[R-es] help
estoy intentado descargar el archivo con la siguiente linea:
install_course_github("ifunam", "programacion-estadistica-r")
Pero cada vez que lo intento me sale lo siguiente:
Error in file(con, "wb") : cannot open the connection
In addition: Warning message:
In file(con, "wb") :
cannot open file
'C:/Users/CCC/Documents/R/win-library/4.1/swirl/Courses/temp.zip': No such
file or directory
Qué debo hacer?
--
Cordial Saludo
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La
información contenida en este mensaje y en los archivos adjuntos es
confidencial y reservada y está dirigida exclusivamente a su destinatario,
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conformidad con las normas legales vigentes, su interceptación,
sustracción, extravío, reproducción, lectura o uso esta prohibido a
cualquier persona diferente. Se les exige expresamente a la comunidad ECCI
(Administrtivos, Docentes y Estudiantes) que no realicen declaraciones
difamatorias, no infrinjan ni autoricen ninguna infracción de las leyes de
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[R-es] alguien me ayuda con algo particular del wordclud
Hola Estimados R Amigos, queria contarles que he aplicado una rutina para la creacion de un wordcloud, el proceso super bien pero ahora estoy intentando buscar la rutinas que me faltan, y esto es lo que deseo si me pueden ayudar. ahora con el wordcloud quiisiera destacar en él algunas palabras especificas que estoy buscando en los dicursos de unos niños con los que trabajo en ciencias y conocer en que en frecuecia y/o porcentaje se encuentran. eso mis queridos amigos, espero puedan ayudarme, Gracias Alex Biologo Marino, msc en Epidemiologia --- [[alternative HTML version deleted]] ___ R-help-es mailing list R-help-es@r-project.org https://stat.ethz.ch/mailman/listinfo/r-help-es
[R] WARNING: Method convertPointFromBase
Hello, I’ve noticed a Warning, while working with both the R console and a document window in R: 2016-05-23 22:16:17.507 R[2985:466607] *** WARNING: Method convertPointFromBase: in class NSView is deprecated on 10.7 and later. It should not be used in new applications. I have Version R 3.3.0 GUI 1.68 Mavericks build (7202) installed on my Macbook Pro running Mac OS X 10.11.5 (15F34). Is this something to worry about? Can anyone help me out? I downloaded R from the official servers. Kind regards and thank you in advance ! Fabian __ R-help@r-project.org mailing list -- To UNSUBSCRIBE and more, see https://stat.ethz.ch/mailman/listinfo/r-help PLEASE do read the posting guide http://www.R-project.org/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R-es] cambiar un valor por NA en data frame
ESTIMADA COMUNIDAD R, Tengo un data frame de datos de salud sobre Enfermedades de Notificacion Obligatoria. Algunas variables tienen una codificacion 99, 999, y para asiganr los valores perdidos. LAs variables que tienen esta codificacion son la EDAD, COMUNA_RESIDENCIA y la REGION_RESIDENCIA, respectivamente. Me gustaria poder editar esos valores a NA, sin tener que hacerlo uno por uno con la opcion fix() ya que son muchos es una base datos muy grande. Conoce alguno de esutedes amigos, alguna funcion para realizar dicha edicion les agradeceria mucho, sus comentarios y ayudas. Envio una extracto de la base de datos en .txt para mostrar el problemita por por si alguno le sirviera mirarla se llama ENO2011 Un gran abrazo desde Chile Alex Mellado V Ms(c) en Epidemiología Universidad Católica de Chile SERVEDADSEXOCOMUNA_RESIDENCIA REGION_RESIDENCIA DIAG FECHNOT 3 14 2 21012 A01012-01-2011 17 16 2 84178 A01011-04-2011 16 14 1 71027 A01012-05-2011 15 15 1 61106 A01024-05-2011 19 15 2 81108 A01029-06-2011 24 16 1 10108 10 A54917-01-2011 2 16 1 11011 A54924-02-2011 33 16 1 10209 10 A54903-03-2011 10 999 1 13117 13 A54923-02-2011 3 16 2 21012 A54918-02-2011 3 16 1 21012 A54906-04-2011 3 15 1 21012 A54918-04-2011 19 16 1 81108 A54919-05-2011 3 16 1 21012 A54924-05-2011 6 15 1 56035 A54905-06-2011 25 16 2 11101 11 A54910-06-2011 13 15 1 13105 13 A54905-04-2011 5 16 1 41024 A54916-06-2011 3 16 1 21012 A54923-06-2011 14 15 2 13110 13 A54907-04-2011 14 16 1 13110 13 A54905-04-2011 24 16 2 10101 10 A54928-06-2011 19 16 1 81108 A54915-07-2011 28 14 1 82038 A54921-07-2011 10 15 1 13117 13 A54923-06-2011 24 14 1 10101 10 A54909-08-2011 7 16 1 58015 A54905-08-2011 1 14 2 15101 15 A54918-08-2011 7 15 1 51075 A54926-08-2011 1 14 2 15101 15 A54907-09-2011 12 14 1 13122 13 A54904-08-2011 24 16 1 10101 10 A54922-08-2011 33 16 1 10209 10 A54903-03-2011 1 15 2 15101 15 A54914-11-2011 2 15 2 11011 A54911-11-2011 13 15 1 13402 13 A54907-10-2011 24 16 1 10101 10 A54922-11-2011 5 16 2 41024 A54901-12-2011 1 16 2 15101 15 A63004-01-2011 1 15 2 15101 15 A63007-01-2011 1 14 1 15101 15 A63021-01-2011 1 16 1 15101 15 A63021-04-2011 1 16 2 15101 15 A63007-06-2011 1 15 2 15101 15 A63008-07-2011 1 16 1 15101 15 A63019-07-2011 1 14 2 15101 15 A63005-08-2011 1 15 2 15101 15 A63029-08-2011 1 16 2 15101 15 A63024-10-2011 1 16 2 15101 15 A64X11-07-2011 1 15 2 15101 15 A64X20-09-2011 1 15 2 15101 15 A64X14-11-2011 1 16 2 15101 15 B15919-01-2011 1 16 1 15101 15 B15918-01-2011 1 16 1 15101 15 B15921-03-2011 12 15 1 13123 13 B15924-03-2011 1 14 1 15101 15 B15911-04-2011 1 14 1 15101 15 B15911-04-2011 1 16 2 15101 15 B15906-04-2011 1 16 2 15101 15 B15919-05-2011 16 15 1 73047 B15917-05-2011 12 14 1 13122 13 B15901-06-2011 13 14 2 13402 13 B15908-05-2011 1 16 1 15101 15 B15913-06-2011 21 16 1 91019 B15927-05-2011 1 14 2 15101 15 B15925-06-2011 13 16 2 13116 13 B15924-06-2011 1 16 1 15101 15 B15915-07-2011 20 16 2 83018 B15910-08-2011 20 14 2 83018 B15908-08-2011 1 15
[R] [R-pkgs] New R package ``bvarsv''
Dear R users, please let me draw your attention to my new R package bvarsv (on CRAN since August 28) which implements the Primiceri (Review of Economic Studies, 2005) vector autoregressive model. The model is popular in macroeconomic analysis as it allows to model instabilities (e.g. in the mean and volatility dynamics) that are often observed in macroeconomic time series like inflation. The package provides functionality for Bayesian analysis of the Primiceri model. To the best of my knowledge, it is the only publicly available R code to do so. The underlying Markov Chain Monte Carlo algorithm is computationally challenging, since each iteration requires multiple calls of a recursive Kalman filter type algorithm. Therefore, bvarsv relies on C++ code ported to R via Rcpp and RcppArmadillo. The clear focus of the package is on forecasting, as opposed to structural analysis of economic relationships. The package allows to compute posterior predictive distributions, which can then be plotted and analyzed using forecast evaluation techniques. More detailed documentation and examples is available here: https://sites.google.com/site/fk83research/code Your feedback on any aspects of the package, as well as possible improvements or extensions, would be highly appreciated. Thank you and best wishes, Fabian Krüger ___ R-packages mailing list r-packa...@r-project.org https://stat.ethz.ch/mailman/listinfo/r-packages __ 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/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] data.frame(1)*1:4 = 1?
This is because your vector is recycled: data.frame(1)*1:4 = data.frame(1)*c(1,2,3,4) only the first element is needed since the data frame has nothing else to multiply with c(2,3,4) (x-data.frame(1:2, 3:4)) X1.2 X3.4 113 224 (y-x*5:7) y[1,1] = x[1,1] * 5 y[2,1] = x[2,1] * 6 y[1,2] = x[1,2] * 7 y[2,2] = x[2,2] * 5 since you have e vector with length 3, for the 4th entry in the data.frame the first element in the vector is recycled. hope this helps On 03-04-2014 08:42, Spencer Graves wrote: Hello, All: What's the logic behind data.frame(1)*1:4 producing a scalar 1? Or the following: data.frame(1:2, 3:4)*5:7 X1.2 X3.4 15 21 2 12 20 I stumbled over this, because I thought I was multiplying a scalar times a vector, and obtaining a scalar rather than the anticipated vector. I learned that my scalar was in fact a data.frame with one row and one column. What am I missing? Thanks, Spencer __ 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/posting-guide.html and provide commented, minimal, self-contained, reproducible code. __ 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/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] mcr mcreg: Why are BCa and quantile se values calculated, stored, then the stored values set to NA?
This piece of code is indeed confusing. Generally quantile and BCa
bootstrap do not estimate a global SE, so the glob.sigma slot you want to
access is not really meaningful for those methods.
The result object of a BCa bootstrap as calculated by the mc.bootstrap
function will contain a glob.sigma slot that is set to the analytical SE
estimates of the complete data set for regression methods where analytical
estimates of SE are available (LinReg, WLinReg, Deming). For the other
regression methods (WDeming, PaBa) glob.sigma will already be NA. Since the
analytical SE estimates are not relevant when you perform quantile or BCa
bootstrap we decided at some point to always set them NA for these
bootstrap methods - independent of regression method. The code copying the
glob.sigma results is pointless and we'll remove it in the next release.
Now, if you want to calculate the mean or the sd of the individual
bootstrap estimates, you can directly access the MCResultResampling and
MCResultBCa objects. They store the estimates of regression coefficients
for each bootstrap sample in slots B0 (intercept) and B1 (slope). Example:
library(mcr)
data(creatinine)
m - mcreg(as.matrix(creatinine), method.bootstrap.ci = quantile, na.rm =
TRUE)
mean.slope.est- mean(m@B1)
sd.slope.est- sd(m@B1)
mean.intercept.est - mean(m@B0)
sd.intercept.est - sd(m@B1)
However, note that these estimates have no direct relationship to the BCa
confidence intervals.
Best regards,
Fabian
On Tuesday, February 25, 2014 3:54:11 PM UTC+1, Peter Crowther wrote:
In mcreg, there's the following snippet for bootstrap CI method of BCa
(and a very similar one for quantile):
else if (method.bootstrap.ci == BCa) {
bootB0 - mc.calc.bca(Xboot = B0, Xjack = B0jack,
xhat = glob.coef[1], alpha)
bootB1 - mc.calc.bca(Xboot = B1, Xjack = B1jack,
xhat = glob.coef[2], alpha)
bootB0$se - glob.sigma[1]
bootB1$se - glob.sigma[2]
bootB0$se - NA
bootB1$se - NA
}
[...]
mc.res - mc.make.CIframe(b0 = glob.coef[1], b1 = glob.coef[2],
se.b0 = bootB0$se, se.b1 = bootB1$se, CI.b0 = bootB0$CI,
CI.b1 = bootB1$CI)
I'd quite like to use in my code the SE values that were calculated in
mcreg. Can any kind soul shed light on why they're calculated,
stored, then the stored values set to NA?
Thanks in advance for any insight!
- Peter
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Re: [R] function doesn't change the variable in the right way
Hello
The function does exactly what you tell it to do; first you substitute
all NA with the mean; then you subtract the mean; for the NA this
meand: mean-mean=0; and this is what you get.
the problem is not the function but the z-score of means.
lg fabian
On 07-03-2014 12:17, David Croll wrote:
Dear R users and friends,
I would like to ask you about the weird behaviour of a function I just
wrote. This little function should take a vector, find NAs and
substitute them for the mean of the vector, and return the normalized
value of that vector.
I've tried both - and - for changing the variables.
That's what I do:
# just a vector:
b - c(1,1,1,NA,3,2)
# my function:
normalize - function(x) {
# copy x into new vector
xn - x
# index of NAs
nas - which(is.na(xn) ==TRUE)
m - mean(xn, na.rm=TRUE)
# insert mean for NAs
xn[nas] - m
# normalize
return((xn - mean(xn))/sd(xn))
}
# run...
normalize(b)
# here's what I get:
# [1] -0.75 -0.75 -0.75 0.00 1.75 0.50
The 4th value should be 1.6, but is 0.
I believe the answer to my problem is pretty obvious, but I can't see
it...
Best regards,
David
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[R] Negative binomial distribution mixture model
Dear R community I'm new to R and I'd like to ask for hints how to approach the following problem: I have two vectors of count data 'observed_S' and 'observed_A'. Whereas 'observed_S' follows a neg. binomial distribution but 'observed_A' is a mixture of an unknown variable 'A' (also from a neg. binomial distribution) and a shadow of 'observed_S', so to say. Therefore my model can be expressed by: observed_A = factor*observed_S + A I d like to find the factor explaining my two data vector most coherently. but I m overwhelmed by the different packages dealing with neg. binomial distribution regression. Can someone please point me which package is most useful for the above described purpose? Thank you very much in advance Regards Fabian __ 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/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Problem with metaMDS in vegan
Thank you very much! This is exactly the problem, I am in. I ran the NMDS for only one gender and it shows me the results, I was expecting. I was told to use bray-curtis, because there are a lot of 0s in my data set and in e.g. manhattan these would have to much influence on the result. But in nature, chemicals that are only weakly represented may not have very much influence on the other individuals. For the moment, I won't question that decision, because it's quite a convention in chemical ecology, as I heard. Regards, Fabian -- View this message in context: http://r.789695.n4.nabble.com/Problem-with-metaMDS-in-vegan-tp4685022p4685134.html Sent from the R help mailing list archive at Nabble.com. __ 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/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Problem with metaMDS in vegan
Hello, I am relatively new to vegan and need some help with metaMDS. My R is Version 3.0.2. I am analysing cuticular profiles of flys from different locations an different gender. My data is in a 48 x 70 matrix. It is about percentages and so, there are a lot of 0s included. If I run metaMDS, it just gives me: Warning message: In metaMDS(data1, distance = bray, k = 2, trymax = 100, autotransform = F) : Stress is (nearly) zero - you may have insufficient data no matter, what I change. The error is always around 9.241406e-05, which seems to be far to low for my data. By inspecting the data sheet, I clearly see that there must be two clusters – but there are still some differences in these (meaning: Not all female flies show the same pattern than the others and there are slight differences in the locations …) - so I expect to see two clusters for the sexes and in these, an accumulation of the different locations. But metaMDS just gives me 2 points (with ordiplot), where the 24 males and the 24 females are completely stacked. That's just some kind of worthless for me. So is there any chance to get some nice clusters? I tried to limit the number of iterationswith maxit, but had to go down to 10 to reach a suitable result. And I don't think, I may limit the number of iterations that much. What could possibly be the problem? It would be great if somebody could help me. Thaks! Fabian -- View this message in context: http://r.789695.n4.nabble.com/Problem-with-metaMDS-in-vegan-tp4685022.html Sent from the R help mailing list archive at Nabble.com. __ 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/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Problem with metaMDS in vegan
Thanks for the fast response! My data is actually just the percentages of the substances found in the cuticular layer of the flies. The quantitative and qualitative composition of this layer differs in gender, locality and so on. My 70 „species“ are the chemical compounds we identified and my 48 „sites“ are the individuals (location and gender are encrypted in the individuals' recognition code). Therefore, I have the percentages of one chemical compound over all tested individuals in one row (as numeric data). You could say that these are the abundances of the species (the compounds) at the different sites (individuals). So my data is quite similar to community structure. I have done this before and it worked just fine. I used: metaMDS(data, distance=“bray“, k=2, trymax=1, autotransform=F) As I mentioned before, this time the stress is very low (around 9.5 * 10^-5) and R gives me that warning message. Though I have always had data in the same structure, I've never been in this situation. The NMDS just stops after the first run, because of the low stress. But there are obviously some differences in my data – I can see them with my eye. If I discard the 'distance=”bray”', nothing changes. May it be possible that these are simply just to little to be recognised correctly? How could I change that? If I add 'maxit=10', I get quite what I assumed. Two completely separated clusters with cloudlike appearance (stress around 0.02). The clusters are the two sexes and in each cluster, the localities group together. But I'm not quite sure, if I may set a maxit-level that low – or if I just create some false relations with that. I also recognised , that the stress gets lower, the higher I set 'maxit' and the lower I set 'trymax'. What could be going wrong? I really have no clue … Thanks for your help! Regards, Fabian -- View this message in context: http://r.789695.n4.nabble.com/Problem-with-metaMDS-in-vegan-tp4685022p4685043.html Sent from the R help mailing list archive at Nabble.com. __ 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/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] comparison of bivariate normal distributions
Hello Petr and thanks for your help! Thanks also for the correction on the code, of cause it is better to use the real mean and covariance than those estimated by mean() and cov(). What I am after is that if I have the two two-dimensional probability density functions of the distribution of my parameters, these functions should have an intersect. And if I know the intersect I should be able to say that a certain volume under both functions is shared by both comparable to the surface shared by two one-dimensional normal distributions. From there on I was hoping to be able to say something about how high the probability is that my samples come from two different populations (in the logic of a t-test). Would this be possible or reasonable at all? Thank again, I really appreciate your help! best Fabian Would it be possible, too to On 26 Apr 2012, at 10:56, Petr Savicky wrote: On Wed, Apr 25, 2012 at 08:43:34PM +, Fabian Roger wrote: sorry for cross-posting Dear all, I have tow (several) bivariate distributions with a known mean and variance-covariance structure (hence a known density function) that I would like to compare in order to get an intersect that tells me something about how different these distributions are (as t-statistics for univariate distributions). In order to visualize what I mean hear a little code example: library(mvtnorm) c-data.frame(rnorm(1000,5,sd=1),rnorm(1000,6,sd=1)) c2-data.frame(rnorm(1000,10,sd=2),rnorm(1000,7,sd=1)) xx=seq(0,20,0.1) yy=seq(0,20,0.1) xmult=cbind(rep(yy,201),rep(xx,each=201)) dens=dmvnorm(xmult,mean(c),cov(c)) dmat=matrix(dens,ncol=length(yy),nrow=length(xx),byrow=F) dens2=dmvnorm(xmult,mean(c2),cov(c2)) dmat2=matrix(dens2,ncol=length(yy),nrow=length(xx),byrow=F) contour(xx,yy,dmat,lwd=2) contour(xx,yy,dmat2,lwd=2,add=T) ## Is their an easy way to do this (maybe with dmvnorm()?) and could I interpret the intersect (shared volume) in the sense of a t-statistic? Hello: I am not sure, what is exactly the question. The parameters of a bivariate normal distribution are the covariance matrix and the mean vector. For the distributions above, these are mean1 - c(5, 6) cov1 - diag(c(1, 1)) mean2 - c(10, 7) cov2 - diag(c(2, 1)^2) These parameters may be used in the code above instead of mean(c), cov(c) and mean(c2), cov(c2). The curves of equal density are ellipses, whose equations may be derived from the mean vector mu and covariance matrix Sigma using the formula for the exponent in the bivariate density of normal distribution. For any fixed value of the density, the formula has the form (x - mu)' Sigma^(-1) (x - mu) = const where (x - mu)' is the transpose of (x - mu), (x - mu) is a column vector and const is some constant. The value of const may be derived using the full formula for the bivariate density, which is at http://en.wikipedia.org/wiki/Multivariate_normal_distribution In order to compute the area of this ellipse, we have to specify a required density, more exactly a lower bound on the density. The area of an ellipse is \pi a b, where a, b, are its axes. If we have two such ellipses, it is possible to compute the area of their intersection, but again, for each ellipse, a lower bound on the density is needed. Is the area of the intersection of two ellipses for some specified lower bounds on the density, what you want to compute? Petr Savicky. __ R-help@r-project.orgmailto: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/posting-guide.html and provide commented, minimal, self-contained, reproducible code. Fabian Roger, Ph.D. student Dept of Biological and Environmental Sciences University of Gothenburg Box 461 SE-405 30 Göteborg Sweden Tel. +46 31 786 2933 [[alternative HTML version deleted]] __ 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/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] comparison of bivariate normal distributions
sorry for cross-posting Dear all, I have tow (several) bivariate distributions with a known mean and variance-covariance structure (hence a known density function) that I would like to compare in order to get an intersect that tells me something about how different these distributions are (as t-statistics for univariate distributions). In order to visualize what I mean hear a little code example: library(mvtnorm) c-data.frame(rnorm(1000,5,sd=1),rnorm(1000,6,sd=1)) c2-data.frame(rnorm(1000,10,sd=2),rnorm(1000,7,sd=1)) xx=seq(0,20,0.1) yy=seq(0,20,0.1) xmult=cbind(rep(yy,201),rep(xx,each=201)) dens=dmvnorm(xmult,mean(c),cov(c)) dmat=matrix(dens,ncol=length(yy),nrow=length(xx),byrow=F) dens2=dmvnorm(xmult,mean(c2),cov(c2)) dmat2=matrix(dens2,ncol=length(yy),nrow=length(xx),byrow=F) contour(xx,yy,dmat,lwd=2) contour(xx,yy,dmat2,lwd=2,add=T) ## Is their an easy way to do this (maybe with dmvnorm()?) and could I interpret the intersect (shared volume) in the sense of a t-statistic? Thanks a lot for your help! _ Fabian Roger, Ph.D. student Dept of Biological and Environmental Sciences University of Gothenburg Box 461 SE-405 30 Göteborg Sweden Tel. +46 31 786 2933 [[alternative HTML version deleted]] __ 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/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] compare bivariate normal distributions
Dear all, I have tow (several) bivariate distributions with a known mean and variance-covariance structure (hence a known density function) that I would like to compare in order to get an intersect that tells me something about how different these distributions are (as t-statistics for univariate distributions). In order to visualize what I mean hear a little code example: library(mvtnorm) c-data.frame(rnorm(1000,5,sd=1),rnorm(1000,6,sd=1)) c2-data.frame(rnorm(1000,10,sd=2),rnorm(1000,7,sd=1)) xx=seq(0,20,0.1) yy=seq(0,20,0.1) xmult=cbind(rep(yy,201),rep(xx,each=201)) dens=dmvnorm(xmult,mean(c),cov(c)) dmat=matrix(dens,ncol=length(yy),nrow=length(xx),byrow=F) dens2=dmvnorm(xmult,mean(c2),cov(c2)) dmat2=matrix(dens2,ncol=length(yy),nrow=length(xx),byrow=F) contour(xx,yy,dmat,lwd=2) contour(xx,yy,dmat2,lwd=2,add=T) ## Is their an easy way to do this (maybe with dmvnorm()?) and could I interpret the intersect (shared volume) in the sense of a t-statistic? Thanks a lot for your help! sincerely, _ Fabian Fabian Roger, Ph.D. student Dept of Biological and Environmental Sciences University of Gothenburg Box 461 SE-405 30 Göteborg Sweden Tel. +46 31 786 2933 __ 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/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] (no subject)
#subject: type III sum of squares - anova() Anova() AnovaM() #R-version: 2.12.2 #Hello everyone, #I am currently evaluating experimental data of a two factor experiment. to illustrate de my problem I will use following #dummy dataset: Factor T1 has 3 levels (A,B,C) and factor T2 has 2 levels E and F. The design is #completly balanced, each factor combinations has 4 replicates. #the dataset looks like this: T1-(c(rep(c(A,B,C),each=8))) T2-(c(rep(rep(c(E,F),each=4),3))) RESPONSE-c(1,2,3,2,2,1,3,2,9,8,8,9,6,5,5,6,5,5,5,6,1,2,3,3) DF-as.data.frame(cbind(T1,T2,RESPONSE)) DF$RESPONSE-as.numeric(DF$RESPONSE) DF T1 T2 RESPONSE 1 A E1 2 A E2 3 A E3 4 A E2 5 A F2 6 A F1 7 A F3 8 A F2 9 B E7 10 B E6 11 B E6 12 B E7 13 B F5 14 B F4 15 B F4 16 B F5 17 C E4 18 C E4 19 C E4 20 C E5 21 C F1 22 C F2 23 C F3 24 C F3 library(biology) replications(RESPONSE ~ T1*T2,data=DF) T1T2 T1:T2 812 4 is.balanced(RESPONSE ~ T1*T2,data=DF) [1] TRUE #Now I would like to know whether T1, T2 or T1*T2 have a significant effect on RESPONSE. As far as I know, the #theory says that I should use a type III sum of squares, but the theory also says that if the design is completely #balanced, there is no difference between type I,II or III sum of squares. #so I first fit a linear model: my.anov-lm(RESPONSE~T1+T2+T1:T2) #then I do a normal Anova anova(my.anov) Analysis of Variance Table Response: RESPONSE Df Sum Sq Mean Sq F valuePr(F) T1 2 103.0 51.500 97.579 2.183e-10 *** T2 1 24.0 24.000 45.474 2.550e-06 *** T1:T2 2 12.0 6.000 11.368 0.000642 *** Residuals 189.5 0.528 #When I do the same with the Anova() function from the car package I get the same result Anova(my.anov) Anova Table (Type II tests) Response: RESPONSE Sum Sq Df F valuePr(F) T1 103.0 2 97.579 2.183e-10 *** T2 24.0 1 45.474 2.550e-06 *** T1:T2 12.0 2 11.368 0.000642 *** Residuals9.5 18 #(type two sees to be the default and type=I produces an error (why?)) #yet, when I specify type=III it gives me something completely different: Anova(my.anov,type=III) Anova Table (Type III tests) Response: RESPONSE Sum Sq Df F valuePr(F) (Intercept) 16.0 1 30.316 3.148e-05 *** T184.5 2 80.053 1.100e-09 *** T2 0.0 1 0.000 1.00 T1:T2 12.0 2 11.368 0.000642 *** Residuals 9.5 18 #an the AnovaM() function from the biology package does the same for type I and II and produces the following #result: library(biology) AnovaM(my.anov,type=III) Df Sum Sq Mean Sq F value Pr(F) T1 2 84.5 42.250 80.053 1.10e-09 *** T2 1 24.0 24.000 45.474 2.55e-06 *** T1:T22 12.0 6.000 11.368 0.000642 *** Residuals 189.5 0.528 #Is type 3 the Type I should use and why do the results differ if the design is balanced? I am really confused, it would #be great if someone could help me out! #Thanks a lot for your help! #/Fabian #University of Gothenburg [[alternative HTML version deleted]] __ 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/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] [R-pkgs] spikeSlabGAM
spikeSlabGAM_0.9-6 (initial public release) spikeSlabGAM implements Bayesian variable selection, model choice, and regularized estimation in (geo-)additive mixed models for Gaussian, binomial, and Poisson responses. Its purpose is (1) to choose an appropriate subset of potential covariates and their interactions, (2) to determine whether linear or more flexible functional forms (P-splines, tensor product splines) are required to model the (joint) effects of the respective covariates, and (3) to fit these regularized effects and return (model-averaged) estimates. Selection and regularization of the model terms is based on a novel spike-and-slab-type prior on coefficient groups associated with parametric and semi-parametric effects. Inference is fully Bayesian with an underlying MCMC sampler implemented in C and can take advantage of multi-core processors via multicore or snow. The package uses standard formula syntax so that complex models can be specified very concisely. It features powerful and user friendly visualizations using ggplot2. -- Fabian Scheipl Department of Statistics Ludwig-Maximilians-University Munich Ludwigstr. 33, room 239 80539 Munich Germany Phone: +49-89-2180-2284 http://www.statistik.lmu.de/~scheipl/ ___ R-packages mailing list r-packa...@r-project.org https://stat.ethz.ch/mailman/listinfo/r-packages __ 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/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] lme:correlationstructure AR1 and random factor
Dear helpers, I tried these models to run in the package nlme, but allways got the same error message... I have a correlation in 5 sessions within a field (n=12) with ten traps in one field. res2a - lme(response~x+y+z+treatment),correlation = corARMA(form = ~ session|trapfield, p = 1, q = 0), random=~1|field, na.action=na.omit, data=plates, method=ML) res2a - lme(response~x+y+z+treatment,correlation = corCAR1(form = ~ session|trapfield), random=~1|field, na.action=na.omit, data=plates, method=ML) Error: Incompatible formulas for groups in random and correlation What is the problem? thanks a lot! Yvonne -- View this message in context: http://r.789695.n4.nabble.com/lme-correlationstructure-AR1-and-random-factor-tp3416248p3416248.html Sent from the R help mailing list archive at Nabble.com. __ 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/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] binary data with correlation
Dear posters, I have a question concerning binary data analysis. I have presence absence data of 5 sampling sessions within 3 years, of 12 fields. Each field had 12 traps. I would like to analyse the data with a Generalized Estimating Equations (GEE) Model in R. For the abundance data I used a gls with the function: correlation = corARMA(form = ~ session|trapfield, p = 1, q = 0) But now I want to use presence absence data…I know about the problem with correlated -binary data- but maybe somebody already created a solution? Something like that, which is not working…. rgee1 - geeglm(Alus01 ~ treatment + scale(LAI) + scale(log(Tot_Sp)) + scale(AveH) , data = plates, family = binomial, waves = session, id = trapfield, corstr = ar1) Any help is appreciated a lot…. Thanks Yvonne -- View this message in context: http://r.789695.n4.nabble.com/binary-data-with-correlation-tp3393884p3393884.html Sent from the R help mailing list archive at Nabble.com. __ 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/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] merge table rows (\multirow)
You can also automate it with this:
do.multirow-function(df, which=1:ncol(df)){
for(c in which){
runs - rle(as.character(df[,c]))
if(all(runs$lengths1)){
tmp - rep(, nrow(df))
tmp[c(1, 1+head(cumsum(runs$lengths),-
1))] -
paste(\\multirow{,runs$lengths,}{*}{,df[c(1,
1+head(cumsum(runs$lengths),-1)),c],},sep=)
df[,c] - tmp
}
}
return(df)
}
This will replace the which-columns of data.frame df that have
only repeated entries with the appropriate \multirow commands.
You then have to use print.xtable with sth like
print(xtable(do.multirow(df)),
sanitize.text.function = function(x){
return(x)
}
)
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[R] Modify the plotting parameters for Vennerable obj.
Dear List, I would like to modify the settings for plotting a Vennerable object, but I don't know how...so if anyone has an idea I would be really graetfull. best, Fabian some R code to illustrate my problem: library(Vennerable) ven - compute.Venn(Venn(SetNames=c(A, B), Weight=c(0,111,106, 26))) # now my problem is that whenever I plot the object, the plot appears in box, and for cosmetic reasons I would like to get rid of that. plot(ven) sessionInfo() R version 2.10.0 (2009-10-26) i386-apple-darwin8.11.1 locale: [1] C attached base packages: [1] grid stats graphics grDevices utils datasets methods [8] base other attached packages: [1] Vennerable_2.0 RColorBrewer_1.0-2 lattice_0.18-3 RBGL_1.22.0 [5] graph_1.24.0 ggplot2_0.8.5 digest_0.4.2 reshape_0.8.3 [9] plyr_0.1.9 proto_0.3-8limma_3.2.1 loaded via a namespace (and not attached): [1] tools_2.10.0 __ 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/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] selection of optim parameters
Hi all, I am trying to rebuild the results of a study using a different data set. I'm using about 450 observations. The code I've written seems to work well, but I have some troubles minimizing the negative of the LogLikelyhood function using 5 free parameters. As starting values I am using the result of the paper I am rebuiling. The system.time of the calculation of the function is about 0.65 sec. Since the free parameters should be within some boundaries I am using the following command: optim(fn=calculateLogLikelyhood, c(0.4, 2, 0.4, 8, 0.8), lower=c(0,0,0,0,0), upper=c(1, 5, Inf, Inf, 1), control=list(trace=1, maxit=1000)) Unfortunately the result doesn't seem to be reasonable. 3 of the optimized parameters are on the boundaries. Unfortunately I don't have much experience using optimizatzion methods. That's why I am asking you. Do you have any hints for me what should be taken into account when doing such an optimization. Is there a good way to implement the boundaries into the code (instead of doing it while optimizing)? I've read about parscale in the help-section. Unfortunately I don't really know how to use it. And anyways, could this help? What other points/controls should be taken into account? I know that this might be a bit little information about my current code. But I don't know what you need to give me some advise. Just let me know what you need to know. Thankds __ 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/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Error when callin g C-Code
Hi
when I call the function below in R, i get the error: Object 'pairlist'
can't be converted to 'double'.
#include R.h
#include Rdefines.h
#include Rmath.h
SEXP CSimPoisson(SEXP lambda, SEXP tgrid, SEXP T2M, SEXP Ni, SEXP NT)
{
double sign, EVar;
double *xlambda, *xtgrid, *xT2M, *xNi, *xNT, *xtau;
SEXP tau;
int ltgrid =0;
int i = 0;
int j = 0;
sign = 0;
EVar = 0;
ltgrid = LENGTH(tgrid);
PROTECT(lambda = AS_NUMERIC(lambda));
PROTECT(tgrid = AS_NUMERIC(tgrid));
PROTECT(T2M = AS_NUMERIC(T2M));
PROTECT(Ni = AS_NUMERIC(Ni));
PROTECT(NT = AS_NUMERIC(NT));
PROTECT(tau = NEW_NUMERIC(1));
xlambda = NUMERIC_POINTER(lambda);
xtgrid = NUMERIC_POINTER(tgrid);
xT2M = NUMERIC_POINTER(T2M);
xNi = NUMERIC_POINTER(Ni);
xNT = NUMERIC_POINTER(NT);
xtau = NUMERIC_POINTER(tau);
GetRNGstate();
if(xlambda[0] != 0)
{
while(1)
{
EVar = rexp(xlambda[0]);
sign = sign + EVar;
if (sign xT2M[0])
{
break;
}
xtau = Realloc(xtau, i+1, double);
xtau[i] = sign;
i = i+1;
for(j; j ltgrid;j++)
{
if (xtgrid[j] sign)
{
xNi[j] = xNT[0];
}
{
break;
}
}
xNT[0] = xNT[0] + 1;
}
for(j;j ltgrid;j++)
{
xNi[j] = xNT[0];
}
}
PutRNGstate();
UNPROTECT(6);
return tau;
}
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[R] Simulating a Poisson Process in R by calling C Code over .Call
Hi
I want to write a C function for the R Code below and call it with .Call:
SimPoisson - function(lambda,tgrid,T2M)
#Simulation eines Poissonprozesses
{
NT - 0
Ni - rep(0,length(tgrid))
tau - 0
sign - 0
if(lambda != 0)
{
i=1
j=1
while (1)
{
EVar - rexp(1,lambda)
sign - sign + EVar
if (sign T2M)
{
break
}
tau[i] - sign
i = i+1
for (j in j:length(tgrid))
{
if (tgrid[j] sign)
{
Ni[j] - NT
}else
{
break
}
}
NT - NT + 1
}
for (j in j:length(tgrid))
{
Ni[j] - NT
}
}
return(list(NT=NT,Ni=Ni,tau=tau))
}
I read the manual writing R extensions over and over again, but i have
no idea, how to solve the problem with tau because i dont no the length
of tau at the begining of the function
Fabian
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[R] Help with R graphics
I need to make a plot illustrating main characterisitig of river drainage data. For this I have 2 questions: how can I rotate a histogram -90° (or 270°) (like the horizontal=TRUE with plot)? how can I use split.screen to produce 3 plot with uneuqal size (1/5, 2/5, 2/5 of the screen width)? thank you very much in advance for your help fabian __ 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/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] [R-pkgs] amer: generalized additive mixed models with lme4
Dear R-users, I'd like to announce the release of the amer-package that adds the capability to fit generalized additive mixed models to lme4. It includes a vignette with real data examples and a brief summary of the theory behind the implementation. Best, Fabian [[alternative HTML version deleted]] ___ R-packages mailing list r-packa...@r-project.org https://stat.ethz.ch/mailman/listinfo/r-packages __ 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/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
Re: [R] Function match.call or saving all initial parameters
This does what you want:
expand.call - function(definition = NULL, call = sys.call(sys.parent()),
expand.dots = TRUE)
# like match.call, but with all formal args instead of only the specified
ones
{
ans - as.list(match.call(definition, call, expand.dots))
frmls - formals(deparse(ans[[1]]))
add - which(!(names(frmls) %in% names(ans)))
return(as.call(c(ans, frmls[add])))
}
Best,
Fabian
[[alternative HTML version deleted]]
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Re: [R] Function match.call or saving all initial parameters
it doesn't actually- this should work:
expand.call - function(call=sys.call(sys.parent(1)), expand.dots = TRUE)
# similar to match.call, but with all formal args not the specified ones
only
{
ans - as.list(call)
frmls - formals(deparse(ans[[1]]))
add - which(!(names(frmls) %in% names(ans)))
return(as.call(c(ans, frmls[add])))
}
On Thu, Jul 30, 2009 at 11:22 AM, Fabian Scheipl
fabian.sche...@stat.uni-muenchen.de wrote:
This does what you want:
expand.call - function(definition = NULL, call = sys.call(sys.parent()),
expand.dots = TRUE)
# like match.call, but with all formal args instead of only the specified
ones
{
ans - as.list(match.call(definition, call, expand.dots))
frmls - formals(deparse(ans[[1]]))
add - which(!(names(frmls) %in% names(ans)))
return(as.call(c(ans, frmls[add])))
}
Best,
Fabian
[[alternative HTML version deleted]]
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and provide commented, minimal, self-contained, reproducible code.
[R] nlme weighted
Dear R-expert I'm fitting a non linear model (energy allocation model to individual growth data) using your nlme routine. For each individual I have thus a number of observations (age and size) to which I fit the nonlinear function, with random effects for the individuals on the estimated parameters (individual as the grouping factor). The sampling of these individuals was stratified (size stratified) and the observations are thus not representative for the population. But as we know the true size distributions over the strata, we can compute a statistical weight for each individual, given by the frequency of the size at age of that individual in the true population distribution. To obtain representative estimates in the nlme, I would therefore preferably weight the fitting by these statistical weights. In each group (which is the individual) a different weighting factor would apply (I guess that the individual estimation will not be much affected by these weights, but the population mean). I don't quite see how to do this weighting by nlme-group. I think what I need is something that multiplies these weights to the residual variance. My first hint would be something as it is described by the function varIdent or varFixed, but it is not quite clear to me what is being done by these (e.g. what is meant by variance covariate etc.?). I thank you very much in advance if you could briefly comment on that and point out the function for the weighting that should be applied. All the best Fabian Mollet [[alternative HTML version deleted]] __ 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/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Survival Regression with multiple records per subject
Dear R users! I reformulate the question with another example perhaps my question will be more clearly now. I have several subjects. One subject has multiple records. Only a starting point exists the end point is vague. Here is an example: itm ID exercise time 1.40186910 1 1.32439010 2 1.32439010 3 1.3810 4 1.34676110 5 1.31544111 6 1.33781220 1 1.31991520 2 1.35123521 3 itm is the covariate; ID is the subject Id; exercise indicates if the subject is dead=1 or alive=0 How can I allocate the multiple records to one subject (for example record 1-6 are part of subject with ID 1 record 7-9 are part of subject with ID2) and process a survival regression. the survRegData - survreg(formula=Surv(time,exercise)~itm, data=Data, dist=weibull) command doesn't take into account that multiple records are part of one subject. Many thanks! Fabian Hefner __ 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/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Survival Regression with multiple events per subject
Dear R users! I want to process a maximum likelihood estimation for a parametric regression survival time model with multiple events per subject. the STATA command for this survival regression is: use survreg stset failure(exercise), id(optionid) local regressors itm posret negret streg `regressors', distribution(weibull) explanation: stset declares data to be survival-time data; exercise is the indicator variable, which shows if the subject is dead or alive; optionid is the multiple-record ID variable which means every subject has a unique id and one subject can have multiple events. (see example below) streg computes a maximum likelihood estimation for parametric regression survival time models with multiple record data now I search an equivalent command in R I found the survival package but I have no solution for the use with multiple records per subject. library(survival) survRegData - survreg(formula=Surv(time,exercise)~itm+posret+negret, data=Data, dist=weibull) summary(survRegData) My Question is: how can I modify the above command for the use of multiple events per subject when the optionid is used for indicating the subject? The dataset look like: data | exercisedate | itm| posret | negret | optionid| exercise | time 1996 | 1996 | 1.4518 | 0.05487 |-0.4485 |1| 0 | 1 1997 | 1997 | 2.4535 | 0.00385 |-0.2525 |1| 1 | 5 1998 | 1998 | 1.2523 | 0.04486 |-0.1482 |2| 1 | 8 1999 | 1999 | 3.4257 | 0.15287 |-0.8615 |3| 0 | 4 2000 | 2000 | 1.1457 | 0.07487 |-0.5485 |3| 1 | 5 2001 | 2001 | 2.4418 | 0.09553 |-0.3772 |3| 0 | 2 Thank you, Fabian Hefner __ 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/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] How do I use as.Date when day values are missing?
I have a data frame which contains some valuable date information. But for a few of the dates, the day information missing . Viz: interesting.data$date [1] 1/22/93 1/22/93 1/23/93 1/00/93 1/28/93 1/31/93 1/12/93 i.e. for dates where the day info is missing, the %d part of the %m/%d/%yy format is simply represented as 00. When I apply as.Date to the date information, the dates which don't contain exact day information are converted to NA. Viz: as.Date(interesting.data$date) [1] 1993-01-22 1993-01-22 1993-01-23 NA 1993-01-28 1993-01-31 1993-01-12 Is there a way of using the as.Date function when I only have partial dates (eg missing day information which is represented as 00, as above) such that the date isn't represented as NA? Thanks, Anupa Never miss a thing. Make Yahoo your home page. [[alternative HTML version deleted]] __ 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/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
[R] Storing Variables of different types
Hi there, I have an ixjxk array where I want to store dates in the first column of all sub-matrices (i.e. j=1 is a column with dates) and real numbers in the rest of the columns...I have been trying many things, but I am not getting anywhere. Thank you very much for your help, Fabian This message and any attachment are confidential and may be privileged or otherwise protected from disclosure. If you are not the intended recipient, please telephone or email the sender and delete this message and any attachment from your system. If you are not the intended recipient you must not copy this message or attachment or disclose the contents to any other person. Nothing contained in the attached email shall be regarded as an offer to sell or as a solicitation of an offer to buy any services, funds or products or an expression of any opinion or views of the firm or its employees. Nothing contained in the attached email shall be deemed to be an advise of, or recommendation by, the firm or its employees. No representation is made as to accuracy, completeness, reliability or appropriateness of the information contained in the attached email. __ 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/posting-guide.html and provide commented, minimal, self-contained, reproducible code.
