Plantky wrote:
Hi all,
Can anyone tell me how I can make 0,0 start at the top left hand
corner of a graph, instead of the typical lower left hand corner? I've
tried to plot with axes=F and then putting on the axes later, but I
want the points to correspond to the axes.
Thanks,
Kang Min
Does t
On Nov 18, 2008, at 12:50 AM, kayj wrote:
Hi All,
I have a column that contains values between 0 and 1. I would like
to make
a table that consists of the number of elements in each category.
For example , how many elements have values between 0 and 0.1, 0.1
to 0.2,
0.2 to 0.3,etc……..0
On Nov 18, 2008, at 12:57 AM, kayj wrote:
I have a huge txt file and I only want to get out of it column 3
and 7.
I tried to read the whole file and then extract the two columns, but
I ran
into a memory problem since the file is huge.
Is it possible just to tell R to read these two col
Hi Daniel,
I'm sorry I wasn't clear. All my values are positive, and I'm plotting
the map of a rectangular plot. I have all the x and y values of
individual points, but the 0,0 point is at the top left hand corner of
the plot. I have no problems plotting it with the 0,0 at the lower left
hand
I have a huge txt file and I only want to get out of it column 3 and 7.
I tried to read the whole file and then extract the two columns, but I ran
into a memory problem since the file is huge.
Is it possible just to tell R to read these two columns without reading the
whole file?
Thanks
-
Dear all,
If the question is too easy, please forgive me since I am only few weeks old in
R.
I have worked on this question a few days and still cannot figure it out.
Here I have a folder with more than 50 tab-delimited files. Each file has a few
hundreds of thousands rows/subjects, and th
Hi All,
I have a column that contains values between 0 and 1. I would like to make
a table that consists of the number of elements in each category.
For example , how many elements have values between 0 and 0.1, 0.1 to 0.2,
0.2 to 0.3,etc……..0.9 to 1.
Is there an easy way to do this?
Thanks
Hi Ernesto,
This is Eleanor from NOHS. Ikaw ni? Nadula ko ang e-mail address mo sang
pag change ko sang provider.
Pero indi ko sure kon ikaw ni. Kon indi, please ignore this e-mail.
Regards,
Eleanor
[[alternative HTML version deleted]]
__
Hi All,
Thank all for the input. These different solutions rock!.
Hao
--
View this message in context:
http://www.nabble.com/how-to-calculate-another-vector-based-on-the-data-from-a-combination-of-two-factors-tp20532749p20553226.html
Sent from the R help mailing list archive at Nabble.com.
__
Hi, I don't understand the question. If your data is in the fourth quadrant
(all positive Xs, all negative Ys) this will happen automatically as the
standard R plot plots only in the range where there is data, e.g.
x=rnorm(100,0,1)
e=rnorm(100,0,1)
y=-abs(x)-abs(e)
plot(y~x,bty="n")
Please be mor
## Use model.matrix
## Data is the same
## continue
m <- model.matrix(lm(rep(0,length(y)) ~ disease + drug +disease:drug));
## Model.matrix(lm(y~...)) will drop is.na(y) rows. That result will
be Type II rather than III
## for residuals of (disease:drug) will be orthogonal to disease and
drug with
Hi all,
Can anyone tell me how I can make 0,0 start at the top left hand
corner of a graph, instead of the typical lower left hand corner? I've
tried to plot with axes=F and then putting on the axes later, but I
want the points to correspond to the axes.
Thanks,
Kang Min
Thank you, that took care of it!
Stephanie
On Mon, Nov 17, 2008 at 12:33 PM, Jorge Ivan Velez <[EMAIL PROTECTED]
> wrote:
>
> Dear Sephanie,
> Try the following:
>
> # Data set including some NA values
> set.seed(123)
> X=cbind(c(3, 2, 2, 1, 2, 5, 5, 4, 1, 1),matrix(rnorm(100),ncol=10))
> colname
wcyee gmail.com> writes:
>
[snip]
>
> How does the function rgamma work in the instance with the rate
> specified as a vector of values? My understanding is that rgamma
> returns m random values from the gamma distribution for a given shape,
> rate. But I don't understand what the resultin
On Mon, 17 Nov 2008, Stavros Macrakis wrote:
Agreed. In fact, the man page for 'for' seems to explicitly specify
that the iteration variable is not mutable: "The variable var... is
read-only". However, the implementation doesn't seem to enforce this:
for (i in 1:5) print(i<-i) # n
On Mon, Nov 17, 2008 at 7:33 PM, Wacek Kusnierczyk
<[EMAIL PROTECTED]> wrote:
> Gabor Grothendieck wrote:
>> The R Language Definition manual that comes with R has a section on promises.
Certainly.
> on promises yes, but the question was whether the behaviour discussed
> before is obvious and int
On Mon, 17 Nov 2008, jim holtman wrote:
You can use the 'local' function to make sure you create a value of
'i' that is defined when the function is defined:
funcs = lapply(1:5, function(i)local({i; function(){ i}}))
You can use pretty much any other function, too
funcs = lapply(1:5, functi
Hi Kitty,
You can scale the grid cells of a levelplot using the 'shrink'
argument; see ?panel.levelplot
As for your mosaic question, are you using the 'mosaicplot' function
in the graphics package, or 'mosaic' in the vcd package, or something
else? I suggest you provide a minimal example if you w
Hi,
It's probably a simple issue but I'm struggling with that. I'll use the
example shown in the help page.
head(Indometh)
wide <- reshape(Indometh, v.names="conc", idvar="Subject",
timevar="time", direction="wide")
head(wide)
reshape(wide, idvar="Subject", varying=list(2:12),
You can use the 'local' function to make sure you create a value of
'i' that is defined when the function is defined:
> funcs = lapply(1:5, function(i)local({i; function(){ i}}))
> funcs[[3]]()
[1] 3
> funcs[[2]]()
[1] 2
On Mon, Nov 17, 2008 at 4:28 PM, Wacek Kusnierczyk
<[EMAIL PROTECTED]> wrot
Hi everybody,
Since I upgrade R to the version 2.7.1 the function boot.sem in the
package sem doesn't work. When I run the function and then call the
summary for the result I obtain:
> summary(BM2w)
Error in lower[i] <- ci[low] : nothing to replace with
I try to solve the problem searching in t
Gabor Grothendieck wrote:
> On Mon, Nov 17, 2008 at 5:42 PM, Wacek Kusnierczyk
> <[EMAIL PROTECTED]> wrote:
>
>> i haven't seen any docs page that would explain such cases, so it's hard
>> to judge for me. this might be an unforeseen side effect of the
>> intended behaviour of the promise mecha
On Mon, Nov 17, 2008 at 5:42 PM, Wacek Kusnierczyk
<[EMAIL PROTECTED]> wrote:
> i haven't seen any docs page that would explain such cases, so it's hard
> to judge for me. this might be an unforeseen side effect of the
> intended behaviour of the promise mechanism.
The R Language Definition manua
Hi All, thank you so very much for your help, i have now got it
working! I thought that i had replied already but i don't think it got
through so this is a repost of it for anyone who does a search on this
topic...
After adding the directory to the path variable, i should have
restarted my laptop.
Fair enough. But I find my interactive data analysis jobs quickly get
big enough (data manipulation, a series of model fits, some customised
output) for the analysis script to turn into something that looks like a
program. Of course, YMMV. I also get annoyed at code that uses = for
assignment outsi
## I got it. IV(s) of interaction should be orthogonal to main effect IV(s).
## Type III ANOVA / Interaction alone
x_interaction<-cbind(
(drug==2)&(disease==2)
,(drug==3)&(disease==2)
,(drug==4)&(disease==2)
,(drug==2)&(disease==3)
,(drug==3)&(disease==3)
On Mon, Nov 17, 2008 at 3:49 PM, Etches Jacob <[EMAIL PROTECTED]> wrote:
> In lattice
>
> #toy data
> library(ggplot2)
> library(lattice)
> x <- rnorm(100)
> y <- rnorm(100)
> k <- sample(c("Weak","Strong"),100,replace=T)
> j <- sample(c("Tall","Short"),100,replace=T)
> w <- data.frame(x,y,j,k)
>
>
To solve the immediate problem of quitting R, you could try to define
a dummy of the missing function, e.g.
validServerIsRunning <- function(...) FALSE;
and then quit(). This does of course ignore any potential problems of
not "finalizing" the 'fame' package correctly. Alternatively, since
the
On Mon, Nov 17, 2008 at 3:42 PM, steve <[EMAIL PROTECTED]> wrote:
> Thank you. Here's my version, using melt instead of do.call(make.groups...
>
> library(reshape)
> fgl2 = melt(fgl[,-10])
> fgl2$type = fgl$type
> bwplot(value ~ type | variable, data = fgl2)
Or even more succintly:
fgl2 <- melt(f
Stavros Macrakis wrote:
> Wacek,
>
> I think when people say that R semantics are derived from Scheme, all
> they mean is that R supports lexical closures. But R has other
> features which are very un-Scheme-like, and when they interact with
> lexical closures, you get behavior you don't find in o
On Mon, 17 Nov 2008, Wacek Kusnierczyk wrote:
the following is a trivialized version of some functional code i tried
to use in r:
(funcs = lapply(1:5, function(i) function() i))
# a list of no-parameter functions, each with its own closure environment,
# each supposed to return the correspondin
Wacek,
I think when people say that R semantics are derived from Scheme, all
they mean is that R supports lexical closures. But R has other
features which are very un-Scheme-like, and when they interact with
lexical closures, you get behavior you don't find in other functional
languages.
R passe
> "N" == Nick <[EMAIL PROTECTED]>
> on Sat, 15 Nov 2008 13:39:44 -0800 (PST) writes:
N> Code: # svm is an S3 class, so: setOldClass("svm")
N> # Create the super class model: setClass("model")
N> # For svm from e1071 this works (well gives no error):
N> setIs("svm", "
Hello list member:
I've recently had a problem in that I'm unable to quit an R Session. I
noticed this after the update to 2.8.0, but I believe I also noticed it
on another machine, in the previous version. It occurs on both linux
and Mac platforms. It only occurs when I start R in some parti
In lattice
#toy data
library(ggplot2)
library(lattice)
x <- rnorm(100)
y <- rnorm(100)
k <- sample(c("Weak","Strong"),100,replace=T)
j <- sample(c("Tall","Short"),100,replace=T)
w <- data.frame(x,y,j,k)
xyplot(y~x|j+k,scales=list(y=list(relation="free")))
will give you a scale in each subplot w
so just to show how scheme would do that:
(map
(lambda (function) (function))
(map
(lambda (index) (lambda () index))
'(1 2 3 4 5)))
;; *this* is the scheme semantics
vQ
xxx wrote:
> I agree, something is wrong here, especially since the R community
> likes to say that R semant
Thanks a lot. That works. I had no clue that the '\n' was still active and
I could use cat to see the result.
On Mon, Nov 17, 2008 at 3:58 PM, Oliver Bandel <[EMAIL PROTECTED]>wrote:
> slurpy gmail.com> writes:
>
> >
> >
> > Win xp sp2, R v2.7.1
> > Hi. If I have two numeric columns in a data
Thank you. Here's my version, using melt instead of do.call(make.groups...
library(reshape)
fgl2 = melt(fgl[,-10])
fgl2$type = fgl$type
bwplot(value ~ type | variable, data = fgl2)
Steve
Deepayan Sarkar wrote:
On Mon, Nov 17, 2008 at 11:15 AM, Chuck Cleland <[EMAIL PROTECTED]> wrote:
On 11
On Mon, 17 Nov 2008, Michael Friendly wrote:
[Using R 2.8.0 / Win XP / ]
I just added a CITATION file to the heplots package--- appended below.
From the document ion for ?CITATION, there can be *one or more* calls to
citEntry() within the CITATION file, and each should produce an object
of clas
the following is a trivialized version of some functional code i tried
to use in r:
(funcs = lapply(1:5, function(i) function() i))
# a list of no-parameter functions, each with its own closure environment,
# each supposed to return the corresponding index when applied to no
arguments
sapply(func
Hello,
Does anyone know of R code that exists to correct for sample selection
with panel data as in:
J.M. Wooldridge (1995), “Selection Corrections for Panel Data Models
Under Conditional Mean Independence Assumptions,” Journal of
Econometrics 68, 115-132.
Many thanks for your consideratio
slurpy gmail.com> writes:
>
>
> Win xp sp2, R v2.7.1
> Hi. If I have two numeric columns in a data frame, I can use the paste
> command to combine them into a new column separated by a comma.
> c3=paste(c1,c2,sep=',')
> gives: 1 1 -> "1,1"
> Is there any way I can use a new line (\n) as a sepa
Dear Dr. Carbon,
Perhaps (not lattice):
barplot(t(foo),beside=TRUE,col=rainbow(6),ylim=c(0,max(foo)),names.arg=paste('Question',1:3),
ylab='Your title here',xlab='Your title here')
legend('topleft',colnames(foo),pch=15,col=rainbow(6),ncol=2,cex=1.1)
HTH,
Jorge
On Mon, Nov 17, 2008 at 3:40 PM
Dear slurpy,
Perhaps:
cat('', 1,'\n', 1,'\n')
See ?cat for more information.
HTH,
Jorge
On Mon, Nov 17, 2008 at 3:03 PM, slurpy <[EMAIL PROTECTED]> wrote:
>
> Win xp sp2, R v2.7.1
> Hi. If I have two numeric columns in a data frame, I can use the paste
> command to combine them into a new c
I have a data.frame containing survey data that is already organized
like a table:
> foo
Excellent Very Good Good Fair Poor Very Poor
Question 1 8 7300 0
Question 2 5 5710 0
Question 3 71010
Alina
I suspect you want match (or %in%):
?match
HTH
Peter Alspach
> -Original Message-
> From: [EMAIL PROTECTED]
> [mailto:[EMAIL PROTECTED] On Behalf Of Alina Sheyman
> Sent: Tuesday, 18 November 2008 9:04 a.m.
> To: r-help@r-project.org
> Subject: [R] looking for matches
>
Dear Sephanie,
Try the following:
# Data set including some NA values
set.seed(123)
X=cbind(c(3, 2, 2, 1, 2, 5, 5, 4, 1, 1),matrix(rnorm(100),ncol=10))
colnames(X)=c('day',paste('X',1:10,sep=""))
X[2,2]<-X[5,10]<-X[3,8]<-NA
X
# mean excluding NA values
# in parenthesis are the number of observati
I'm not sure I understand what you're looking for in the result. What
exactly do you mean by a "match"? What do you want in the third table
besides the class names?
Do you just want a list (not a data frame) of those class names which
table A and table B have in common? Then how about
intersect
[Using R 2.8.0 / Win XP / ]
I just added a CITATION file to the heplots package--- appended below.
From the document ion for ?CITATION, there can be *one or more* calls to
citEntry() within the CITATION file, and each should produce an object
of class "citation".
With just a single citEntry(), ci
My question is probably pretty basic, but since I'm really new to R, here it
goes
I have two separate data frames that include class names and various other
information on classes. I'm trying to create a match based on class names
and if a match exists to create a third data frame with the c
On Sun, Nov 16, 2008 at 7:27 PM, Stavros Macrakis <[EMAIL PROTECTED]> wrote:
> Thank you for the pointer to your paper "Getting Started with Lattice
> Graphics" -- it looks very useful.
>
> I think we're talking past each other on the question of the semantics
> of formula operators, and it's proba
Win xp sp2, R v2.7.1
Hi. If I have two numeric columns in a data frame, I can use the paste
command to combine them into a new column separated by a comma.
c3=paste(c1,c2,sep=',')
gives: 1 1 -> "1,1"
Is there any way I can use a new line (\n) as a separator?
i.e.
1 1 -> 1
1
I tried paste
Thank you very much for the suggestions, however I do not think they will
help. Please let me know if I misunderstand.
1) Just using "colSums" without the "(!is.na)" portion provides:
Error in FUN(X[[1L]], ...) :
'x' must be an array of at least two dimensions
2) In that context, sum appears
Hi Ann,
>> Specifically it is for a MNL ...
>> How are you definng "response" and "predictor"?
You really are still leaving me, and I imagine other people on the list who
read your email, largely in the dark. (But since you don't seem to be
prepared to address anyone, or even the list, I don't s
The help page for the subplot function in the TeachingDemos package shows an
example of adding an image to an existing plot.
--
Gregory (Greg) L. Snow Ph.D.
Statistical Data Center
Intermountain Healthcare
[EMAIL PROTECTED]
801.408.8111
> -Original Message-
> From: [EMAIL PROTECTED] [
Hi Susan,
This is one of the things you are required to do yourself. Head over to:
http://stat.ethz.ch/mailman/listinfo/r-help
and look towards the bottom of the page for unsubscribe options.
If that doesn't work for you then contact the list admins, email
addresses for them are located towards
On Mon, Nov 17, 2008 at 11:15 AM, Chuck Cleland <[EMAIL PROTECTED]> wrote:
> On 11/17/2008 1:50 PM, steve wrote:
>> Using the data set fgl in MASS the following code
>>
>> layout(matrix(1:9,3,3))
>> for(i in 1:9){
>> boxplot(fgl[,i] ~ type, data = fgl,main=dimnames(fgl)[[2]][i])}
>>
>> produces a 3
This gives a data frame rather than a matrix, but you can convert it:
x <- do.call( expand.grid, rep( list( c(-1,1) ), 20 ) )
hope this helps,
--
Gregory (Greg) L. Snow Ph.D.
Statistical Data Center
Intermountain Healthcare
[EMAIL PROTECTED]
801.408.8111
> -Original Message-
> From: [E
## Question1: How to define IV with interaction alone, without main effects?
## Question2: Should Type III ANOVA in package car be independent of
the factor level order?
## data from http://www.otago.ac.nz/sas/stat/chap30/sect52.htm
drug <- c(t(t(rep(1,3)))%*%t(1:4));
disease <- c(t(t(1:3)) %*% t
Hi, does anybody know if it is possible to use an image (rather than
text) to label the leaves of a dendrogram? I realize that this will
not always lead to a useful on-screen display, but ideally I'd like
to directly go from the dendrogram to a PDF and on-screen viewing is
not to too import
How about
!!0 # FALSE
!!1 # TRUE
Although its one more char, its very easy to press ! twice and when you
look at it its more obvious since 0 is associated with FALSE and
1 with TRUE.
On Mon, Nov 17, 2008 at 12:42 PM, Henrik Bengtsson <[EMAIL PROTECTED]> wrote:
> To save my fingers and still bein
On 11/17/2008 1:50 PM, steve wrote:
> Using the data set fgl in MASS the following code
>
> layout(matrix(1:9,3,3))
> for(i in 1:9){
> boxplot(fgl[,i] ~ type, data = fgl,main=dimnames(fgl)[[2]][i])}
>
> produces a 3 by 3 array of plots, each one of which consists of six
> boxplots.
>
> Is it pos
Hello;
I have a function with 3 variables. I would like to take integration with
respect to just one variable. How can I write this function? You can find my
function above. (
I am trying to find \int_ZL^ZU
rho.bisquare(rho*z+u*sqrt(1-rho^2)*dnorm(u)du.Integrand has three variables, I
know the
I need help generating all possible combination of a vector of randomly
assigned values of 1 and -1. For example, a vector of 20 randomly placed 1s
and -1s, and all possibilities (which would amount to 2^20 total vectors). I
am able to generate one such vector (via sample(c(-1,1),20,replace=TR
Hi Søren,
http://jekyll.math.byuh.edu/other/howto/R/Rcmdr.shtml
Søren Højsgaard wrote:
>
> Dear all,
>
> On Windows XP with R 2.8.0, I get the error message below when I try
> install a package from the command line.
> (Installing as a local zip-file from the menu in the GUI works fine.)
Using the data set fgl in MASS the following code
layout(matrix(1:9,3,3))
for(i in 1:9){
boxplot(fgl[,i] ~ type, data = fgl,main=dimnames(fgl)[[2]][i])}
produces a 3 by 3 array of plots, each one of which consists of six
boxplots.
Is it possible to do this in lattice?
Steve
"R version 2.7.
Susan Ofner
Biostatistician II
Division of Biostatistics
Indiana University School of Medicine
410 West 10th St., Suite 3045
Indianapolis, IN 46202
(317) 278 - phone
(317) 274 - 2678 FAX
http://www.biostat.iupui.edu/
[[alternative HTML version deleted]]
___
Specifically it is for a MNL I just want to drop temporarily to run a
test. I have tried upData (dataset, g=as.numeric (g) but then I am not sure
how I could apply na.omit to omit the variable. How are you definng
"response" and "predictor"?
On Mon, Nov 17, 2008 at 3:42 AM, Mark Difford <[EMAIL
Hi. I have built a levelplot with 3 variables, X, Y, and Z where X and Y are
the two axes and Z represents the intensity (i.e. Z~X*Y).
Now I want to adjust the size of the grid (like a mosaic plot) where the size
of each grid is weighted by a variable, W.
Just wonder how can I do that with lev
Hi All,
I am making a R package which is using a boost c++ library. Can some
suggest how can I compile the dll in windows using boost. It makes a
package but during installation in fails to link boost and give
errors
Any help will be useful
Thanks
Vidhu
_
To save my fingers and still being on the safe side, I always do:
> !0
[1] TRUE
> !1
[1] FALSE
;) ...still hackable though.
/Henrik
On Mon, Nov 17, 2008 at 5:25 AM, Duncan Murdoch <[EMAIL PROTECTED]> wrote:
> On 17/11/2008 8:03 AM, hadley wickham wrote:
>>
>> On Sun, Nov 16, 2008 at 7:41 PM, S
On Nov 17, 2008, at 12:30 PM, David Winsemius wrote:
On Nov 17, 2008, at 11:01 AM, paul murima wrote:
heatmap.2(m, margins= c(9,9), col = bluered(64), trace=c("none"),
breaks=c(seq(-60,0,60/20)), symkey=TRUE, density.info="histogram",
cexRow=1,)
Error in image.default(1:nc, 1:nr, x, xlim = 0
On Mon, Nov 17, 2008 at 5:00 AM, megh <[EMAIL PROTECTED]> wrote:
>> lapply(1:5, function(i) c(1,2,3)^i)
> [[1]]
> [1] 1 2 3
...
> This is fine. However my goal is : each element of this list should depend
> on previous element like :
>
> lis # List name
> then,
>
> lis[[i]] = lis[[i-1]] + c(1,2,3)^
On Nov 17, 2008, at 11:01 AM, paul murima wrote:
heatmap.2(m, margins= c(9,9), col = bluered(64), trace=c("none"),
breaks=c(seq(-60,0,60/20)), symkey=TRUE, density.info="histogram",
cexRow=1,)
Error in image.default(1:nc, 1:nr, x, xlim = 0.5 + c(0, nc), ylim =
0.5 + :
must have one more br
Hi,
Does anyone know of an R ORM (Object Relational Mapper)? I'm thinking of
something similar to sqlalchemy (http://www.sqlalchemy.org/).
Alternatively or additionally, can people offer suggestions about managing
relational databases from R. I'm currently using postgresql, but would
like a
On 17 Nov 2008, at 13:00, hadley wickham wrote:
Hi Dave,
On Mon, Nov 17, 2008 at 6:35 AM, Dave Murray-Rust
<[EMAIL PROTECTED]> wrote:
I'm trying to plot multiple lines using different colours/symbols to
distinguish them. If I try to plot more than 6 lines, I get an error:
ggplot( dat, aes(x
On Mon, 2008-11-17 at 07:20 -0800, P.Branco wrote:
> Sorry, it does not work.
>
> If I do a rnorm I lose the original values of my vectors, and the equation
> result must be attained by the use of the original values.
Dimitris was generating some dummy data to test that the function
worked. He ha
On Mon, 2008-11-17 at 15:12 +0100, Wacek Kusnierczyk wrote:
> Duncan Murdoch wrote:
> >
> > paramValue <- 15
> > source("myRfile.R")
> >
> > The quotes are necessary, because source(myRfile.R) would go looking
> > for a variable named myRfile.R, rather than using "myRfile.R" as the
> > filename.
>
Lathouri, Maria imperial.ac.uk> writes:
[...]
> Until here it works fine, but I have the graph plotting only the points. What
I want is to have a line (join
> these points) so to have a time plot. I have tried different commands such as
lines(DATE,pH) or with(DF,
> lines(DATE,pH) but nothing work
Also if this is a time series you may wish to represent it as such
to facilitate other computations as well. This assumes that the
DATE column is first and the remaining columns are numeric:
library(zoo)
mypath <- "/whatever/myfile.csv"
z <- read.zoo(mypath, sep = ",", header = TRUE, format = "%d
Have a look at ?lockBinding
Best,
luke
On Mon, 17 Nov 2008, Lucke, Joseph F wrote:
The TRUE/FALSE vs. T/F issue brings up a related one. Can one assign a
variable a value during an R session that cannot be re-assigned any new
value during the session? That is, the variable is `protected' f
On Mon, 17 Nov 2008 15:30:14 - "Lathouri, Maria"
<[EMAIL PROTECTED]> wrote:
LM> I want to do some time plots and actually the dates are in the
LM> format of dd/mm/. So first I input my dataframe in R in a csv
LM> form. What I do is DF<-read.csv("C:/Documents and Settings/DF.csv")
LM> DATE<
On Mon, Nov 17, 2008 at 07:20:22AM -0800, P.Branco wrote:
>
> Sorry, it does not work.
>
> If I do a rnorm I lose the original values of my vectors, and the equation
> result must be attained by the use of the original values.
He didn't want you to actually do the rnorm part - he just used
that
heatmap.2(m, margins= c(9,9), col = bluered(64), trace=c("none"),
breaks=c(seq(-60,0,60/20)), symkey=TRUE, density.info="histogram",
cexRow=1,)
Error in image.default(1:nc, 1:nr, x, xlim = 0.5 + c(0, nc), ylim = 0.5 + :
must have one more break than colour
Hie all.
I am working on scaling my d
try
plot(DATE, pH, type='l')
On Mon, Nov 17, 2008 at 10:30 AM, Lathouri, Maria
<[EMAIL PROTECTED]> wrote:
> Dear all,
> I want to do some time plots and actually the dates are in the format of
> dd/mm/. So first I input my dataframe in R in a csv form. What I do is
>
> DF<-read.csv("C:/Docum
Hello,
I have written a mail to gmane
so that they can correct their page (eventually throwing out that link).
Btw, the framed version of the link to the gmane-page (from r-cran project
pages)
appears like being part of the r-cran-pages; that was the reason, why
I thought the gmane-archive-serve
Dear David,
Thank you very much for your help! I think I have solved the issue.
Lucho
--- On Fri, 11/14/08, David Winsemius <[EMAIL PROTECTED]> wrote:
From: David Winsemius <[EMAIL PROTECTED]>
Subject: Re: [R] Epicalc package
To: "Luciano La Sala" <[EMAIL PROTECTED]>
Cc: "r help"
Date: Fri
"Lucke, Joseph F" <[EMAIL PROTECTED]> writes:
>
> The TRUE/FALSE vs. T/F issue brings up a related one. Can one assign a
> variable a value during an R session that cannot be re-assigned any new
> value during the session? That is, the variable is `protected' from
> change during the session.
"paul murima" <[EMAIL PROTECTED]> writes:
> Hello R-Community,
>
> I am a rookie in R and I am fascinated with the power of bio computing
> by R. I am analysing gene expression data from Real time PCR. I have
> used absolute gene quantitation to measure gene copy number in all my
> transcripts. Al
Dear all,
I want to do some time plots and actually the dates are in the format of
dd/mm/. So first I input my dataframe in R in a csv form. What I do is
DF<-read.csv("C:/Documents and Settings/DF.csv")
DATE<-as.Date(DATE, "%d/%m/%Y") # to tell R that DATE column is indeed dates
with(DF,
Thanks to you both
--
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Sent from the R help mailing list archive at Nabble.com.
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R-help@r-project.org maili
Hi,all
I am doing a nonlinear regression and I am going to use nls.
To Minimize:sum{(Y-BS(s,x,r,t,v)) ^2}
a<-nls(Y~BS(s,x,r,t,v))
BS is the black-scholes model.
BS<-function(s,x,r,t,v){
d1=(ln(s/x)+v^2*t/2)/(v*t^(0.5))
d2=(ln(s/x)-v^2*t/2)/(v*t^(0.5))
c=e^(rt)*(s*N(d1)-x*N(d2))
}
where s,x,
Sorry, it does not work.
If I do a rnorm I lose the original values of my vectors, and the equation
result must be attained by the use of the original values.
Thanks,
P.Branco
Dimitris Rizopoulos-4 wrote:
>
> try this (presented only for two pairs):
>
> r1 <- rnorm(49)
> r2 <- rnorm(49)
> s1
The TRUE/FALSE vs. T/F issue brings up a related one. Can one assign a
variable a value during an R session that cannot be re-assigned any new
value during the session? That is, the variable is `protected' from
change during the session. `Session' here is not precisely defined, so
that will i
Roland,
I don't know of the specific data you are asking for but the official
website of the 'National Hot Rod Assoc' (?) is www.nhra.com.
DaveT.
*
Silviculture Data Analyst
Ontario Forest Research Institute
Ontario Ministry of Natural Resources
[EMAIL PROTECTE
Rainer,
here is a quick workaround:
layout.svd3 <- function (graph, d = shortest.paths(graph), ...)
{
if (!is.igraph(graph)) {
stop("Not a graph object")
}
l <- svd(d, 3)$u
l[, 1] <- l[, 1]/dist(range(l[, 1]))
l[, 2] <- l[, 2]/dist(range(l[, 2]))
l[, 3] <- l[, 3]/dist(range(l[, 3]
try this (presented only for two pairs):
r1 <- rnorm(49)
r2 <- rnorm(49)
s1 <- rnorm(300)
s2 <- rnorm(300)
dd <- sqrt(0.723523 * outer(r1, s1, "-")^2 + 0.215518 * outer(r2, s2,
"-")^2)
dd
I hope it helps.
Best,
Dimitris
P.Branco wrote:
Dear all,
I am for the first time trying to work with
Dear all,
I am for the first time trying to work with R, but I have bumped into a
problem.
I have four vectors:
r1 (49 values)
r21 (49 values)
r22 (49 values)
r3 (49 values)
s1 (300 values)
s21 (300 values)
s22 (300 values)
s3 (300 values)
And I would like to calculate the distances from all re
On 17/11/2008 9:14 AM, Brigid Mooney wrote:
Is there a better command to use rather than source which would take
command arguments?
I ask because I currently have 6 parameters, will likely have additional
paramaters later, and would like to be able to have default values for
each, if I do no
Is there a better command to use rather than source which would take command
arguments?
I ask because I currently have 6 parameters, will likely have additional
paramaters later, and would like to be able to have default values for each,
if I do not specify new values.
Thanks so much!
On Mon, No
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