I have a folder full of pngs and jpgs, and would like to consolidate them into
a pdf with appropriate title and labels. Can this be done via R ?
_
Easily publish your photos to your Spaces with Photo Gallery.
[[alternative
Hi R,
I need to generate 50 random numbers (preferably poisson), such that
their sum is equal to 100. How do I do this?
Thank you,
Shubha
This e-mail may contain confidential and/or privileged i...{{dropped:13}}
__
R-help@r-project.org
...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 allocations(i.e.,
5 8 56 12 etc without any decimals)...
Thank you,
Shubha
-Original
Hi,
I am afraid there is no other way except using brute force, that
is , loop until their sum reaches your expectation.
it is easy to figure out this probability by letting their sum to be
a new random variable Z and Z = X_1 + \ldots + X_n
where X_i ~ Poisson({\lambda}_i) . By
Farley, Robert wrote:
I found and loaded the survey package. ?rake and ?postStratify seem
promising. Are there other packages/procedures I've missed? Are there
online references that an R newbie could use to feel comfortable
applying these procedures to a survey? How about a reference
For arbitrary lambda_i it can take years until the sum of 50 such random
variables is 100!
But if one makes lambda_i = 2, then the probability that the sum of 50 of them
equals 100 is about 1/sqrt(2*pi*100), so on average that sequence of 50 numbers
must be generated about sqrt(2*pi*100)) ~ 25
what you could (what I suggest) is not technically clean because your draw
is not strictly random any more. But if you want to distribute 100 on 50
units with a Poisson-distributed variable x, then your lambda must be
100/50=2. You could then sample Poisson distribution ( rpois(50,2) ), sum
over
sum.x=sum(x)
while(sum.x!=100)
x=rpois(50,2)
I have tried this, but this can really take some time. In fact, I would
suggest an similarly fraudulent activity, draw 49 values, and set the last
nonrandomly to the difference between 100 and the sum of the 49 values if
this difference is a.)
For some reason, the while-loop I sent did not work. It was running forever
although I don't know why. Anyway, using repeat works fast:
repeat{x=rpois(50,2)
if(sum(x)==100) break
}
-
cuncta stricte discussurus
-
-Ursprüngliche
Shubha
Does table(ceiling(runif(100,0,50))) give you something like you want?
Richard.
Shubha Vishwanath Karanth 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
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))
apply(a,2,sd,na.rm=T)
In R2.6.0 this gives (which is
Hi everyone
I am trying to add a percent sign to my labels in a hist() plot. labels =
TRUE gives me the values, but I don't know how to add the percent sign. I
would prefer to annote the plot after drawing it, e.g. using text(), but
don't know how to address the positions in a standard
x - rmultinom(1, 100, rep(1/50, 50))
as.vector(x)
[1] 4 0 5 4 2 2 4 4 3 2 4 1 1 1 2 0 0 0 2 1 0 4 3 3 2 4 2 2 2 0 1 1 4 2 2 2 0
1 1 1 2 3 2 2 4 1
[47] 3 1 3 0
sum(x)
[1] 100
Bill Venables
CSIRO Laboratories
PO Box 120, Cleveland, 4163
AUSTRALIA
Office Phone (email preferred): +61 7 3826
Richard Pearson wrote:
Shubha
Does table(ceiling(runif(100,0,50))) give you something like you want?
That's a neat(-ish) solution of Moshe's multinomial formulation.
More generally, if p is a vector of probabilities you can simulate
independent indicators with cut(runif(N), c(0,cumsum(p))
Hi , dear R experts .
I use the package mclust to do some cluster analysis . doing the
example that the document provides
faithfulModel - Mclust( faithful )
plot( faithfulModel )
I remember in R 2.7.0 I can get several figures using the plot
command, but after I update R, using the
I am trying to add a percent sign to my labels in a hist() plot. labels =
TRUE gives me the values, but I don't know how to add the percent sign. I
would prefer to annote the plot after drawing it, e.g. using text()
This is probably a good starting point:
x - rnorm(100)
h - hist(x)
Hi, while using package segmented (version 0.2-4) by Vita Muggeo to
investigate a possible change point (around time = 222) in admissions
for a specific medical condition I had the following error message:
fit2.seg-segmented(fit2, seg.Z=~time,psi=222)
Error in segmented.lm(fit2, seg.Z =
So really you want random portfolios.
While your specification may satisfy the demands
of your current application, in general generating
random portfolios is more complicated. There can
be a large number of constraints required to make
the portfolios conform to a realistic situation.
Two
On Tue, Jul 8, 2008 at 6:59 AM, Daren Tan [EMAIL PROTECTED] wrote:
I have a folder full of pngs and jpgs, and would like to consolidate them
into a pdf with appropriate title and labels. Can this be done via R ?
I do not know whether R can do that. However, you can accomplish that
easily with
If this is about more than a handful files, then it is really painful
to do it with OpenOffice.org or LyX, I guess.
You can use imagemagick, this is fairly standard on Linux. Then it is
something like this, assuming you have bash:
for f in *.png; do convert $f ${f%png}pdf; done
for f in *.jpg;
Hi Daren,
Can R (out)do Emacs? I think you just need to ?Sweave a little.
Mark.
Daren Tan wrote:
I have a folder full of pngs and jpgs, and would like to consolidate them
into a pdf with appropriate title and labels. Can this be done via R ?
Ooops, please ignore my previous mail, I did not read the
question carefully enough.
Gabor
On Tue, Jul 08, 2008 at 02:27:51AM -0700, Mark Difford wrote:
Hi Daren,
Can R (out)do Emacs? I think you just need to ?Sweave a little.
Mark.
Daren Tan wrote:
I have a folder
Hi everybody,
I've been looking around, but can't seem to find the answer.
I get a list of names (which vary, so I never know them in advance), and
transform them into variables, each of which is a dataframe, using
assign:
polyList - c(rs123, rs124, rs555, rs000)
numPoly -
Hi Daniel,
Thanks for the info., I read the wiki link and it made sense
Chibisi
On Tue, Jul 8, 2008 at 1:42 AM, Daniel Malter [EMAIL PROTECTED] wrote:
If that is so, i.e. x1=-x2, then they do not convey different meaning and
cannot be estimated. Think about it that way you leave the house 8
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 impossible of find the best initial value for every step. also i am not
exactly sure how
Dear all,
I am fitting an arimax (arima with some extra explanatory variables)
model to a time series. Say, I have a Y (dependent variable) and an X
(explanatory).
Y is 100 observations (time series) and X is 100 + 20 (20 to use for the
forecast horizon).
I can not make xreg work with
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
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 to
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
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
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,
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
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
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
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
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,
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
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
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
Francisco Javier Santos Alamillos fsantos at 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
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 binding
Joerg van den Hoff j.van_den_hoff at 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
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,
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
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
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 want
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, 5:40 AM
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
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]]
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 Everyone,
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 automatically manipulate the
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
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
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
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( polyList,
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
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
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.
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
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 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 want
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 PROTECTED]
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_mean,
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
given
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 carlendar
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 impossible
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 will
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, but
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 measurements
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
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:
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 13 33 32 21 12
[31] 22 24 44 42 21 12 23 31 12
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
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
?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 counting a
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$R.2-as.matrix(dat[,c(1:5)])
#this is a subset of a larger data frame and I am okay with subsetting it as
there are redundant time stamps, but I would like to create a zoo object out
of this and I am having a hard #time figuring out how to do this the date
structure is year and then month
x - structure(list(Yearmonth =
are any of the subsets all NA?
On Tue, Jul 8, 2008 at 1:39 PM, Paul Adams [EMAIL PROTECTED] wrote:
Hello everyone,
I am trying to plot an MvA plot with the following code:
dat-read.table(file=C:\\Documents and Settings\\.txt,header=T)
file.show(file=C:\\Documents and
Hello,
The quick version of my question is how can I extract a matrix instead of
a vector using tapply()? I would like to be able to access both the results
of tapply() and also the index variables.
In case further explanation would help: I am analyzing a large (3million
rows x 9 columns)
Dear R-Users,
I am looking for a way to get legends placed automagically in an empty
spot on a graph. Additional complication comes through my useage of
multiple graphs on the same plot through mfrow.
Is there a way to achieve this in R ? I have legends for each of the
sub-plots.
Many
On Tue, Jul 8, 2008 at 10:25 AM, collonil [EMAIL PROTECTED] wrote:
hello,
i cant find a solution on this (might be) easy problem:
i have a time serie by carlandar weeks, so for every carlendar week i have a
value. now i would like to use the functions for time series, so i change
structur
There is no data in your data frame, just index info, so I assume you
want a zero width time series:
zoo(, as.yearmon(x$Yearmonth, %Y-%m))
This also works but then you are left with a character date which you
may not want:
zoo(, x$Yearmonth)
On Tue, Jul 8, 2008 at 1:43 PM, stephen sefick
On Tue, Jul 8, 2008 at 2:43 PM, Gabor Grothendieck
[EMAIL PROTECTED] wrote:
There is no data in your data frame, just index info, so I assume you
want a zero width time series:
zoo(, as.yearmon(x$Yearmonth, %Y-%m))
This also works but then you are left with a character date which you
may
#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,
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
x.zoo - zoo(x,as.yearmon(as.character(x$Yearmonth), %Y-%m))
plot(x.zoo[,25])
#Error in plot.window(...) : invalid 'ylim' value
#there are values
On Tue, Jul 8, 2008 at 2:55 PM, Gabor Grothendieck [EMAIL PROTECTED]
wrote:
On Tue, Jul 8, 2008 at 2:43 PM, Gabor Grothendieck
[EMAIL PROTECTED]
That worked fine- now one more question-
plot(x.zoo[,25])
produces a graph with True as the first label on the x-axis
1. why?
2. is it wrong to assume this is february 2006?
thanks
stephen
R2.7.1 Windows XP (I updated zoo last week when I installed 2.7.1)
On Tue, Jul 8, 2008 at 3:17 PM, Gabor
On Tue, Jul 8, 2008 at 2:59 PM, stephen sefick [EMAIL PROTECTED] wrote:
x.zoo - zoo(x,as.yearmon(as.character(x$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 -
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 in
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
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, but
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=485root=zoo;)
plot(x.zoo[, 25])
axis.zoo uses the same algorithm as axis.Date in R and so it gives
similar results:
#
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,
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?
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
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
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 useage
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
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))
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
1 - 100 of 123 matches
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