similar to: Problem with legend

Hi weekend R helpers, *Is it possible to get a ylabel on the right hand side axis ==> axis(4)? Thankx, Jan opar <- par(mfrow = c(2,1)) plot(ts.Origi[,1],ylab='NDII', main=name) # Add the legend text in the right order legend.txt <- c("NDII", "Inverse KBDI") legend(2001.8,0.29, legend.txt,col=c(1,2), lty=1)

Hello~   Did any one have used the package 'orcutt' ?   I find that it can not work smoothly in a single variable regression. I use the example following, it function very well.   But when I regress "cons" on "price" (use the "reg1<-lm(cons~price+income+temp)") , then  use "reg11<-cochrane.orcutt(reg1) ". There is an error message “Error in

Dear R specialists, I'm working with time series and want to investigate the relationship between two time series by correlation analysis or by fitting a gen. lineair model to the plot of x(timeserie1) and y(timeserie2). Lin1 <- data.frame( Nr = c(1:lengte), NDII = window(ts.mNDII,c(1998,10),c(2003,11)), InvERC = window(Inv.ERC,c(1998,10),c(2003,11)) )

Hi all R users I did a logistic regression with my binary variable Y (0/1) and 2 explanatory variables. Now I try to draw my nomogram with predictive value. I visited the help of R but I have problem to understand well the example. When I use glm fonction, I have a problem, thus I use lrm. My code is: modele<-lrm(Y~L+P,data=donnee) fun<- function(x) plogis(x-modele$coef[1]+modele$coef[2])

Hi everybody, I try to do an exponential model with decreasing. In my data, there is no data missing but there negative values. I adapted this following code whose the origin code comes from in this forum: Regress<-read.table("C:\\Users\\Regression.csv",sep=";",dec=",",header=TRUE) Regress f <- function(x,a,b) {a * exp(b * x)} x<-Regress$EWT

Hi, I try to build a loop difficultly. I have in a folder called Matrices several files (.csv) called Mat2002273, Mat2002274 to Mat2002361. I want to calculate for each file the mean of the column called Pixelvalues. I try this code but as result, I have this message: Mat2002273 not found

HI, Sorry Carl, I received your message in my spam folder. Sarah proposed me a good example of code. Thank you Momadou  ________________________________ De : Carl Witthoft [via R] <ml-node+s789695n4188375h96@n4.nabble.com> Envoyé le : Mardi 13 Décembre 2011 3h34 Objet : Re: Polygon Please read the posting guide and provide a (small) reproducible example of your data. The statement

Hi, I created a figure with R and I want to save it in .pdf. I used this code: > pdf("res.pca.pdf",width=10,height=8) > library(FactoMineR) > res.pca<-PCA(acp) > res.pca > dev.off() When I go in my folder, I find an empty file ( 0 Ko). Do you know where is the problem. Thank you in advance -- View this message in context:

Hi everybody, I have a matrix with 3 columns (Date, MeanArea and SdArea). I want to draw a figure showing the variable MeanArea in terms of the Date. But instead to use the variable SdArea as bar error, I want to use ?polygon error?. I use this code but the output does not seem good.

Hi, Instead to put colour in my histogram, I want to put symbols like lines, dots etc. Do you know the function that does it? Thank you in advance -- View this message in context: http://r.789695.n4.nabble.com/Not-colour-but-symbols-tp4490030p4490030.html Sent from the R help mailing list archive at Nabble.com.

Hi I want to overlay my raster and its boundary which is a shapefile. When I used thise code separately, all is ok: # Open raster >Image<-read.table("C:\\Users\\Documents\\Nouveau\\Frequence.txt",sep="",dec=",",header=TRUE) >testo<-rasterFromXYZ(Image) >plot(testo) >testo2 <- aggregate(testo,fact=10, fun=mean) >plot(testo2) # open

Hi, I used simple linear regression with the R software and EXCEL on the same data. Although , I find the same R2=0.84, I find different estimated values (intercept and slope). For the R software (slope =0.0009, Intercept = -0.1478), for EXCEL (slope =927.7, Intercept = 154,41). When I use the estimated values from the R software, the results seem bad, however the results of Exel seem correct.

Dear R users, When running the program below I receive the following error message: fit <- optim(parm, objective, yt = tyield, hessian = TRUE) Error in as.vector(data) : no method for coercing this S4 class to a vector I can't figure out what the problem is exactly. I imagine that it has something to do with "tyield" being a matrix. Any help on explaining what's going on

I have data on the proportion of clutches experiencing different fates (e.g., 4 different sources of mortality) for 5 months . I need to test 1) if the overall proportion of these different fates is different over the entire study and 2) to see if there are monthly differences within (and among) fate types. Thus, I am pretty sure this is an RM analysis -( I measure the same quadrats each month).

Hi, I'm fitting a standard nonlinear model to the luminances measured from the red, green and blue guns of a TV display, using nls. The call is: dd.nls <- nls(Lum ~ Blev + beta[Gun] * GL^gamm, data = dd, start = st) where st was initally estimated using optim() st $Blev [1] -0.06551802 $beta [1] 1.509686e-05 4.555250e-05 7.322720e-06 $gamm [1] 2.511870 This works fine but I

Hello, In one of my functions, I need to extract the hour from a date. For example: as.POSIXlt("2010-03-27 02:00:00")$hour gives 2 as expected. It works for all the dates I've tested except the following one which is in my list of dates: as.POSIXlt("2010-03-28 02:00:00")$hour which gives 0. I don't understand why it doesn't work with this particular date. Any

Hello, I'm still trying to make the life of my colleagues easier. Nice, isn't it ? At the moment, I'm looking for a way to generate multiple "single report". In fact I have a .rnw file which send a query to a MySQL database (rs<-dbSendQuery(con, statement="select * from treatdata where name='Device1'") But of course my colleagues have many devices and

Hello, I would like to replace a for loop with lapply in order to speed up the treatment of my data (I've read lapply can help to do it). At the end of the message, you will find a simple example (function is more complex and data frames is more than 1 million of rows) of what I would like to do, that is applying the same calculation to all the rows of the data frame (knowing that I cannot

Hi, I recently discovered the R program and I thought it could be useful to me. I have to analyse data saved as .Px file (x between 0 and 8 - .P0 files have 18 lines at the beginning that I have to skip). New files are generated everyday. This is my strategy : In order to analyse the data, I first want to copy the new data in a database in MySQL (which already contains the previous data). So

Hello, I'm trying to fit some points with a 8-degrees polynom (result of lm is stored in pfit). In most of the case, it is ok but for some others, some coefficients are "NA". I don't really understand the meaning of these "NA". And the problem is that I can't perform a derivation (pderiv<-as.function((deriv(polynomial(pfit$coefficients))))) on pfit due to the