sem packa
Dear Joshua,
If I understand correctly what you want to do, the sem package won't do it.
That is, the sem() function won't do what often is called FIML estimation
for models with missing data. I've been thinking about implementing this
feature, and don't think that it would be too difficult, but I
?deparse
And please include context -- relatively few of us use Nabble.
Michael
On Jul 12, 2012, at 4:16 AM, purushothaman wrote:
> Hi,
>
> readLines read a file and retrun as vector.
>
> i need to read particular function in Rfile.
>
> Thanks
> B.Purushothaman
>
> --
> View this message i
You could read the .png (or some other formats) in then use the
rasterImage function to put that into the current plot.
On Mon, Jul 9, 2012 at 8:25 PM, Jie Tang wrote:
> hi R-users
>Now I have a figure in emf or png or tiff format that have been drawn
> by other tool and I want to insert thi
On Thu, Jul 12, 2012 at 9:41 AM, scstrein wrote:
> Thanks! For some reason, example 5 in that documentation doesn't work for me.
> It runs the query, but I'm getting zero results. If I substitute the
> variable with the actual value, I do get the results I want. Is there a
> reason for this?
Re
?subset
John Kane
Kingston ON Canada
> -Original Message-
> From: jubil...@live.com.sg
> Sent: Wed, 11 Jul 2012 20:53:51 -0700 (PDT)
> To: r-help@r-project.org
> Subject: [R] plot only a same variable timing graph
>
> hi all.
> for example :
>
> Table 1 variable:
> BERTH TIME
I really am not sure of the question but perhaps ?order for a start?
John Kane
Kingston ON Canada
> -Original Message-
> From: jubil...@live.com.sg
> Sent: Thu, 12 Jul 2012 01:15:26 -0700 (PDT)
> To: r-help@r-project.org
> Subject: [R] plot graph by first letter
>
> Hi all, may i know i
I think that you have the statements out of order and I know that you are
lacking the last ) to match the replicate(
Try this"
replicate(100, {
x=rep(1:10,10)
y=rnorm(100,x,5)
plot(y~x)
abline(lm(y~x))
})
John Kane
Kingston ON Canada
> -Original Message-
> From: aldoushuxle...@gmail.c
Duncan has given a indication of why nls() has troubles, and you have found a
way to work
around the problem partially. However, you may like to try nlmrt (from R-forge
project
R-forge.r-project.org/R/?group_id=395
It is intended to be very aggressive in finding a solution, and also to deal
wi
Hi,
I have a question on testing spatial autocorrelation on raster data
including NA-values. In particular, I like to calculate Moran´s I and
Geary´s C indices by using inverse distance weighting matrices.
Calculating Moran´s I with moran.test works fine, because the function
contains the o
Thanks! For some reason, example 5 in that documentation doesn't work for me.
It runs the query, but I'm getting zero results. If I substitute the
variable with the actual value, I do get the results I want. Is there a
reason for this?
--
View this message in context:
http://r.789695.n4.nabbl
Post on the r-sig-mixed-models list, not here.*
-- Bert
* Or consult a local statistician, as your problem appears to be
insufficient statistical knowledge, rather than R-related. In particular,
fitting a (mixed effect) linear model without an intercept is almost always
unwise.
On Thu, Jul 12, 2
... and you are unlikely to get a helpful reply until you follow the
posting guide and post code that shows what you did. Unless there's a
claiRvoyant package out there somewhere to figure it out.
-- Bert
On Thu, Jul 12, 2012 at 3:43 AM, yan wrote:
> Dear R users,
>
> if all my numerical variab
Hi Manning,
There are two obvious mistakes:
- close the brackets for replicate() and
- do the things in the correct order
replicate(100, {
x=rep(1:10,10) ## first define x
y=rnorm(100,x,5) ## and then y because it depends on x
plot(y~x) ## then plot, because it depends on x a
Hi Jean,
If we convert the NEWMatrix to dataframe,then headers do appear.
NewMatrix1<-data.frame(NewMatrix)
NewMatrix1
X1 X2 X3 X4
SumMatrix SUM 57 64 65
CountMat COUNT 3 3 3
#assigning headers to NULL
colnames(NewMatrix1)<-NULL
NewMatrix1
NA NA NA NA
SumMatr
Hi,
If I understand it correctly, probably this is what you want:
a<-list("abc","def","ghi")
a1<-unlist(a)
paste(a1[1],a1[2],a1[3],sep=" ")
[1] "abc def ghi"
A.K.
- Original Message -
From: purushothaman
To: r-help@r-project.org
Cc:
Sent: Thursday, July 12, 2012 8:22 AM
Subject: R
Hello,
The reason you are only getting one row is because each time you are replacing
the NEWMatrix with new output.
Try this:
Oldmatrix<-read.table(text="
X1 X2 X3
22 24 23
25 27 27
10 13 15
",sep="",header=TRUE)
NewMatrix1<-rbi
Hello,
I am using read.csv.sql first time for reading the large data file.If I am
ran this code that showns warning “closing unused connection”.
Is it I am missing any argument from my command or how to comeout from this
warning?.
R code:-
Library(sqldf)
ip_dir_path <- “D:/BharatWarule/big_dat
Thanks jean.
From: Jean V Adams [via R]
[mailto:ml-node+s789695n4636289...@n4.nabble.com]
Sent: Thursday, July 12, 2012 6:14 PM
To: Akkara, Antony (GE Energy, Non-GE)
Subject: Re: Add row into a Matrix witout headers from Function
Try using the rbind() function to combine the two vectors f
Hi!
I would like to post the following question:
I was trying to figure out how to do the simulation shown in Fig 10.6 of
John Verzani's book 'Using R for Intro Statistics'. It is on page 290, with
a description on the previous page. It seems like a simple thing... Just
needing to duplicate a pro
On Thu, Jul 12, 2012 at 05:22:45AM -0700, purushothaman wrote:
> hi,
>
> sorry it's not 2 different list all item in same list like this
>
> a[1]="abc"
> a[2]="def"
> ...
> output ="abc def ..."
Hi.
Try this
a <- list("abc", "def", "ghi")
paste(a, collapse=" ")
[1] "abc def ghi"
Hope t
Hi,
What you did is equivalent to, but more complicated (and slower as well
I guess) than:
paste(a, collapse="")
Ivan
--
Ivan CALANDRA
Université de Bourgogne
UMR CNRS/uB 6282 Biogéosciences
6 Boulevard Gabriel
21000 Dijon, FRANCE
+33(0)3.80.39.63.06
ivan.calan...@u-bourgogne.fr
http://biogeo
Try this...
a<-c("abc","def","ghi","klm","nop","qrs","tuv","wxyz")
b<-data.frame()
for (i in 1:length(a)){
b<-paste(b,a[i],sep="")
}
print(b)
Thanks
Arun
---
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S
Try using the rbind() function to combine the two vectors from colSums()
into a matrix. Then assign row names and get rid of column names using
the dimnames() function. For example:
OLDMatrix <- matrix(c(22, 25, 10, 24, 27, 13, 23, 27, 15),
ncol=3, dimnames = list(NULL, c("X1", "X2",
Gary Dong gmail.com> writes:
>
> Dear R users,
>
> I have created a Loess surface in R, in which x is relative longitude by
> miles, y is relative latitude by miles, and z is population density at the
> neighborhood level. The purpose is to identify some population centers in
> the region. I'm w
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