Hi,
Next time give folks code to produce a toy sample of your problem
DF <-data.frame(ID=rep(1:5,each=3),Data=rnorm(15),Stuff=seq(1:15))
DF
ID Data Stuff
1 1 2.0628225 1
2 1 0.6599165 2
3 1 0.5672595 3
4 2 -0.5308823 4
5 2 -0.5358471 5
6 2 -0.1414992
Thank you all for the advice.
I have looked through the Introduction to R pdf and got some pointers but
when I try to implement them it does not work. If someone could clarify a
couple of basic things, I would appreciate it.
When I successfully read in my file, the prompt changed from > to +. The
I need to add a belated acknowledgement to my prior comments. Thomas
Lumley has also been a significant contributer to the survival code.
Until I moved my development from Splus to R he was the primary
maintainer of the R code, he tightened up a lot of the C code to make
sure it would work on mult
Hi,
if I have a dataframe such that
ID Time Earn
1 1 10
1 2 50
1 3 68
2 1 40
2 2 78
24 88
3 1 50
3 2 60
3 3 98
4 1 33
4 2 48
44
Hello Samaire,
I don't know much about eyesight measurements (other than that my own
would probably be 0.1, 0.2) but I'll attempt some suggestions...
For objective (1) you will have to define what you mean by
significantly different. If you had data on left/right values from a
larger population t
Have you done a google search, looked on CRAN, searched the archives,
or looked at the task views? I can't be of any help, but maybe one of
those places could be.
HTH,
Stephen
On Tue, Dec 14, 2010 at 7:57 PM, Jason Rupert wrote:
> By any chance is there an R package that may contain functionali
Dear R world,
Do you know about a function that would capitalize in the correct manner
first and family names?
I found in the cwhmisc only the CapLeading function, but it just does not do
the job, taking care only to capitalize the first letter of a word.
I am looking for a function that would r
By any chance is there an R package that may contain functionality similar to
the frestimate in Matlab/Simulink.
Here is the URL for a description of the frestimate functionality:
http://www.mathworks.com/help/toolbox/slcontrol/ug/frestimate.html
Thank you again for any feedback.
___
Hi all,
Version 0.2.2 of the googVis package has been released on CRAN and
will be available from your local CRAN mirror in due course.
googleVis provides an interface between R and the Google Visualisation
API. The functions of the package allow users to visualise data
stored in R with the Goo
Try this:
l <- sapply(c(4, 3, 2), q.1)
aggregate(V3 ~ V1 + V2, as.data.frame(do.call(rbind, l)), mean)
On Tue, Dec 14, 2010 at 9:19 PM, Worik R wrote:
> I am sure this si a simple problem but the solution is evading me.
>
> I have a list of matrices all with the same number of columns but diffe
Hi Jonathan,thanks a lot! Mine is windows operating system so I will
try other ways .
About your question, do you want to get a matrix like this:
0 1 2 3 ...1817393 5524385
0687
0 64
0 71
0 55
...
Thank you guys I learnt a lot .
But when I tried to run the library(ecodist) function, R says there is no
package called "ecodist".
why?
__
View message @
http://r.789695.n4.nabble.com/from-table-to-matrix-tp3087972p3088184.html
Hi Mike, Hi Uwe,
thanks for your reply.
The main thing is that we have mixed OSs. On the one hand we have Linux,
on the other WinXP. I need the same version of packages and R on both
systems. Otherwise I cant explain my colleagues why error messages are
reproduceable on only one system.
Instal
I have a set of results where I have the eyesight power of both left and right
eyes for each participant (e.g. 0.75, 0.5). Each participant then had to throw
basketballs into a hoop and the number of successful throws was recorded.
I would like to do two things:
1. Test whether the eyesight po
Hello:
I would like to obtain probability of an event for one single patient as a
function of time (from survfit.coxph) object, as I want to find what is the
probability of an event say at 1 month and what is the probability of an
event at 80 months and compare. So I tried the following but it fai
Dear Derek, dear list,
A couple suggestions (and more) :
0) check that your problem can't be expressed in terms of what glmer()
(lme4(a|b)? package(s)) can solve. I use this often, and am quite
impressed with its abilities.
1) Why not modeling that directly in BUGS ? That might not be easy, wi
I figured out how to toggle the legend for the second scenario, but now I
have three different map sizes!
trellis.par.set( peelTheme)
plot( spplot( thumbDf, "pri",
colorkey= list( space= "left")),
split= c( 1, 1, 3, 1), more= TRUE)
plot( spplot( thumbDf, "sec",
co
You ask some good questions.
> I would like to clarify statistics that ridge coxph returns. Here is
> my understanding and please correct me where I am wrong.
> 1) In his paper Gray [JASA 1992] suggests a Wald-type statistics with
> the formula for degree of freedom. The summary function for r
I actually got it to work now using levelplot, and I also worked out how to
get the graph I wanted to display, I had to assign the graph to a reference
and then print it -- yes, that simple!
Now I just need to work out why my levelplot doesn't look quite right, even
though the data is in there...
Hello R helpers:
*My first message didn't pass trough filter so here it's again*
I would like to obtain probability of an event for one single patient as a
function of time (from survfit.coxph) object, as I want to find what is the
probability of an event say at 1 month and what is the probabilit
I have some geospatial data where two layers are thematic and the third is a
percentage, so the maps need to have different themes.
thumbDf <- as( stack( thumb), "SpatialGridDataFrame")
names(thum...@data) <- c("pri", "sec", "pct")
thum...@data$pri <- factor(thum...@data$pri, levels=c(0:8), lab
Wayne,
So far, no one has said the obvious:
Please do work your way through (or at least
skim) "An Introduction to R" which you'll
find right there on your computer under
Help/Manuals. Your questions indicate that
you have not yet done so. Do it, it really
will pay off.
Peter Ehlers
On 2010-12-
> Date: Tue, 14 Dec 2010 23:48:20 +0100
> From: er...@phonetik.uni-muenchen.de
> To: marchy...@hotmail.com
> CC: r-help@r-project.org; lig...@statistik.tu-dortmund.de
>
> Hi Mike, Hi Uwe,
>
> thanks for your reply.
> The main thing is that we have mixed
I am sure this si a simple problem but the solution is evading me.
I have a list of matrices all with the same number of columns but different
number of rows. The first two columns label the row. The labels are
allways the same for the same row numbers, just some matricies have more
rows.
For e
Maybe I can help here:
In OS X at least you can open the R Package Installer from the Packages &
Data menu, select a mirror, etc, then type ecodist into the search bar and
then Install Selected... this will install the ecodist package on your
machine.
Not exactly sure how it would be done under
Hi, Tal,
Here is a quick way of getting around. First create two responses via
dummy variables
y1 <- ifelse(y=="a", 1, 0)
y2 <- ifelse(y=="b", 1, 0)
and then built two separate tree models for y1 and y2 separately.
Hope it helps.
Xiaogang
On Tue, Dec 14, 2010 at 8:33 AM, Tal Galili wrote:
>
OK well I don't mean to "hijack" Jessica's thread with a tangent on graphics
and plots, but I'm still having some trouble. I'm a total newbie here so if
the correct etiquette would be for me to start a new thread at this point
then please do advise me!!
The code I'm now trying to run is:
library
On Tue, 2010-12-14 at 22:36 +, e-letter wrote:
> Readers,
>
> I have been reading 'the r book' by Crawley and think that the
> generalised additive model is appropriate for this problem. The
> package 'gam' was installed using the command (as root)
>
> install.package("gam")
> ...
> library(g
Dear Lurker,
If all you art trying to do is to plot something, isn't all you need something
like the following?
x <- c( 30, 50, 80, 90, 100)
y <- c(160, 180, 250, 450, 300)
sp <- spline(x, y, n = 500)
plot(sp, type = "l", xlab = "x", ylab = "y",
las = 1, main = "A Spline Interpolation")
For contour plots, some of the contour functions in graphics packages
(lattice for one IIRC) are pretty good at understanding that columns in
a matrix correspond to x,y, and z values already.
There are many ways to do this in R. For very simple problems, this one
is convenient:
Readers,
I have been reading 'the r book' by Crawley and think that the
generalised additive model is appropriate for this problem. The
package 'gam' was installed using the command (as root)
install.package("gam")
...
library(gam)
> library(gam)
Loading required package: splines
Loading require
THere's the peaks package which will do some peak detection, or 'peaks'
in the simecol package. Or msc.peaks.find in caMassClass,
I'm madly trying to remember what spectral package I found which has a
tool similar to msextrema but returns a couple more useful lists
describing the data and t
Jonathan -
Same problem, same solution:
Suppose your data frame is called df:
thematrix = matrix(NA,max(df$x),max(df$y))
thematrix[as.matrix(df[,1:2])] = df[,3]
- Phil Spector
Statistical Computing Facility
There are many ways to do this in R. For very simple problems, this
one is convenient:
library(ecodist)
newdata <- crosstab(mydata$x, mydata$y, mydata$z)
For more complicated problems, reshape is very powerful.
Sarah
On Tue, Dec 14, 2010 at 5:13 PM, jonathan wrote:
>
> That's so weird, I just
If the goal is to produce a filled contour plot, then look at the levelplot
function in the trellis package, it takes the data in the form that you already
have it, no need to transform.
--
Gregory (Greg) L. Snow Ph.D.
Statistical Data Center
Intermountain Healthcare
greg.s...@imail.org
801.408
Hi R-Helpers,
I am stuck with my analysis. I wonder if anybody could have with this.
My data set looks like:
Squares
Hours
SppRichness
Feeding_guild
natprop
1
10
1
aq inverts
0.118697
2
25
2
aq inverts
0.605874
3
35
2
fish
0.61255
4
20
4
fish
0.764418
5
40
3
ins
I would like to clarify statistics that ridge coxph returns. Here is my
understanding and please correct me where I am wrong.
1) In his paper Gray [JASA 1992] suggests a Wald-type statistics with the
formula for degree of freedom. The summary function for ridge coxph returns the
degree of free
That's so weird, I just signed up on here to ask exactly the same question!
However, I think my issue is like Jessica's who says that her data is like
that, not actually that...
So the issue is not in generating that data on-the-fly but in transforming
it from a data frame to a matrix.
As a mor
Wayne,
>> I don't know how to assign a name for the df, or what to put for "fac",
>> and what is worse,
>> I get an error message saying that the program cannot find the
>> "discrimin.coa" command.
Before you can use a package you have downloaded you need to "activate" it.
There are different w
Dear R users,
I am looking for a simple way to define a contrast in a linear model.
I have a data frame with two factors (f1, f2) and a dependent y.
x <- data.frame(y =rnorm(30), f1=gl(2, 15), f2=gl(3, 5, 30))
Now I want to specify the following contrast: "f1= 1 or 2 and f2=1" vs. "f1=
1 or
Many thanks Phil. This is perfect. I usually forget about lapply and
try something more complicated. Your solution works really well.
Best, Mark
On Tue, Dec 14, 2010 at 3:45 PM, Phil Spector wrote:
> Mark -
> I believe
>
> lapply(dd,function(m)eval(parse(text=m)))
>
> will do what you want.
Here's one way:
df = data.frame(Date=rep(LETTERS[1:3],each=3),TIME=rep(letters[1:3],3),Q=1:9)
df
Date TIME Q
1Aa 1
2Ab 2
3Ac 3
4Ba 4
5Bb 5
6Bc 6
7Ca 7
8Cb 8
9Cc 9
mat = matrix(0,3,3,dimnames=list(LETTERS[1:3],letters[1:3
Mark -
I believe
lapply(dd,function(m)eval(parse(text=m)))
will do what you want.
- Phil Spector
Statistical Computing Facility
Department of Statistics
I have a table like this:
Date TIME Q
A a1
A b2
A c3
B a4
B b5
B c6
C a7
C b8
C c9
I want use R language to turn it to a matrix like :
a b c
A 1 2 3
B 4 5 6
C 7 8 9
I am new to R , anyone can help? Than
Hi,
I have a run of 5 graphs that I want to place them under the same page.
Everything works fine to place them in a pdf file , or eps file, but
when it comes to have a high quality of
300 dpi these graphs are not good. For example I open the eps file with
Adobe Illustrator (AI) and it shows th
Try
step()
?
On Tue, Dec 14, 2010 at 10:05 PM, Mark Na wrote:
> Hi R-helpers,
>
> I have a character object called dd that has 32 elements each of which
> is a model formula contained within quotation marks. Here's what it
> looks like:
>
>> dd
> [1] "lm(y ~ 1,data=Cement)" "
Hi R-helpers,
I have a character object called dd that has 32 elements each of which
is a model formula contained within quotation marks. Here's what it
looks like:
> dd
[1] "lm(y ~ 1,data=Cement)" "lm(y ~
X,data=Cement)" "lm(y ~ X1,data=Cement)"
[4] "lm(
Mark -
Since regular expressions in R are just character
strings, it's pretty easy to assemble a regular expression
to delete leading or trailing characters. For example:
delchars = function(str,n,lead=TRUE){
+dots = paste(rep('.',n),collapse='')
+pat = if(lead)paste('^',dots,sep=''
Read files, if you're on windows remember to include the path like this:
"C:\\Documents and Settings\\USER\\My Documents\\MyFile.csv"
-Original Message-
From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-project.org] On
Behalf Of Wayne Sawtell
Sent: Tuesday, December 14, 2010 12:3
Hello everyone,
I am totally new to the R program. I have had a look at some pdf documents
that I downloaded and that explain how to do many things in R; however, I
still cannot figure out how to do what I want to do, which is to perform
Discriminant Correspondence Analysis on a rectangular matrix
Try this:
noquote("lm(y ~ X2 + X3 + X4)")
To remove X characters:
gsub("^.|.$", "", "lm(y ~ X2 + X3 + X4)")
On Tue, Dec 14, 2010 at 6:27 PM, Mark Na wrote:
> Hi R-helpers,
>
> I have a character string, for example:
>
> "lm(y ~ X2 + X3 + X4)"
>
> from which I would like to strip off the lead
Hi R-helpers,
I have a character string, for example:
"lm(y ~ X2 + X3 + X4)"
from which I would like to strip off the leading and trailing
quotation marks resulting in this:
lm(y ~ X2 + X3 + X4)
I have tried using gsub() but I can't figure out how to specify the
quotation mark using a regular
Hi,
I have a run of 5 graphs that I want to place them under the same page.
Everything works fine to place them in a pdf file , or eps file, but
when it comes to have a high quality of
300 dpi these graphs are not good. For example I open the eps file with
Adobe Illustrator (AI) and it shows th
Look at the hexbin package (bioconductor).
--
Gregory (Greg) L. Snow Ph.D.
Statistical Data Center
Intermountain Healthcare
greg.s...@imail.org
801.408.8111
> -Original Message-
> From: r-help-boun...@r-project.org [mailto:r-help-boun...@r-
> project.org] On Behalf Of Matthew Vernon
> S
On 2010-12-14 08:43, Sarah Goslee wrote:
You are most likely to get reasonable help if you provide at a minimum
your OS, the version of R you're using, and the error message you get.
And also if you send your replies to the R-help list, not just me.
Sarah
Sarah's code works perfectly well on
Hi,
I've spent a while scrabbling around at this problem, to no avail. I'm
sure there /should/ be a simple answer...
I have a large (14 million rows) data set, with two columns. Each row
contains the number of times an individual moved, and the distance
that individual moved in total. I would lik
What about thoughtless un-nasty comments?
I would suggest the original poster (and others thinking that they want to do
normality testing) read the help page for SnowsPenultimateNormalityTest in the
TeachingDemos package, which I think agrees mostly with what Bert wrote below
(though it goes mo
Or in short, type
install.packages("car")
within R.
Note that you won't get new versions of car for the outdated version of
R. Note also that your zip file might not fit to the R version you are
using.
Hence recommendation is to upgrade to R-2.12.0 patched (or 2.12.1 which
will be release
( hotmail just randomly decides not to prefix original text, my comments below)
Date: Tue, 14 Dec 2010 17:36:11 +0100
From: er...@phonetik.uni-muenchen.de
To: r-help@r-project.org
Hi there,
I have the following problem and I hope somebody mig
... (in addition to the very useful suggestion to plot your data):
(Sounds like a homework question... ?).
Sigh. [mount soapbox]
1. "Data" never deviate from normality. They only provide provide
evidence to challenge ("test" is the formal term) the assumption that
the population from which t
On Fri, Dec 10, 2010 at 5:21 PM, Eleni Rapsomaniki
wrote:
> Dear R-users,
>
> I need to use the aftreg function in package 'eha' to estimate failure times
> for left truncated survival data.
Be careful! This is only possible under strong assumptions about the
(unobserved) covariate vector. This
Thank you. I will look at that also.
On Mon, Dec 13, 2010 at 9:49 PM, Michael Bedward
wrote:
> Hi Joe,
>
> Just for info, I've done this in the past by applying lowess followed
> by diff to a vector, then identifying points with change of sign in
> the diffs.
>
> Michael
>
> On 14 December 2010 1
Hi there,
I have the following problem and I hope somebody might help me.
First of all: I am using WinXP SP3 (english and/or german) with R in
Version 2.10.0.
Now I am trying to install some packages but unfortunately I am getting
a weird error. No matter which package I am trying to install
I figured out how to call coefficients and R² from the summary.
-
fit<-lm(...) #multiple regression function
stepComp <- stepAIC(fit, direction="both")
summary(stepComp)$coef[1,1] #call first coefficient
summary(stepComp)$coef[1,2] #call Std. Error of first coefficient
summary(ste
You are most likely to get reasonable help if you provide at a minimum
your OS, the version of R you're using, and the error message you get.
And also if you send your replies to the R-help list, not just me.
Sarah
On Tue, Dec 14, 2010 at 10:58 AM, Val wrote:
> Thank you Sarah,
>
>
> It worked
nps.gov> writes:
>
> Greetings
> In attempting to create a date variable based on month (e.g.,
> February, April, etc.) and year (e.g., 2006) data, wherein I converted
> Month to a factor with Jan=1...Dec=12, I used the following command:
>
> data$Date<-mdy.date(month=data$Month,day=15,ye
Hello everyone,
I have been doing a binary classification using random forest from the
library "randomForest". One of the predictors is a factor variable, which is
known to be highly related to the binary response I am trying to predict.
Other 80 predictors are numeric. Totally I have 44 subjects.
On Tue, 14 Dec 2010, Matthew Rosett wrote:
How do I determine if my data deviate from the normal distribution?
The sample size is 1000 (weights of people).
See
?qqnorm
and/or
?shapiro.test
and/or
a text book on applied statistics
and/or
google for "testing normality".
Hth,
Gerrit
PS
Tianchan, why aren't you just using
col_no <- cut(r,c(0,2,4,6,8,10,100))
levels(col_no) <- c("<2%","2-4%","4-6%","6-8%","8-10%",">10%")
?
Your use of as.numeric() is nonsensical; check step by step what is
happening with that.
Hth,
Gerrit
On Tue, 14 Dec 2010, Tianchan Niu wrote:
Dear all
unlist(strsplit(output," +"))
On Tue, Dec 14, 2010 at 2:09 PM, Scott Chamberlain
wrote:
> Hello,
>
> I am attempting to manipulate strings in which there are differing amounts of
> whitespace before and after each element taht I want to keep (any word,
> letter, or number). However, after strs
Hi,
I have just recently started to use regular expression, so I'm not sure
that my solution would be valid in every case. But it works in yours:
> unlist(strsplit(output, split=" +"))
[1] "a""b""" "6"
It would split them whatever the number of spaces between them is
(because of t
Tianchan Niu isis.georgetown.edu> writes:
>
> Dear all,
> I would like to use cut() to make numerics to factors, the sample codes are as
follows. However, the result is
> not what I want, since r[3] = 9 should be in the interval of "8-10%" rather
than "2-4%". Maybe cut() is not
> the right funct
Hi:
On Mon, Dec 13, 2010 at 4:28 PM, chandu wrote:
>
> Dear all,
>
> I am relatively new to R. I would like to know how can we write the
> realizations (for example generated through rnorm or runif) in to a data
> file. It would be very inefficient to first generate values and then write
> them
On Tue, 2010-12-14 at 15:40 +, Tianchan Niu wrote:
> Dear all,
> I would like to use cut() to make numerics to factors, the sample codes are
> as follows. However, the result is not what I want, since r[3] = 9 should be
> in the interval of "8-10%" rather than "2-4%". Maybe cut() is not the r
chandu wrote:
>
> I am relatively new to R. I would like to know how can we write the
> realizations (for example generated through rnorm or runif) in to a data
> file. It would be very inefficient to first generate values and then write
> them in to file using "write" function. Instead, is th
Andrew Collier gmail.com> writes:
> i am sure that this is a trivial question but i have not been able to
> find an answer by searching the mailing lists. i want to plot points on
> a graph, joined by lines. the command that i am using is
>
> points(x, y, type = "b", pch = 21)
>
> this plots ni
Hello,
I am attempting to manipulate strings in which there are differing amounts
of whitespace before and after each element taht I want to keep (any word,
letter, or number). However, after strsplit and unlist, I know how to select
specific elements with "[ ]", but I want to select instead all e
Hello,
I am attempting to manipulate strings in which there are differing amounts of
whitespace before and after each element taht I want to keep (any word, letter,
or number). However, after strsplit and unlist, I know how to select specific
elements with "[ ]", but I want to select instead a
How do I determine if my data deviate from the normal distribution?
The sample size is 1000 (weights of people).
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailman/listinfo/r-help
PLEASE do read the posting guide http://www.R-project.org/
Thanks! I got the results! Now I would like to put them in a nice and
readable plot to see if there are outlier means or variances. Using
"plot(by_many$mean)" I got "Error in by_many$mean : object of type 'closure'
is not subsettable".
Another question: I have done the principal components analys
Hi,
Try the following,
plot(1:10,rnorm(10),t="o")
## fill the points in white
plot(1:10,rnorm(10),t="o",pch=21,bg="white")
You could also try this with Grid graphics,
library(gridExtra)
# like type="o"
grid.barbed(space=0)
# like type="b"
grid.barbed(space=1)
# like the example above, but with
hi all,
how can i get correspondence analysis plots (package ca, plot.ca()) with the
inertia on the two dimensions of the map?
thanks,
kat
[[alternative HTML version deleted]]
__
R-help@r-project.org mailing list
https://stat.ethz.ch/mailm
hi,
i am sure that this is a trivial question but i have not been able to
find an answer by searching the mailing lists. i want to plot points on
a graph, joined by lines. the command that i am using is
points(x, y, type = "b", pch = 21)
this plots nice open circles at the data points and draws
Dear all,
I would like to use cut() to make numerics to factors, the sample codes are as
follows. However, the result is not what I want, since r[3] = 9 should be in
the interval of "8-10%" rather than "2-4%". Maybe cut() is not the right
function to use for my situation. Please help. > r <- c(1
Thank you all for your input.
View is quite nice, despite the fact that you need to be in a X session and
it seems(?) difficult to copy paste data out of the view.
page(my_var) works great and is most similar to what I had in mind.
TkListView seems interesting but quite slow for large list or fr
Val,
Here's the complete console output. The graph produced is at:
http://www.functionaldiversity.org/temp/curve.png
R version 2.12.0 (2010-10-15)
Copyright (C) 2010 The R Foundation for Statistical Computing
ISBN 3-900051-07-0
Platform: x86_64-redhat-linux-gnu (64-bit)
R is free software and co
Hi dear R-help memebers,
When building a CART model (specifically classification tree) using rpart,
it is sometimes obvious that there are variables (X's) that are meaningful
for predicting some of the outcome (y) variables - while other predictors
are relevant for other outcome variables (y's onl
> On Sun, 12 Dec 2010, jagdeesh_mn wrote:
>
>> Hi,
>>
>> Suppose i have generated an object using the following :
>> fit <- rpart(Kyphosis ~ Age + Number + Start, data=kyphosis)
>>
>> And when i print fit, i get the following :
>>
>> n= 81
>>
>> node), split, n, loss, yval, (yprob)
>> * denot
On 14/12/2010 8:45 AM, Uwe Wolfram wrote:
Dear R-Users,
I am trying to plot an ellipsoid like function that represents some
physical threshold in its eigenvalue space. I am facing a few problems
generating a figure I need for my thesis. A small example looks as
follwos where the two contour3d pl
Dear R-Users,
I am trying to plot an ellipsoid like function that represents some
physical threshold in its eigenvalue space. I am facing a few problems
generating a figure I need for my thesis. A small example looks as
follwos where the two contour3d plots do NOT overlay as desired so you
may try
Hello everybody!
I´m trying to create figures summarizing the abundance of a given species
(y), by its length classes (x), year (conditional factor) and a grouping
named as "fraction". I am using the lattice´s barchart but some problems
arises:
When I run the function, the si
> From: pda...@gmail.com
> Date: Tue, 14 Dec 2010 13:50:27 +0100
> To: bastiaan.berg...@wdc.com
> CC: r-help@r-project.org
> Subject: Re: [R] multivariate multi regression
>
>
> On Dec 14, 2010, at 03:21 , Bastiaan Bergman wrote:
>
> > That doesn't work
On Dec 14, 2010, at 03:21 , Bastiaan Bergman wrote:
> That doesn't work, one would get two different answers depending on the
> order of execution.
>
> The physics is: Overlay error on a Silicon wafer. One wafer has many flash
> fields, each flash field has multiple locations where the overlay e
Please consider posting to R-sig-geo at
https://stat.ethz.ch/mailman/listinfo/r-sig-geo
https://stat.ethz.ch/mailman/listinfo/r-sig-geo , where maybe someone will
be able to point you to a data source.
Roger
Wang, Kevin (SYD) wrote:
>
> Hi all,
>
> I'm trying to draw a bubble plot over a ma
On Tue, Dec 14, 2010 at 10:45 AM, Wang, Kevin (SYD)
wrote:
> Hi all,
>
> I'm trying to draw a bubble plot over a map -- where the map shows local
> government areas (LGA) in Australia (or other structures under the Australian
> Standard Geographical Classification), but am not sure where I can f
SKATER is included in the spdep package, and was contributed by its original
authors.
Roger
Georg Ruß wrote:
>
> On 10/12/10 23:26:28, dorina.lazar wrote:
>> I am looking for a clustering method usefull to classify the countries in
>> some clusters taking account of: a) the geographical distan
Hi,
I'm trying to understand when and how to use do.call().
In this case, would it work with do.call() instead of lapply(), like this?
vars <- list(names(Tmese[, -(1:2)]))
res <- do.call(idw.cv, vars)
Thanks,
Ivan
Le 12/14/2010 11:48, Liviu Andronic a écrit :
On Tue, Dec 14, 2010 at 10:34 AM,
On Tue, Dec 14, 2010 at 10:34 AM, Annalaura wrote:
>
> Thanks a lot Uwe Ligges, I've abandoned R for a few time but now I'm working
> with it, so I've a question about the last problem that you solved: instead
> to write
> cv_1994<- idw.cv("X01_1994")
> cv_1995<- idw.cv("X01_1995")
> cv_1996<- idw
Hi all,
I'm trying to draw a bubble plot over a map -- where the map shows local
government areas (LGA) in Australia (or other structures under the Australian
Standard Geographical Classification), but am not sure where I can find the
data to plot such maps (e.g. draw the state of Northern Ter
Thanks a lot Uwe Ligges, I've abandoned R for a few time but now I'm working
with it, so I've a question about the last problem that you solved: instead
to write
cv_1994<- idw.cv("X01_1994")
cv_1995<- idw.cv("X01_1995")
cv_1996<- idw.cv("X01_1996")
cv_1997<- idw.cv("X01_1997")
.
1 - 100 of 103 matches
Mail list logo
