Hi I am using Version R 2.15 and I haven't been able read html table. Following
is my code and error message.
Error in htmlParse(doc) :
error in creating parser for
http://en.wikipedia.org/wiki/Brazil_national_football_team
theurl <- "http://en.wikipedia.org/wiki/Brazil_national_football_team
Dear all,
When I use a unicode symbol in the labels for a factor object, the
corresponding level does not display as expected. However, using levels() on
the factor returns the desired output. I noticed the discrepancy when the
legend labels from a call to ggplot() did not display the desired s
0 down vote favorite
I am working on R and trying to draw a clock using a pie chart.
code:
pie(c(25,20,15,10,10,30),labels = c(1,2,3,4,5,6,7,8,9,10,11,12),
col=rainbow(length(lbls)), clockwise = TRUE, init.angle = 90)
but i need all 12 labels to be there independent of no of segments i
Hi,
i am overlapping 2pie chart with same radius.
1. with labels
2. without labels.
Output
pie chart 2 with label of first.
http://r.789695.n4.nabble.com/file/n4639723/Ist_Pie.png
http://r.789695.n4.nabble.com/file/n4639723/IInd_Oie.png
output:
http://r.789695.n4.nabble.com/file/n4639
I am not allowed to connect to Nabble from work, sorry.
Petr
>
> hi peter the pdf folder is in
> http://r.789695.n4.nabble.com/file/n4639434/aj.pdf
>
>
>
> --
> View this message in context: http://r.789695.n4.nabble.com/help-to-
> program-my-function-tp4639434p4639629.html
> Sent from the R
Hi
>
> > Specifically, since it has only a single detection indicator column
> > (ceneq1), it implies that within any single sample either all the
analytes
> > were detected, or all were not. Not what I would expect.
>
> Don,
>
>I have been thinking about this and wondered whether the cast
thank you for your help.
my input data looks like this (tab separated):
Ind.nr. Pop.nr. scm266 rms1280 scm247 rms1107
1 101 305 318 222 135
1 101 305 318 231 135
2 101 305 313 999 96
2 101 305 321 999 130
3
On 08/09/2012 03:43 PM, Manish Gupta wrote:
0 down vote favorite
I am working on R and trying to draw a clock using a pie chart.
code:
pie(c(25,20,15,10,10,30),labels = c(1,2,3,4,5,6,7,8,9,10,11,12),
col=rainbow(length(lbls)), clockwise = TRUE, init.angle = 90)
but i need all 12 labe
On Aug 9, 2012, at 06:53 , Wyatt, Kristin M wrote:
> Dear all,
>
> When I use a unicode symbol in the labels for a factor object, the
> corresponding level does not display as expected. However, using levels() on
> the factor returns the desired output. I noticed the discrepancy when the
> le
Hi,
Part of my program is to calculate the number of time series in a zoo
object. It works well if it has more than one time series, but it fails if
it has only one. How can I access the number of column (i.e. the number of
time series) when I have only one column? Why is the number of an objec
On 08/09/2012 04:28 PM, Manish Gupta wrote:
Hi,
i am overlapping 2pie chart with same radius.
1. with labels
2. without labels.
Output
pie chart 2 with label of first.
http://r.789695.n4.nabble.com/file/n4639723/Ist_Pie.png
http://r.789695.n4.nabble.com/file/n4639723/IInd_Oie.png
output:
On 09/08/12 14:35, Yihui Xie wrote:
The Department of Redundancy Department can "recompensate"...
Ye :-)
cheers,
Rolf
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do re
SORRY it should be:
Fridolin wrote
>
> for (s in 3:6) { #walks though the matrix colum by colum, starting at
> colum 3
> for (z1 in 1:5) { #for each current colum, take one row (z1)...
> for (z2 in 1:5) { #...and compare it to another row (z2) of the
> current colum
>
error is go
On 08/09/2012 04:52 AM, array chip wrote:
Hi, does anyone know how to make decimal points midline when plotting? The
journal to which we are going to submit a manuscript require this particular
formatting, and it gives direction how to do this on PC: hold down ALT key and
type 0183 on the numb
On Thu, 9 Aug 2012, jpm miao wrote:
Hi,
Part of my program is to calculate the number of time series in a zoo
object. It works well if it has more than one time series, but it fails if
it has only one. How can I access the number of column (i.e. the number of
time series) when I have only one
On 08/09/2012 07:17 AM, Rolf Turner wrote:
...
P. S. There is no such word as "recompensate".
Thanks to R, the internet and pretentious, semi-literate public servants
around the world, there probably is now.
Jim
__
R-help@r-project.org mailing list
Hello,
If I understand it well, the following should do it.
dd <- subset(f, sold == 1)
at <- with(dd, bid == purchase)
mean(at) # wanted value
Hope this helps,
Rui Barradas
Em 09-08-2012 01:52, Daisy Englert Duursma escreveu:
Your question is very unclear.Can you provide a better question
all problems solved. thank you for your help!
for the sake of completeness, here my solution:
#1) read in data:
daten<-read.table('K:/Analysen/STRUCTURE/test.txt', header=TRUE, sep="\t")
daten<-as.data.frame(daten)
#2) create empty matrix:
indxind<-matrix(0,nrow=617, ncol=617)
#indxind[1:20,1:19]
Good Day!
I have RasterLayers list with same resolution. (31 items: 720 x 1440) . I
need to get RasterLayer with average values without NA value. How I can
do that?
For example:
We have 3 RasterLayers
1 2 34
2 3 NA 3
NA 1 3NA
1 2 34
2NA NA 3
NA 1 3
As Uwe Ligges-3 guessed, i am working on a Windows Station.
I also tried the link Uwe Ligges-3 posted and I think it worked
Thank you all for your answers!
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http://r.789695.n4.nabble.com/Installing-HDF5-package-tp4639633p4639728.html
Sent from the R help mailin
Dear all,
I have a question about how the function "cm" works? It can be found under
the package actuar.
In particular, I want to use it to apply the Buhlmann-Straub model into some
data.
I am using the example that R has in the help page for cm.
I use the dataset hachemeister and then the code
Thanks David, but this is not easy to me as these string are already in my
data (labels, variable strings). Well, I still can make some script to do
it, but it's messy, maybe I want make a Word file (not a html through Word)
after this. When a HTML page has a UTF-8 charset it works ok with characte
First: Im sorry for my bad english ;-)
I use "portfolio" to create treemaps with "R".
I've got a .csv dataset of all the schools in my city, containing 6
variables each with 1469 lines of data.
When I create a treemap, 3 of these schools were not taken into account by
"R". I dont know why. It do
I'm puzzled by the behaviour of factors in rma models, see example and
comments below. I'm sure there's a simple explanation but can't see it...
Thanks for any input
John Hodgson
- code/selected output -
library(metafor)
##Set up data
Great! It works fine for me. But i have one query. In my clock i have only 10
points so i am using as follows.
pie(c(25,20,15,10,10,30))
floating.pie(0,0,c(25,20,15,10,10,30))
pie.labels(0,0,seq(0,19*pi/10,by=pi/5),c(3,2,1,0,9,8,7,6,5,4)*10,radius=1.1,border=NA)
But format is not correct. I tri
HI everybody!
Has anybody figure out how would be possible to plot several ctree plots
beside each other?
I would really appreciate it
My idea would be like this, but the par() function seem to be lower in
hierarchy
Thanks in advance
#
libra
Please help me to get a function for Kolmogorov distribution.
I've looked through the ks.test code and found that Kolmogorov distribution
is calculating in the following way:
pkstwo <- function(x, tol = 1e-06) {
if (is.numeric(x))
x <- as.double(x)
els
On 12-08-08 9:21 PM, Hillary Ward wrote:
I'm having problems creating an axis label for a plot.
y_label = expression(paste(plain('CPUE
'),plain('(fish'),plain('x'),plain('h'^{-1}),plain(')')))
I'd like to replace the "x" with an interpunct symbol (dot). Any
suggestions how to do this?
The int
On 08/09/2012 08:58 PM, Manish Gupta wrote:
Great! It works fine for me. But i have one query. In my clock i have only 10
points so i am using as follows.
pie(c(25,20,15,10,10,30))
floating.pie(0,0,c(25,20,15,10,10,30))
pie.labels(0,0,seq(0,19*pi/10,by=pi/5),c(3,2,1,0,9,8,7,6,5,4)*10,radius=1.1,
Hi
> thank you for your help.
>
> my input data looks like this (tab separated):
>
> Ind.nr. Pop.nr. scm266 rms1280 scm247 rms1107
> 1 101 305 318 222 135
> 1 101 305 318 231 135
> 2 101 305 313 999 96
> 2 101 305 321 999 130
> 3 101 305 32
Hello,
You have three coordinates but the problem description only envolves two
of them, x and y. The code below is valid for any number of dimensions.
min.dist <- function(p, coord){
which.min( colSums((t(coord) - p)^2) )
}
# Use set.seed to have reproducible simulations
set.seed(1)
tes
Hi
> all problems solved. thank you for your help!
> for the sake of completeness, here my solution:
> #1) read in data:
> daten<-read.table('K:/Analysen/STRUCTURE/test.txt', header=TRUE,
sep="\t")
> daten<-as.data.frame(daten)
not needed, daten is already data frame
I've never seen this, and have no idea how to reproduce it.
For resloution you are going to have to give me a working example of the
failure.
Also, per the posting guide, what is your sessionInfo()?
Terry Therneau
On 08/09/2012 04:11 AM, r-help-requ...@r-project.org wrote:
I have a couple of
Hi,
You may want to check grid.arrange in gridExtra package.
Weidong
On Thu, Aug 9, 2012 at 7:36 AM, rodrock wrote:
> HI everybody!
>
> Has anybody figure out how would be possible to plot several ctree plots
> beside each other?
> I would really appreciate it
> My idea would be like this, but
Hi All,
My question is more academic than technical.
The question is about interpretation of Poisson curve. I am using it to assess
scale-free nature of degree distribution. Can you please briefly describe what
is meant by the Poisson curve when you give it your range of k-neighbour data
whic
There is a simple explanation:
1) The command:
factor(ftype)
does not actually turn 'ftype' permanently into a factor, since you are not
re-assigning it back to the object 'ftype'. You have to use:
ftype <- factor(ftype)
2) If you want to use the formula interface for specifying moderators, y
> -Original Message-
> From: j...@bitwrit.com.au
> Sent: Thu, 09 Aug 2012 19:55:29 +1000
> To: rolf.tur...@xtra.co.nz
> Subject: Re: [R] R versus SAS
>
> On 08/09/2012 07:17 AM, Rolf Turner wrote:
>> ...
>> P. S. There is no such word as "recompensate".
>>
> Thanks to R, the internet and
Hi David,
Thanks a lot for the reply.
I might not have stated the problem clearly. Let me try again.
Given a set of observations X, I want to find out the estimated density
values for the observations X?
I believe that the values "x" returned from "density" function is not the
observat
Hello all,
I'm just starting to learn R and I heard a good way of doing that was R
Commander. For my work I use a lot of time series, so I installed (and
loaded) R Commander with epack.
When I go to Ts-Models, after loading my data, I click on ARIMA Models tab.
I load my variable D1 Ln Demand (1
HI,
I am using R 2.15 in Ubuntu 12.04. It works fine for me.
library(XML)
theurl <- "http://en.wikipedia.org/wiki/Brazil_national_football_team";
tables <- readHTMLTable(theurl)
A.K.
- Original Message -
From: "Lay, Kiung"
To: "r-help@R-project.org"
Cc:
Sent: Thursday, August 9, 201
Dear All,
I'm starting to use the quantile regression, I would like to know if there is a
way to calculate the fit of the model such as a pseudo R2.
Many thanks.
Best regards.
Vito Ricci
[[alternative HTML version deleted]]
__
R-help@r-proj
On Wed, 8 Aug 2012, Jeff Newmiller wrote:
I took a closer look, and unused factor levels is not the problem... the
problem is defining id variables appropriately.
Jeff,
OK.
1) "sample" is the name of a builtin function, so it is not advisable to use
it as the name of data. I have used "sa
Works fine for me.
sessionInfo()
R version 2.15.1 (2012-06-22)
Platform: i686-pc-linux-gnu (32-bit)
locale:
[1] LC_CTYPE=en_US.UTF-8 LC_NUMERIC=C
[3] LC_TIME=en_CA.UTF-8LC_COLLATE=en_US.UTF-8
[5] LC_MONETARY=en_CA.UTF-8LC_MESSAGES=en_US.UTF-8
[7] LC_P
Hi,
I want to use four independent variables to predict the output of one
dependent variable using a linear model lm. I want to compare all possible
combinations of the 4 independent variables, including singles, pairs and
triples.
I was thinking of using the AIC test to compare all models and pi
You can use approx(d, xout=) (or spline()) on the output of density()
to get density estimates at the points in xout. E.g.,
> X <- c(rnorm(20, -1, 1), rgamma(50,1/2))
> d <- density(X)
> plot(d)
> points(approx(d, xout=-2:2))
Or you could use the functions in package:logspline to fit
I don't think a lot of people here use R Commander so diagnosing something from
there may be difficult.
Probably the first thing to do is to supply us with some sample data. The best
way do do this is usually to use the dput() command and just copy and paste the
output into your email . If
Thanks, I suspected it would be something simple. Best wishes John
On 09/08/2012 14:42, Viechtbauer Wolfgang (STAT)-2 [via R] wrote:
> There is a simple explanation:
>
> 1) The command:
>
> factor(ftype)
>
> does not actually turn 'ftype' permanently into a factor, since you
> are not re-assign
Probably not. See
http://r.789695.n4.nabble.com/Boxplot-Fill-Pattern-td4457209.html on this topic.
What exactly are you doing? There may be a workaround or alternative.
John Kane
Kingston ON Canada
> -Original Message-
> From: meyfa...@uni-potsdam.de
> Sent: Wed, 8 Aug 2012 16:56:05 -
On Aug 9, 2012, at 9:21 AM, John Kane wrote:
> I don't think a lot of people here use R Commander so diagnosing something
> from there may be difficult.
>
> Probably the first thing to do is to supply us with some sample data. The
> best way do do this is usually to use the dput() comman
Hi Uwe,
I’ve tried changing the working directory and made a few tests and it
seems it works perfectly fine. Thank you so much for your suggestion.
I’ve found a minor issue running my code. Although I’m using:
unlink(temporary.folder, recursive=TRUE)
, it seems that many folders are not unlin
Sorry, between x and X I got confused about what you were trying to do. The
quickest route is approx() or approxfun():
set.seed(42)
X <- rnorm(100)
den0 <- density(X, bw="SJ")
XY <- approx(den0$x, den0$y, X)
XY$y will differ slightly from your values (you could double the number of
points in the
On 09.08.2012 13:56, Duncan Murdoch wrote:
On 12-08-08 9:21 PM, Hillary Ward wrote:
I'm having problems creating an axis label for a plot.
y_label = expression(paste(plain('CPUE
'),plain('(fish'),plain('x'),plain('h'^{-1}),plain(')')))
I'd like to replace the "x" with an interpunct symbol (d
Dear Billpete002,
This looks to me like it's possibly a bug in RcmdrPlugin.epack -- that is,
you may have done something wrong, but I'd expect in this case that the
timeseries plug=in would intercept the error and return an informative error
message. You might want to correspond directly with the
> Rather dput(head(my data, 50))
Argh!!! I think that in the third time in three weeks .
John Kane
Kingston ON Canada
> -Original Message-
> From: michael.weyla...@gmail.com
> Sent: Thu, 9 Aug 2012 10:29:07 -0500
> To: jrkrid...@inbox.com
> Subject: Re: [R] R Commander - Time Series
>
Hello everyone,
I have two sets of data, with the following structure:
DataSet1
Location PartSample 1 Sample 2
A 1 value value
A 2 value value
A 3 value value
B
Sorry for the late answer.
You both put me iin the correct way.
Problem is solved
-
Mario Garrido Escudero
PhD student
Dpto. de Biología Animal, Ecología, Parasitología, Edafología y Qca. Agrícola
Universidad de Salamanca
--
View this message in context:
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john benson hotmail.com> writes:
> I've been using gamm4 to build GAMMs for exploring environmental
> influences on genetic ancestry. Things have gone well and I have 2
> very straightforward questions: 1. I've used method=REML. Am I
> correct that this is an alternative method for estimating
On 09.08.2012 10:18, sihemsaad wrote:
As Uwe Ligges-3 guessed, i am working on a Windows Station.
I also tried the link Uwe Ligges-3 posted and I think it worked
Actually, my surname is "Ligges" rather than "Ligges-3". And you failed
to cite the contents of the former part of the thread. P
Hi everyone.
I'm trying to customize a biplot (RDA). I would like to remove all axes
ticks. Therefore, I was using *ann=FALSE* (see following example). However,
it only clear the axis 1 and 2 leaving ticks on axis 3 and 4. Any
suggestions to get rdi of the remaining ticks?
data(varespec)
data(va
Try a look at this:
http://stat.ethz.ch/R-manual/R-patched/library/MASS/html/stepAIC.html
Regards,
Phil
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Sent from the R help mailing list archive at Nabble.
On 09/08/2012 11:52 AM, Uwe Ligges wrote:
On 09.08.2012 13:56, Duncan Murdoch wrote:
> On 12-08-08 9:21 PM, Hillary Ward wrote:
>> I'm having problems creating an axis label for a plot.
>>
>> y_label = expression(paste(plain('CPUE
>> '),plain('(fish'),plain('x'),plain('h'^{-1}),plain(')')))
>>
>
Dear R users,
I have a microRNA (similar to microarray) data set with gene expressions and
group variables. I'd like to perform (class) prediction analysis to see which
genes or gene sets predict the group variable well.
I'm at the beginner level. Is there any free software in R or in point/cli
On 09.08.2012 18:05, Duncan Murdoch wrote:
On 09/08/2012 11:52 AM, Uwe Ligges wrote:
On 09.08.2012 13:56, Duncan Murdoch wrote:
> On 12-08-08 9:21 PM, Hillary Ward wrote:
>> I'm having problems creating an axis label for a plot.
>>
>> y_label = expression(paste(plain('CPUE
>> '),plain('(fish'
I might suggest you move this question to the Bioconductor help list.
(And look into bioconductor more generally)
Best,
Michael
On Thu, Aug 9, 2012 at 10:26 AM, Dai, Hongying, wrote:
> Dear R users,
>
> I have a microRNA (similar to microarray) data set with gene expressions and
> group variabl
Perhaps load them both and ?merge can show you the way.
Michael
On Thu, Aug 9, 2012 at 9:54 AM, JenniferH wrote:
> Hello everyone,
>
> I have two sets of data, with the following structure:
>
> DataSet1
> Location PartSample 1 Sample 2
> A 1 value va
Hello,
I have access to my database via command line and through workbench, and
have access on the grant tables:
mysql> SELECT host,user,password,select_priv,insert_priv FROM user;
+--+---+---+-+-+
| host | user | passwor
Thanks for the help - both solutions work fine.
On Thu, Aug 9, 2012 at 9:20 AM, Uwe Ligges
wrote:
>
>
> On 09.08.2012 18:05, Duncan Murdoch wrote:
>
>> On 09/08/2012 11:52 AM, Uwe Ligges wrote:
>>
>>>
>>> On 09.08.2012 13:56, Duncan Murdoch wrote:
>>> > On 12-08-08 9:21 PM, Hillary Ward wrote:
>>
Thank you. I saw these postings, but I don't want to learn lattice for
this reason (was afraid to have to change then everything else in my
graph). Anyway, I now tried with different shades of greyscale (4
shades). I'm not fully satisfied with it, but it's ok. It's for a
publication and dependi
Hello,
Ditto, Windows 7.
> library(XML)
> theurl <- "http://en.wikipedia.org/wiki/Brazil_national_football_team";
> tables <- readHTMLTable(theurl)
>
> sessionInfo()
R version 2.15.1 (2012-06-22)
Platform: i386-pc-mingw32/i386 (32-bit)
locale:
[1] LC_COLLATE=Portuguese_Portugal.1252 LC_CTYPE=Po
You might want to have a look at RColorBrewer. If I remember correctly some
of their palettes should work better than a standard R greyscale but I must
admit I have not tried them.
http://www.decisionstats.com/color-palettes-in-r-using-rcolorbrewer-rstats/
for a start.
John Kane
Kingston ON
Hi Jen,
It's generally best to keep cc'ing R-help so others can lend a hind
when I step away from my computer:
On Thu, Aug 9, 2012 at 11:49 AM, Jennifer Hobbs wrote:
> Hi Michael -
>
> thanks for the advice - I did find merge() just after posting but I'm having
> difficulty with using it. I've
On Aug 8, 2012, at 2:16 PM, greatest.possible.newbie wrote:
Ok I see that point with the quotes. But what I want to do still
doesn't
work:
a <- matrix(1:15,ncol=3)
b <- paste( paste("a[," ,paste(1:3), "]",sep="")
,"^",1:3,sep="",collapse="+")
b
#[1] "a[,1]^1+a[,2]^2+a[,3]^3"
#instead of (wh
On Wed, Aug 8, 2012 at 6:18 PM, Eberle, Anthony wrote:
> I have a question about multiple cores and CPU's for running R. I've
> been running various tests on different types of hardware and operating
> systems (64 bit, 32 bit, Solaris, Linux, Windows, RV.10, .12, .15,
> .15.1.) Generally speakin
It's not too hard to use rect() to add shading to the boxplots. The boxes
are centered on consecutive integers and the width is +/- .4. The boxplot()
function returns the quartiles of each box.
set.seed(42)
DF <- data.frame(val=rnorm(150), grp=rep(letters[1:6], 25))
outstat <- boxplot(val~grp, DF
On Aug 9, 2012, at 8:55 AM, John Kane wrote:
Rather dput(head(my data, 50))
There's an Internet name for the misspelling a correction of a
spelling error. Something like Corollary. This is what I
found, but it's not the name I (don't) remember:
http://rationalwiki.org/wiki/Skitt%27s_
On Wed, Aug 8, 2012 at 10:37 AM, alijk1989 [via R]
wrote:
>
>
> Hi Michael,
>
> Thanks for your response. Here is a simple example of what I am trying to
> do:
>
> w=rep(0.02,10)
> Q=rep(0.02,10)
> rho=matrix(0.5,nrow=10,ncol=10)
> m=10
> LGD=0.45
>
> M1=sum(sapply(1:m,
> function(k){sum(sapply(1:
HI,
I hope this helps you,
set.seed(1)
dat1<-data.frame(X1=rnorm(25,15),X2=rnorm(25,5),X3=runif(25,0.4),X4=rnorm(25,12),Y=rnorm(25,35))
ColNam<-names(dat1)
ColNam
#[1] "X1" "X2" "X3" "X4" "Y"
ColNam<-ColNam[!ColNam %in% "Y"]
n<-length(ColNam)
ColNam
#[1] "X1" "X2" "X3" "X4"
id1<-unlist(
HI,
I forgot about the AIC.
resAIC<-list()
for(i in 1:length(res1)){
resAIC[[i]]<-list()
resAIC[[i]]<-AIC(res1[[i]])
}
unlist(resAIC)
# [1] 71.25981 65.22991 71.32024 71.29489 67.20616 73.15101 73.13823 66.17742
#[9] 66.96219 73.27309 67.78183 68.85621 75.03196 68.00660 69.39852
A.K.
I have a researcher who is running a large R job The job is currently on
3725 iterations. The total number of iterations is 27500. Is there a way to
kill this and recover the data that has been generated thus far?
Thanks,
Eric
--
Eric Kaufmann | Application Support Analyst - Advanced Technolo
Hello Guys,
I would like help in script R when I try to plot the histogram appears the
error:
Erro em hist.default some 'x' not counted; maybe 'breaks' do not span range
of 'x'
I using a file with more 1000 datas (vector)
hist(dados[[1]], seq(0, 20, 0.5), prob=TRUE, xlab="chuva
(mm/dia)",
Hi. I'm using American Housing Survey (AHS) data with replicate weights. I
want subpopulation estimates. When I try to subset the survey design, I get
an error message. I don't get the error message when not using the replicate
weights (using the regular svydesign function).
Any help would be ap
On 09/08/2012 21:13, David Winsemius wrote:
On Aug 9, 2012, at 8:55 AM, John Kane wrote:
Rather dput(head(my data, 50))
There's an Internet name for the misspelling a correction of a spelling
error. Something like Corollary. This is what I found, but
it's not the name I (don't) remember:
Use summary(dados[[1]]). The error message is telling you that the data
include values outside the 0 to 20 range you have specified.
--
David L Carlson
Associate Professor of Anthropology
Texas A&M University
College Station, TX 77843-4352
> -Origi
On Aug 8, 2012, at 6:21 PM, Hillary Ward wrote:
I'm having problems creating an axis label for a plot.
y_label = expression(paste(plain('CPUE
'),plain('(fish'),plain('x'),plain('h'^{-1}),plain(')')))
I'd like to replace the "x" with an interpunct symbol (dot). Any
suggestions how to do this?
On 12-08-09 3:41 PM, Eric Kaufmann wrote:
I have a researcher who is running a large R job The job is currently on
3725 iterations. The total number of iterations is 27500. Is there a way to
kill this and recover the data that has been generated thus far?
Not likely, unless the process is writi
I'm still not fully understanding exactly how R is handling data frames, but am
getting closer. Any help with this one will likely go a long way in getting me
there. Let's say I have a data frame, let's call it "a". Within that data frame
i have two variables, let's call them "b" and "c", where
Hi,
I have a dataframe (try.1) with date/time and temperature columns, and the
date/time is in POSIXct fomat. Sample included below.
I would like to to try decompose () or stl() to look at the trends and
seasonality in my data, eventually so that I can look at autocorrelation.Â
The s
Have you tried: options(digits=9)
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
Behalf Of Rich Shepard
Sent: Wednesday, 8 August 2012 5:51a
To: r-help@r-project.org
Subject: [R] Setting Number of Displayed Digits
Some chemicals have co
Hi,
After running a regression on a factor variable, summary() reports the
coefficients 'nicely,' ie, labelled with a string that is a concatenation
of the variable name and the factor label.
However, changing the base case a la
contrasts(variable)<-contr.treatment(N, base=x)
results in the coeff
You have not defined a data frame since data frames cannot contain lists,
but lists can contain data frames so you are asking about how to process a
list. I'm changing your object names to a, b, and d because c() is the
concatenation function and it can cause all kinds of problems to use it as
an o
On Thu, Aug 9, 2012 at 3:30 PM, Mary Ann Middleton wrote:
>
> Hi,
>
> I have a dataframe (try.1) with date/time and temperature columns, and the
> date/time is in POSIXct fomat. Sample included below.
>
> I would like to to try decompose () or stl() to look at the trends and
> seasonality in my
I would like to plot some lat-lon data in a filled contour, and then
overlay a map of the globe on top. I am trying something like this:
filled.contour(lons, lats, glb.data,
plot.axes={axis(1);axis(2);map(projection='rectangular',parameters=0,add
=T)}
)
The filled contour plots fine, and
On Aug 9, 2012, at 2:43 PM, David L Carlson wrote:
You have not defined a data frame since data frames cannot contain
lists,
Not true:
> dput(a)
structure(list(b = c(1988, 1989),
c = list(c(1985, 1982, 1984),
c(1988, 1980))), .Names = c("b", "c"))
> a
On 12-08-09 6:14 PM, Cable, Sam B Civ USAF AFMC AFRL/RVBXI wrote:
I would like to plot some lat-lon data in a filled contour, and then
overlay a map of the globe on top. I am trying something like this:
filled.contour(lons, lats, glb.data,
plot.axes={axis(1);axis(2);map(projection='rectangular
Amazing. You have to create the data frame and then add a variable
containing the list to keep R from checking the number of rows and
objecting:
This does not work
data.frame(b = c(1988, 1989),c = list(c(1985, 1982, 1984),c(1988, 1980)))
Nor this
data.frame(a$b, a$c)
---
David
> -Origin
Does any one know how to calculate/predict the value at some time point (say
90%) based on dose response curve using "drc' in R? I know how to fit the
curve with drc. But need to calculate 90% value at some time point. Many
thanks.
--
View this message in context:
http://r.789695.n4.nabble.com
HI,
If "b" is also an "element of list, then it would be much easier.
b<-c(1988,1989,1990,1991,1992)
c<-list(c(1985,1982,1984),c(1988,1980),c(1983,1984),c(1988,1998,1993),c(1993,1994,1998))
a<-list(b,c)
names(a)<-c("b","c")
lapply(1:length(a$c),function(x) a$b[x]-a$c[[x]])
[[1]]
[1] 3 6 4
[[
HI,
You can get the results you wanted by:
c=list(c(1985,1982,1984),c(1988,1980),c(1983,1984),c(1988,1998,1993),c(1993,1994,1998))
b1<-list(1988,1989,1990,1991,1992)
for(i in 1:length(b1)){
anew[[i]]<-list()
anew[[i]]<-b1[[i]]-c[[i]]
}
anew
[[1]]
[1] 3 6 4
[[2]]
[1] 1 9
---
---
But, i
Duncan,
I agree that my second code doesn't make sense. Sorry for the red herring.
I was grasping at straws.
My first code, however, differs from your suggestion only in that I am
asking for a rectangular projection, AFAICT. As it turns out, I do get
quite different results when asking for the
HI,
In the reply I sent, I forgot to add,
anew<-list()#before,
for(i in 1:length(b1)){
anew[[i]]<-list()
anew[[i]]<-b1[[i]]-c[[i]]
}
A.K.
- Original Message -
From: jimi adams
To: r-help@r-project.org
Cc:
Sent: Thursday, August 9, 2012 4:42 PM
Subject: [R] indexing in data frames
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