similar to: x,y vs row,column

I am using R on unix version 63.0 I am doing an image plot of the following data file: ================================ lag1 lag2 cif2d 0.000 0.000 NaN 0.000 1.000 0.500000 0.000 2.000 0.489831 0.000 3.000 0.492986 0.000 4.000 0.493409 0.000 5.000 0.492727 0.000 6.000 0.494485 1.000 0.000 0.500000 1.000 1.000 NaN 1.000 2.000 0.495098 1.000 3.000 0.489831 1.000 4.000 0.492986 1.000 5.000

I am using R 63.0. Now let's try this simple image plot. Here is the data file: ============================ lag1 lag2 cif2d 1 1 11 1 2 12 1 3 13 2 1 21 2 2 22 2 3 23 3 1 31 3 2 32 3 3 33 ==================== data<-read.table("~/r/rt/data/unif/junk.out",header=TRUE) x<-unique(data$lag1) y<-unique(data$lag2) z<-matrix(data$cif2d,length(y),length(x)) At this point, see

Hi all, I have run a ridge regression as follows: reg=lm.ridge(final$l~final$lag1+final$lag2+final$g+final$g+final$u, lambda=seq(0,10,0.01)) Then I enter : select(reg) and it returns: modified HKB estimator is 19.3409 modified L-W estimator is 36.18617 smallest value of GCV at 10 I think it means that it is

Full_Name: Ben Cooper Version: 1.5.0 OS: linux Submission from: (NULL) (134.174.187.90) The help page for glm.nb (in MASS package) says that it takes "Any other arguments for the glm() function except family" One such argument is start "starting values for the parameters in the linear predictor." However, when called with starting values glm.nb returns: Error in

Hello, I have a question on unit root test with urca toolbox. First, to run a unit root test with lags selected by BIC, I type: > CPILD4UR<-ur.df(x1$CPILD4[5:nr1], type ="drift", lags=12, selectlags ="BIC") > summary(CPILD4UR) The results indicate that the optimal lags selected by BIC is 4. Then I run the same unit root test with drift and 4 lags:

Hi all, I have run a ridge regression as follows: reg=lm.ridge(final$l~final$lag1+final$lag2+final$g+final$u, lambda=seq(0,10,0.01)) Then I enter : select(reg) and it returns: modified HKB estimator is 19.3409 modified L-W estimator is 36.18617 smallest value of GCV at 10 I think it means that it is advisable to

Hi, how to estimate a the following model in R: y(t)=beta0+beta1*x1(t)+beta2*x2(t)+...+beta5*x5(t)+beta6*y(t-1)+beta7*y(t-2)+beta8*y(t-3) 1) using "lm" : dates &lt;- as.Date(data.df[,1]) selection&lt;-which(dates&gt;=as.Date("1986-1-1") &amp; dates&lt;=as.Date("2007-12-31")) dep &lt;- ts(data.df[selection,c("dep")]) indep.ret1

Hi all, I have run a ridge regression on a data set 'final' as follows: reg=lm.ridge(final$l~final$lag1+final$lag2+final$g+final$u, lambda=seq(0,10,0.01)) Then I enter : select(reg) and it returns: modified HKB estimator is 19.3409 modified L-W estimator is 36.18617 smallest value of GCV at 10 I think it

Hello guys .. I would like to have some help about as.table . I made a table with the autocorrelations of the returns whit 10 lags and i get this : autocorrelazione2 <- as.table(c((cor(r2[-1151,],lag(r2))),(cor(r2[- c(1151,1150),],lag(r2, k=2))),(cor(r2[- c(1151,1150,1149),],lag(r2, k=3))),(cor(r2[- c(1151,1150,1149,1148),],lag(r2, k=4))),(cor(r2[- c(1151,1150,1149,1148,1147),],lag(r2,

Hello all, I need to figure out a way to lag a variable in by a number of days without using the zoo package. I need to use a remote R connection that doesn't have the zoo package installed and is unwilling to do so. So that is, I want a function where I can specify the number of days to lag a variable against a Date formatted column. That is relatively easy to do. The problem arises when I

Dear Members, Greetings! I would like to know how to create the lag variable for my data. # Load data and create time series object ---- oil <- read_xlsx("crudefinal.xlsx") pricet=ts(oil$price, start = c(2020, 22), frequency = 365) roilt=ts(diff(log(oil$price))*100,start=c(2020,22),freq=365) # Fit MSW model ---- roilt.lag0 =

You appear to be trying to write C code in R. Don't do this. If you can trade off space for efficiency, the calculation can be easily vectorized (assuming I correctly understand what you want to do, of course). set.seed(135) ## for reproducibility D1<-data.frame(x=1:5,y=6:10,z=rnorm(5)) D2<-data.frame(x=19:30,y=41:52,z=rnorm(12)) D.all <-merge(D1,D2, by.x=NULL,by.y=NULL) ##

Hi, I'm looking for a function that will take a list of columns or data.frame and corvert it to cartesian pairlist. For example for this data.frame (see below), I'd like to get a list of all possible pairs: > sound cs rs ns 7 5 2 4 5 4 6 3 1 8 4 1 6 4 2 6 7 1 2 2 5 9 2 5 how can I get this? > mylist ((cs,rs), (cs,ns), (rs,ns))

Is there packaged code to convert geographical coordinates (e.g., longitude, latitude, elevation) to Cartesian coordinates in 3-space? I can see how to do this using 1. a spherical-to-Cartesian conversion like pracma::sph2cart(tpr) http://cran.r-project.org/web/packages/pracma/ 2. a geographical-to-spherical conversion. This seems to involve (in roughly increasing order of difficulty or

Hi all, I'm trying to plot this two curves in a single cartesian plane, but when I plot the first one, the plot appears with no negative "y" value. When I plot the second curve, it almost does not apear in the graph. I was trying the plot.window but with no success. Can someone help me with this? If possible, I'd like to plot this curves in a perfect cartesian plane. f =

Hi all,I had a question last week on asking for a function that will help me draw three different circles on x,y,z axis based on polar coordinates (Each X,Y,Z circle are coming from three independent measurements of 1-360 polar coordinates). It turned out that there ?is no such function in R and thus I am trying to write my own piece of code that hopefully I will be able to share. I have spent

I'm working in Rosario, Argentina, trying to plot some contours. I found the function "draw.contour" created by James Forester at R-bloggers. If you take a look at that function you'll see it's very useful! But I have a problem with it and James is trying to help me (but also asked me to seek for more help) I hope you can help me (don't worry if you can't): Using

Please look at what I see in your code below (run-on code mush) to understand part of why it is important for you to send your email as plain text as the Posting Guide indicates. You might find [1] helpful. [1] https://wiki.openstack.org/wiki/MailingListEtiquette -- Sent from my phone. Please excuse my brevity. On June 25, 2017 2:42:26 PM EDT, Alaios via R-help <r-help at r-project.org>

Hello List, I am looking for efficient code to produce the Cartesian product of two or more data.frames. I'd like to be able to do this without resorting to looping. I have searched the FAQ, web, etc without luck. That being said, the help page for merge says that the function can produce what I'm looking for if the by vectors are of zero length. Would someone be so kind as to

Hi Does anyone know how I might pick out diagonal elements of a matrix using a vector? If I create a matrix a: a <- matrix(c(1:16), 4, byrow=TRUE) and I want to pick out the elements (1,1),(2,2),(3,3), or another arbitrary diagonal (upper or lower), is there any way I can use a vector to do this? So if I want a diagonal of size 3, I could create a vector like x <- c(0:2) and then pick