Hello,
I would like to change the font size of the x axis annotations.
cex.axis changes the y axis annotations only.
Does anyone know how to change the x axis annotations?
With thanks,
Pavel.
--
View this message in context:
http://www.nabble.com/cex.axis-for-the-x-axis-tp18353453p18353453.html
I am installing a package but got the following error. How can I install this
package ?
R_HOME is /home/daren/MyHome/lib64/RAttempting to determine R_ARCH...R_ARCH
isAttempting to detect how R was configured for Fortran 90Unsupported
Fortran 90 compiler or Fortran 90compilers unav
Hi Rers,
I have a silly question. I don't know how to express the loglikelihood
function of
1/(x!) where x=x1,x2,xn in R.
Could anyone give me a hint?
Thank you in advance.
Chunhao Tu
__
R-help@r-project.org mailing list
https://stat.ethz.ch/ma
Dear Michael,
Is this what you are looking for?
ex.dat$rn=as.numeric(rownames(ex.dat))
ex.dat
# Are new3 and ex.dat equals?
all.equal(new3,ex.dat)
[1] TRUE
HTH,
Jorge
On Tue, Jul 8, 2008 at 10:58 PM, Michael Rennie <[EMAIL PROTECTED]> wrote:
> Hi there,
>
> I'm sure there's an easy answer to
Can't you just do
newdat <- newdat[order(row.names(newdat)),]
Or am I missing something?
cheers,
Rolf Turner
On 9/07/2008, at 2:58 PM, Michael Rennie wrote:
Hi there,
I'm sure there's an easy answer to this, and I can't wait to see it.
The question: is the
Hi there,
I'm sure there's an easy answer to this, and I can't wait to see it.
The question: is there an easy way to sort a data frame by it's row names?
My dilemma:
I've had to pull apart a data frame, run it through a loop to do some
calculations and generate new variables, and then re-constr
Can you perform regression by steps, yes. Will results be the same as those you
would obtain from a multiple variable regression? No not if the independent
variables are non-orthogonal and you don't take into consideration the
correlation of the independent variables. Should you take into consid
someone else can blast me if this is not correct but i think that 2
step procedure only gives the same answer as the regular regression if
X1 and
X2 perfectly uncorrelated. If they are at all correlated, then what john
pointed out messes the procedure up. i was asked that question on
an intervi
Thanks for the reply.
I am awared of the difference, but can I do regression by steps at all? I
am not feeling comfortable about it.
John Sorkin wrote:
>
> Be very careful!
> When regression is performed by steps, you often will not get the same
> results as you would get from a single multiv
thanks Jorge. This is great!
regards,
Dhruv
--- On Tue 07/08, Jorge Ivan Velez < [EMAIL PROTECTED] > wrote:
From: Jorge Ivan Velez [mailto: [EMAIL PROTECTED]
To: [EMAIL PROTECTED]
Date: Tue, 8 Jul 2008 20:45:06 -0400
Subject: Re: [R] question on lm or glm matrix of coeficients X tes
Working code would help. I would probably use 'lapply' since it
appears that you want to return a variable number of items for each
condition.
On Tue, Jul 8, 2008 at 2:23 PM, hesicaia <[EMAIL PROTECTED]> wrote:
>
> Hello,
> The quick version of my question is how can I extract a matrix instead o
On Tue, 8 Jul 2008, hesicaia wrote:
Hello,
The quick version of my question is how can I extract a matrix instead of
a vector using tapply()? I would like to be able to access both the results
of tapply() and also the index variables.
In case further explanation would help: I am analyzing a
Hi,
I found some of what I was looking for.
using the following I can get a matrix of regression coefficient multiplied out
by the variable data.
g<-predict(comodel,type='terms',data4)
m<-cbind(data4,g)
What remains is how do I pick the 3-4 rows for each data row with the highest
va
Duncan Murdoch wrote:
On 08/07/2008 5:01 PM, Rolf Turner wrote:
On 8/07/2008, at 7:38 PM, Fiona Johnson wrote:
Hi
I have just upgraded from R2.6.0 to R2.7.1 (running on Windows) and
a part
of my code that previously ran ok now gives an error. The following
is a
simple example to demonstr
Hi Daniela,
There may be other more elegant ways of doing it, but here is one:
> myD <- data.frame(code = sample(3, 10, rep = T), prev = rnorm(10), Pr_mean =
> 0)
> myD
codeprev Pr_mean
1 3 -0.06710968 0
2 2 -1.43422034 0
3 1 0.22717580 0
4 3 0.32
Be very careful!
When regression is performed by steps, you often will not get the same results
as you would get from a single multivariable regression. The explanation for
this is not simple, but a simplified explanation is that when you do your first
regression,
y=f(x1)
all the total variance
That may have something to do with that you have "empty" groups. In your
example, ALL Hour=0 have Y2=NA. The following example may illustrate the
point. The first 2 aggregate commands perform the function on data that
contain NAs. However, the NAs are not perfectly collinear with any level by
which
On 08/07/2008 5:01 PM, Rolf Turner wrote:
On 8/07/2008, at 7:38 PM, Fiona Johnson wrote:
Hi
I have just upgraded from R2.6.0 to R2.7.1 (running on Windows) and
a part
of my code that previously ran ok now gives an error. The following
is a
simple example to demonstrate my problem.
a <-
Hi,
FYI, *the* NEWS file containing updates for all R versions is
available at http://cran.r-project.org/ - see the link 'new features
and bug fixes'. This links to the URL (which I think is rather
stable):
https://svn.r-project.org/R/trunk/NEWS
The NEWS file also comes with your R installati
quote: In the present case use rmultinom(1, 100, rep(1/50,50))
At least in my version, rmultinom normalizes the last argument,
so rep(1,50) works just as well.
Back to the original problem: either assign 49 randoms and give the
remainder to the 50th slot, or do the right thing :-) :
You hav
I've usually done my own calculations of entropy. Suppose p is a vector of
values (counts of outcomes, probabilities, etc.). Here's a simple function
to compute Shannon entropy:
shannon.entropy <- function(p)
{
if (min(p) < 0 || sum(p) <= 0)
return(NA)
p.norm <- p[p
jpmorgan.com> writes:
> I am looking for a way to get legends placed automagically in an empty
> spot on a graph. Additional complication comes through my useage of
> multiple graphs on the same plot through mfrow.
Take a look in Hmisc package. There is function for this task.
Gregor
_
On Tue, 08 Jul 2008, squall44 wrote:
> Hi,
>
> I've got the following edf:
>
> ***
>
> x = c(1.6,1.8,2.4,2.7,2.9,3.3,3.4,3.4,4,5.2)
> F2.5 <- ecdf(x)
> plot(F2.5,
> verticals= TRUE,
> do.p = TRUE,
> lwd=3,
> ylab = "",
> xlab = "",
> xlim = c(1,5.5))
> abline(h= (0:5)*
On Tue, Jul 8, 2008 at 9:53 AM, Shubha Vishwanath Karanth <
[EMAIL PROTECTED]> wrote:
> ...actually I need to allocate certain amount of money (here I mentioned
> it as 100) to a randomly selected stocks(50 stocks)... i.e., 100 being
> divided among 50 stocks and preferably all are integer allocat
On 8/07/2008, at 7:38 PM, Fiona Johnson wrote:
Hi
I have just upgraded from R2.6.0 to R2.7.1 (running on Windows) and
a part
of my code that previously ran ok now gives an error. The following
is a
simple example to demonstrate my problem.
a <- array(c(1,2,3,4,5,6,rep(NA,6)),dim=c(6,2))
All,
I've been using aggregate() to compute means and standard deviations at
time/treatment combinations for a longitudinal dataset, using na.rm = TRUE
for missing data.
This was working fine before, but now when I re-run some old code it isn't.
I've backtracked my steps and can't seem to find
You may be able to use the coordinates returned from emptyspace() in plotrix
package as coordinates for the legend in legend().
tolga.i.uzuner wrote:
>
> ...
> I am looking for a way to get legends placed automagically in an empty
> spot on a graph. Additional complication comes through my use
Allan Clark wrote:
hello all
i havnt had a chance to read through the references provided for the
"nls" function (since the libraries are closed now).
can anyone shed some light on how the "plinear" algorithm works? also,
how are the fitted values obtained? also, WHAT DOES THE ".lin" below
REP
On 7/8/2008 2:33 PM, Allan Clark wrote:
hello all
i havnt had a chance to read through the references provided for the
"nls" function (since the libraries are closed now).
can anyone shed some light on how the "plinear" algorithm works?
Nonlinear regression is least squares. There are stan
The following message is provided by Erik
Please provide the reproducible code to do this. Generate a sample data
set using the random data generating functions and show us what you'd
like, we can then more easily help.
[EMAIL PROTECTED] wrote:
Hi,
How about using "subset"?
x1<-tapply(subs
i want to calculate shannon entropy which is
H1,H2,H3upto H7?
if there is any function or any package in
which i can find this entropy directly. do you have any information
please share this and i will be very thankful to you.
Regards,
+
Its a bug in axis.zoo. I have just fixed it in the svn repository so try this:
source("http://r-forge.r-project.org/plugins/scmsvn/viewcvs.php/*checkout*/pkg/R/yearmon.R?rev=485&root=zoo";)
plot(x.zoo[, 25])
axis.zoo uses the same algorithm as axis.Date in R and so it gives
similar results:
# u
Hi!
I am Marta Colombo, student in Mathematical Engineering at "Politecnico di
Milano". For my master degree thesis I have to analyze some categorical data.
My dataset is composed by 327 individuals and 16 variables. I am using Fisher
exact test to test independence on IxJ contingency tables, bu
You may want to look at locator(1) for manual placements;
legend(locator(),...)
BW
Marco
-Oorspronkelijk bericht-
Van: [EMAIL PROTECTED] namens [EMAIL PROTECTED]
Verzonden: di 8-7-2008 20:31
Aan: r-help@r-project.org
Onderwerp: [R] Automatic placement of Legends
Dear R-Us
hello all
i havnt had a chance to read through the references provided for the
"nls" function (since the libraries are closed now).
can anyone shed some light on how the "plinear" algorithm works? also,
how are the fitted values obtained? also, WHAT DOES THE ".lin" below
REPRESENT?
thanking you
On Tue, Jul 8, 2008 at 2:59 PM, stephen sefick <[EMAIL PROTECTED]> wrote:
> x.zoo <- zoo(x,as.yearmon(as.character(x$Yearmonth), "%Y-%m"))
> plot(x.zoo[,25])
1. You are trying to pass data frame to zoo whereas it must be a numeric vector,
matrix or a factor. See ?zoo and try this:
x.zoo <- zoo(
Hi,
How about using "subset"?
x1<-tapply(subset(years, length(area)>20), function(x) length(unique(x)))
I hope this works
Chunhao
Quoting hesicaia <[EMAIL PROTECTED]>:
Hello,
The quick version of my question is how can I extract a matrix instead of
a vector using tapply()? I would like to
That worked fine- now one more question-
plot(x.zoo[,25])
produces a graph with True as the first label on the x-axis
1. why?
2. is it wrong to assume this is february 2006?
thanks
stephen
R2.7.1 Windows XP (I updated zoo last week when I installed 2.7.1)
On Tue, Jul 8, 2008 at 3:17 PM, Gabor Gr
x.zoo <- zoo(x,as.yearmon(as.character(x$Yearmonth), "%Y-%m"))
plot(x.zoo[,25])
#Error in plot.window(...) : invalid 'ylim' value
#there are values
On Tue, Jul 8, 2008 at 2:55 PM, Gabor Grothendieck <[EMAIL PROTECTED]>
wrote:
> On Tue, Jul 8, 2008 at 2:43 PM, Gabor Grothendieck
> <[EMAIL PROTECT
legend will accept locator() which would not automate it but would get
closer
On Tue, Jul 8, 2008 at 2:31 PM, <[EMAIL PROTECTED]> wrote:
> Dear R-Users,
>
> I am looking for a way to get legends placed automagically in an empty
> spot on a graph. Additional complication comes through my useage of
#this is the whole data frame and I tried the suggested and it looks like it
is working but will not plot. thanks agian
x <- structure(list(Yearmonth = structure(c(12L, 24L, 1L, 13L, 14L,
3L, 15L, 4L, 16L, 5L, 17L, 6L, 18L, 7L, 19L, 8L, 20L, 9L, 21L,
10L, 22L, 11L, 23L), .Label = c("2006-02", "200
On Tue, Jul 8, 2008 at 2:43 PM, Gabor Grothendieck
<[EMAIL PROTECTED]> wrote:
> There is no data in your data frame, just index info, so I assume you
> want a zero width time series:
>
> zoo(, as.yearmon(x$Yearmonth, "%Y-%m"))
>
> This also works but then you are left with a character date which yo
There is no data in your data frame, just index info, so I assume you
want a zero width time series:
zoo(, as.yearmon(x$Yearmonth, "%Y-%m"))
This also works but then you are left with a character date which you
may not want:
zoo(, x$Yearmonth)
On Tue, Jul 8, 2008 at 1:43 PM, stephen sefick <[E
On Tue, Jul 8, 2008 at 10:25 AM, collonil <[EMAIL PROTECTED]> wrote:
>
> hello,
>
> i cant find a solution on this (might be) easy problem:
>
> i have a time serie by carlandar weeks, so for every carlendar week i have a
> value. now i would like to use the functions for time series, so i change
>
Dear R-Users,
I am looking for a way to get legends placed automagically in an empty
spot on a graph. Additional complication comes through my useage of
multiple graphs on the same plot through mfrow.
Is there a way to achieve this in R ? I have legends for each of the
sub-plots.
Many thanks
Hello,
The quick version of my question is how can I extract a matrix instead of
a vector using tapply()? I would like to be able to access both the results
of tapply() and also the index variables.
In case further explanation would help: I am analyzing a large (3million
rows x 9 columns) spa
are any of the subsets all NA?
On Tue, Jul 8, 2008 at 1:39 PM, Paul Adams <[EMAIL PROTECTED]> wrote:
> Hello everyone,
> I am trying to plot an MvA plot with the following code:
> dat<-read.table(file="C:\\Documents and Settings\\.txt",header=T)
> file.show(file="C:\\Documents and Settings\\O
#this is a subset of a larger data frame and I am okay with subsetting it as
there are redundant time stamps, but I would like to create a zoo object out
of this and I am having a hard #time figuring out how to do this the date
structure is year and then month
x <- structure(list(Yearmonth = str
?unique
--- On Tue, 7/8/08, muhammad faisal <[EMAIL PROTECTED]> wrote:
> From: muhammad faisal <[EMAIL PROTECTED]>
> Subject: [R] How I count the all possible samples??
> To: r-help@r-project.org
> Received: Tuesday, July 8, 2008, 1:23 PM
> Dear Members,
>
> I am facing a problem during countin
Hello everyone,
I am trying to plot an MvA plot with the following code:
dat<-read.table(file="C:\\Documents and Settings\\.txt",header=T)
file.show(file="C:\\Documents and Settings\\Ow...txt")
library(sma)
data(MouseArray)
dat.o<-as.list(dat)
dat.o$R.2<-as.matrix(dat[,c(1:5)])
dat.o$G.2<-a
Dear Muhammad,
Does it work for you?
x=scan()
41 12 22 23 32 22 23 32 22 21 12 22 23 31 12
21 11 11 14 43 32 22 23 34 41 13 33 32 21 12
22 24 44 42 21 12 23 31 12 23 33 32 22 22 21
13 31 11 12 22 21 14 43 33 32 23 34 43 32 23
33 34 41 13 34 44 42 23 33 32 24 44 44 43 33
31 14 43 32 22 22 21 12 22
Dear Daniela,
Try this:
set.seed(123)
myD<-data.frame(code=sample(letters[1:5],200,replace=T),value=rnorm(200))
tapply(myD$value,myD$code,mean)
a b c d e
0.04401465 0.07813648 0.07018791 -0.14508544 -0.02369875
See ?tapply for more information
Dear Members,
I am facing a problem during counting a sample.
I have 4 characters i.e. A,C,G,T of DNA sequence and code this sequence by
1,2,3,4 respectively.
[1] "41" "12" "22" "23" "32" "22" "23" "32" "22" "21" "12" "22" "23" "31" "12"
[16] "21" "11" "11" "14" "43" "32" "22" "23" "34" "41"
Hello,
Could you let me know if there any R packages available for performing Gage
R & R studies.
Thank you!
Sincerely,
Isabella
Isabella R. Ghement, Ph.D.
Ghement Statistical Consulting Company
301-7031 Blundell Road, Richmond, B.C., Canada, V6Y 1J5
Tel: 604-767-1250
Fax: 604-270-3922
E-mail:
I noticed a problem using R 2.7.1 on Windows XP SP2 with the precompiled
Atlas Rblas.dll. Running the code below causes R to crash. I started R
using Rgui --vanilla and am using the precompiled Atlas Rblas.dll from
cran.fhcrc.org dated 17-Jul-2007 05:04 for Core2 Duo.
The code that causes the cr
Would ave() do what you want?
Rashid
On Tue, 8 Jul 2008, Daniela Ottaviani wrote:
> Dear All,
>
> I have a database of 200 observations named myD.
> In the dataframe there are a column named code (with codes varying from 1 to
> 77), a column named "prevalence" with some quantitative measur
Hi!
I am Marta Colombo, student in Mathematical Engineering at "Politecnico di
Milano". For my master degree thesis I have to analyze some categorical data.
My dataset is composed by 327 individuals and 16 variables. I am using Fisher
exact test to test independence on IxJ contingency tables, bu
Or look at the fame package, which has ti (TimeIndex) and tis
(TimeIndexedSeries) classes that handle this kind of problem. I think it's
simpler and faster than the zoo stuff, but then I would say that, since I
wrote it.
Jeff
"stephen sefick" <[EMAIL PROTECTED]> writes:
> I don't know if this wi
On Tue, Jul 8, 2008 at 11:28 AM, Kanak Choudhury <[EMAIL PROTECTED]> wrote:
> i have a function like
>
> 1+sin(a+bx) where -pi/2<=a+bx<=pi/2
>
> i made a progrom using constrOptim() function but it is not giving good
> result. it depends on the initial value. but when i am doing simulation it
> is
I don't know if this will help, but look at the zoo, chron, and Posix Date
Time packages/classes.
On Tue, Jul 8, 2008 at 10:25 AM, collonil <[EMAIL PROTECTED]> wrote:
>
> hello,
>
> i cant find a solution on this (might be) easy problem:
>
> i have a time serie by carlandar weeks, so for every ca
On Tue, Jul 8, 2008 at 3:18 PM, Daniela Ottaviani <[EMAIL PROTECTED]> wrote:
> Dear All,
>
> I have a database of 200 observations named myD.
> In the dataframe there are a column named code (with codes varying from 1 to
> 77), a column named "prevalence" with some quantitative measurements are
>
On 08-Jul-08 13:18:13, Daniela Ottaviani wrote:
> Dear All,
> I have a database of 200 observations named myD.
> In the dataframe there are a column named code (with codes varying
> from 1 to 77), a column named "prevalence" with some quantitative
> measurements are given and an column named Pr_mea
Try:
> myD <- transform(myD, Pr_mean = ave(prevalence, codes))
See ?ave and ?transform for details.
--
Gregory (Greg) L. Snow Ph.D.
Statistical Data Center
Intermountain Healthcare
[EMAIL PROTECTED]
(801) 408-8111
> -Original Message-
> From: [EMAIL PROTECTED]
> [mailto:[EMAIL PROTEC
Hi all,
I want to draw dot plots for 2 rows in a matrix separately but want to
have them close to each other.
For example, I have a matrix a:
> dim(a)
[1] 319 2
and then transpose it to matrix b
> b<-t(a)
> dim(b)
[1] 2 319
So I want to draw dot plots for these two rows separately but wan
On Monday 07 July 2008, Philippe Grosjean wrote:
> Hello,
>
> As far as I know, Rcmdr is already translated in French. It is thus just
> a question of switching R to French.
Alternatively, you could switch the language after R starts:
Sys.putenv(LANGUAGE="fr")
After this, Rcmdr will be opened in
hi.
i did some research first, but didn't find what i was looking for...
the thing is:
i generated data with correlated errors and simulated the power with using
aov(). what i wanna do now is something similar while using lme(), so that
the corr-structure will be paid attention to.
i'm not quite
hello,
i cant find a solution on this (might be) easy problem:
i have a time serie by carlandar weeks, so for every carlendar week i have a
value. now i would like to use the functions for time series, so i change
structur to a time serie with
cam <- ts(number,start=c(2001,1),deltat=7/365)
or
Hi thanks for your answer.
> ii) hist() will not show the same frequencies as density() unless hist has
> unit bin sizes. density*length is showing number per unit change in Weight;
> hist shows number per bin width.
I belive this is what is confounding me. I have a bin width of 0.1 in
the hist
Dear All,
I have a database of 200 observations named myD.
In the dataframe there are a column named code (with codes varying from 1 to
77), a column named "prevalence" with some quantitative measurements are given
and an column named Pr_mean, with no values.
I would like to set a cycle to comp
In the long run it will probably be easier/simpler to work in a list (or a new
environment) rather than doing everything in the global environment.
For example you can do some of what you are trying below with code like:
> polyList <- c("rs123", "rs124", "rs555", "rs000")
> mylist <- sapply( pol
I saw this type of models in some of my company projects.
To simplify:
Y is regressed on X1 and X2. But the regression is done by two steps:
First Y is regressed on X1 with intercept, and the residuals from the first
step are used to regress on X2, without the constant. The reason to do so
i
Hi,
Can I save a workspace under a 64bit machine in a way that allows reading it
correctly on a 32bit machine? Will the normal procedure of save.image() or
save(#object#) produce errors here, because the data is stored in a binary
mode?
Regards,
Benjamin
==
On 7/8/2008 10:53 AM, Sergi M.Garrido wrote:
Hello,
I have tried to create a package, and I have got it. I checked the files and
I built the package. Nevertheless, I want a .pdf file with the package's
documentation. Anyone know what I have to do?
In a shell in Windows,
R CMD Rd2dvi.sh --p
Try this:
x <- factor(c("220", "220a", "221", "221b", "B221"))
pat <- "[^0-9]+" # match non-digits
nums <- as.numeric(gsub(pat, "", x))
has.lets <- as.numeric(regexpr(pat, x) > 0)
On Tue, Jul 8, 2008 at 7:11 AM, Kunzler, Andreas <[EMAIL PROTECTED]> wrote:
> Dear Everyone,
>
>
>
> I try to automa
Check out sedit() in the Hmisc package
Cheers!
--- On Tue, 7/8/08, Kunzler, Andreas <[EMAIL PROTECTED]> wrote:
> From: Kunzler, Andreas <[EMAIL PROTECTED]>
> Subject: [R] Manipulate Data (with regular expressions)
> To: r-help@r-project.org
> Date: Tuesday, July 8, 2008, 7:11 AM
> Dear Everyon
Hello,
I have tried to create a package, and I have got it. I checked the files and
I built the package. Nevertheless, I want a .pdf file with the package's
documentation. Anyone know what I have to do?
Thanks in advance,
Sergi Martínez
[[alternative HTML version deleted]]
___
I don't think there can be. The EM algorithm isn't
really an algorithm -- it's an outline for an algorithm.
Patrick Burns
[EMAIL PROTECTED]
+44 (0)20 8525 0696
http://www.burns-stat.com
(home of S Poetry and "A Guide for the Unwilling S User")
Peng Jiang wrote:
Hi, dear R experts .
is there
Does not just
polyList[[1]])[1, 1] <- 5
work?
--- On Tue, 7/8/08, Stephane Bourgeois <[EMAIL PROTECTED]> wrote:
> From: Stephane Bourgeois <[EMAIL PROTECTED]>
> Subject: [R] attributing values to dataframe positions following eval
> To: r-help@r-project.org
> Received: Tuesday, July 8, 2008,
> Given a data frame with a continuous variable and a factor. I would like to
> generate a histogram of the continuous variable, where each bar is filled
> with different colors according to the percentage of factor values falling
> into this region of the continuous variable.
How exactly do you w
Dear R-friends,
I am stuck making an LD plot of a small genotype set:
An exert of my data (genotypes)
>tempped.exert
V27/V28 V33/V34 V39/V40 V41/V42
1 B/B B/B A/A B/B
2 B/A B/B A/B B/B
3 B/B B/B A/A B/B
4 B/A B/A A/B B/A
5
Hi, dear R experts .
is there any package contain an universal EM procedure,
that is , for arbitrary d.f. , not just the one in mclust .
thanks in advance
best regards
---
Peng Jiang 江鹏 ,Ph.D. Candidate
Antai College of Economics & Management
安泰经济管理
Hi, as usual I have a problem with R functions!
I formulated a geostatistical model with geoR.
I decided to include some covariates in my model, and I estimated all the
parameters (linear model and covariance function parameters) by the likfit
function as follows:
mod1<-likfit(geodata, ini=va
Joerg van den Hoff fzd.de> writes:
>
>
> res <- nls( y ~ (x * Bmax) / (x + Kd), start = list(Bmax=, Kd=))
>
> (providing some sensible start values for the free parameters , of course)
>
> cf. `nls' manpage for details
There are also "self-starting" models -- SSmicmen I think?
Maybe in nlm
Check out the drc package.
On Mon, Jul 7, 2008 at 5:15 AM, Thiemo Schreiber
<[EMAIL PROTECTED]> wrote:
> Hello everyone,
>
> I have biological data from a competition experiment where a free ligand is
> titrated against the binding of a protein.
>
> Now, I would like to fit a standard on-site bin
After reading your question again I came up with these plots.
dataset <- data.frame(x = c(rnorm(1000), rnorm(200, mean = 1)), y =
gl(2, 100, labels = LETTERS[1:2]))
ggplot(dataset, aes(x = x, group = y)) + stat_bin(aes(fill = ..count..),
width = 0.2) + scale_fill_gradient(low = "red", high = "gree
Francisco Javier Santos Alamillos ujaen.es> writes:
>
> Hi everybody,
>
> I have some problems with the function eigen. I have a square matrix and I
> want to calculate the eigenvalues and eigenvectors. I apply the function
> eigen and I get it, however when I solve the same problem in Statisti
No; thanks for your try, but this is not what I want.
Here each bar has one single color. I would like to render each bar with
several colors according to the distribution of a factor.
I now learned that this is called "stacked histogram" (damned Excel). In
the following entry
https://stat
One of possible solutions is:
generate all the numbers.
num <- rpois(...)
num <- round(num/sum(num)*100,0)
sum(num)
I don't know if it is the best solution, but is one!
Leandro Marino
-Mensagem original-
De: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED]
nome de Peter Dalgaard
Enviada e
Hi Peng,
Does this help?
sum(diff(c(0,sort(sample(seq(1,99,1),50,replace=T)),100)))
Regards
Alex
-Original Message-
From: [EMAIL PROTECTED] [mailto:[EMAIL PROTECTED] On Behalf Of Peng Jiang
Sent: July 8, 2008 3:57 AM
To: [EMAIL PROTECTED]
Cc: [EMAIL PROTECTED]
Subject: Re: [R] Sum(Random N
Hi everybody,
I have some problems with the function eigen. I have a square matrix and I
want to calculate the eigenvalues and eigenvectors. I apply the function
eigen and I get it, however when I solve the same problem in Statistica
software, I realise that some eigenvectors are the opposite. How
>>> Lord Yo <[EMAIL PROTECTED]> 07/08/08 9:00 AM >>>
>I am trying to add a percent sign to my labels in a hist() plot.
?hist says "labels: logical or character. " This should be a clue;
labels could be a character vector.
Try
x<-rlnorm(128, 1)
h<-hist(x, plot=F)
plot(h, labels=paste(round(100*
Is this what you want?
dataset <- data.frame(x = c(rnorm(100), runif(100), rchisq(100, 1)), y =
gl(3, 100, labels = LETTERS[1:3]))
ggplot(dataset, aes(x = x, fill = y)) + geom_histogram()
ggplot(dataset, aes(x = x, fill = y)) + geom_histogram(position =
"dodge")
HTH,
Thierry
-
Dear Everyone,
I try to automatically manipulate the data of a variable (class =
factor) like
x
220
220a
221
221b
B221
Into two variables (class = numeric) like
x y
220 0
220 1
221 0
221 1
221 1
y has to carry the information about the class (number or st
Two thoughts:
i) If you have a narrow distribution, the density can be higher than 1. The
area comes out at 1 for density, and n for the frequency.
ii) hist() will not show the same frequencies as density() unless hist has unit
bin sizes. density*length is showing number per unit change in Weig
Hi,
I'm using the mfrow parameter in par() to plot several timeseries with a
common time x axis as a sequence of plots one below the other. I reduced
the top and bottom margins to zero to get a very nice looking plot but
sometimes the labels on the y axes from one plot overlap with the y axis
la
Hi,
I've got the following edf:
***
x = c(1.6,1.8,2.4,2.7,2.9,3.3,3.4,3.4,4,5.2)
F2.5 <- ecdf(x)
plot(F2.5,
verticals= TRUE,
do.p = TRUE,
lwd=3,
ylab = "",
xlab = "",
xlim = c(1,5.5))
abline(h= (0:5)*0.2)
#mean
abline(v=mean(x), lwd=2)
mtext(text=expression(bar(x
Hi!
Sorry for bothering you again but I can't seem to get it right.
When i multiply the density with the number of observations it seems
to be way to high, The reference curve is drawn at maybe 20 times
higher frequency count than it should be.
I use the following code where "weights$Weight" is
Given a data frame with a continuous variable and a factor. I would like to
generate a histogram of the continuous variable, where each bar is filled
with different colors according to the percentage of factor values falling
into this region of the continuous variable.
I looked into packages like
On Mon, Jul 07, 2008 at 11:15:57AM +0200, Thiemo Schreiber wrote:
> Hello everyone,
>
> I have biological data from a competition experiment where a free ligand is
> titrated against the binding of a protein.
>
> Now, I would like to fit a standard on-site binding curve to this data in
> order
At first sorry for the possibly dumb question of a newbie.
I am using the hydrogeo package to visualize approx. 300 data within the
Piper-Hill diagrams. The package is using a dataframe with 6 columns,
first of them is containing the row.names, next four with the data
itself. The last column c
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