similar to: extract parts of a list before symbol

You could have negative indices. There are two ways to do this. 1) provide a large offset. Offset <- 30 for (i in -29 to 120) { print(df[i+Offset])} 2) use absolute values if all indices are negative. for (i in -200 to -1) {print(df[abs(i)])} Tim -----Original Message----- From: R-help <r-help-bounces at r-project.org> On Behalf Of Peter Dalgaard via R-help Sent: Monday, April 22,

Please see fortunes::fortune(285). cheers, Rolf Turner -- Honorary Research Fellow Department of Statistics University of Auckland Stats. Dep't. (secretaries) phone: +64-9-373-7599 ext. 89622 Home phone: +64-9-480-4619

See fortunes::fortune(36). cheers, Rolf Turner -- Honorary Research Fellow Department of Statistics University of Auckland Stats. Dep't. (secretaries) phone: +64-9-373-7599 ext. 89622 Home phone: +64-9-480-4619

I have no real idea what you are trying to do, but if a table is what you want, you can probably get it using the table() function. Or, more likely, the xtabs() function. Using your example from an earlier post (adjusted to make it comprehensible to the human mind): set.seed(1000) time <- factor(rep(c("Pre","Post"),each=200)) treatment <-

Heh. Did anyone bring up negative indices yet? -pd > On 22 Apr 2024, at 10:46 , Rolf Turner <rolfturner at posteo.net> wrote: > > > See fortunes::fortune(36). > > cheers, > > Rolf Turner > > -- > Honorary Research Fellow > Department of Statistics > University of Auckland > Stats. Dep't. (secretaries) phone: > +64-9-373-7599

Hi Rolf, No it is not. I don't know to which question did you want to respond ? I desribed everything in my first email and attached links from SO with pictures included, which are quite understandable. Cheers, Jacek wt., 27 lut 2024 o 02:29 Rolf Turner <rolfturner at posteo.net> napisa?(a): > > I have no real idea what you are trying to do, but if a table is > what you

I have done a scatter plot in R. I want to insert the coefficient of determination R^2 = 0.62 as a text in the plot. I have tried to write R^2 but could not produce R2. I would appreciate it if someone could help me with the syntax. I have tried: expression(paste("", R^2,"=", 0.62)), but it did not produce R squared, rather it gave me error messages. Thanks. Jibrin Alhassan

I am writing my master thesis in which I compared two cultures . So for my statistics I need to compare Age,Sex,Culture as well as have a look at the tasks scores . Anyone familiar with this ? I?d love to share my script so you guide me where I did wrong . Regards

x[?V2?] would retain columns of x headed by V2. What I need is the opposite??I need a data grime with those columns excluded. Steven from iPhone > On Feb 13, 2023, at 9:33 AM, Rolf Turner <r.turner at auckland.ac.nz> wrote: > > ? >> On Sun, 12 Feb 2023 14:57:36 -0800 >> Jeff Newmiller <jdnewmil at dcn.davis.ca.us> wrote: >> >> x["V2"]

Doesn't sound like you got the point. x[-1] normally removes the first element. With 0-based indices, this cannot work. - pd > On 22 Apr 2024, at 17:31 , Ebert,Timothy Aaron <tebert at ufl.edu> wrote: > > You could have negative indices. There are two ways to do this. > 1) provide a large offset. > Offset <- 30 > for (i in -29 to 120) { print(df[i+Offset])} >

Hans, It is a good question albeit R made a conscious decision to have indices that correspond to things like row numbers and thus start with 1. Some others have used a start of zero but often for reasons more related to making use of all combinations of the implementation of integers on many machines where starting with 1 would only allow use of the 255 of the 256 combinations available in 8

Hello Peter, Unless I too misunderstand your point, negative indices for removal do work with the Oarray package (though -0 doesn't work to remove the 0th element, since -0 == 0 -- perhaps what you meant): > library(Oarray) > v <- Oarray(1:10, offset=0) > v [0,] [1,] [2,] [3,] [4,] [5,] [6,] [7,] [8,] [9,] 1 2 3 4 5 6 7 8 9 10 > dim(v)

Hi, I do not want to make a plot, I try to make an output table in R, (in GUI like Stata this is trivially easy task) with regard to SO OP question. As I mentioned, in paper I would not do this, but out of curiosity I use R this time trying to create it. If in R this is trivial task as well, could you please show how to do it ? After googling a lot I did not find solution. regards, Jacek niedz.,

Sorry, didn't cc this to the list. -------- Forwarded Message -------- Subject: Re: [R] Strange results : bootrstrp CIs Date: Sat, 13 Jan 2024 17:37:19 -0500 From: Duncan Murdoch <murdoch.duncan at gmail.com> To: varin sacha <varinsacha at yahoo.fr> You can debug things like this by setting options(error = recover). That will drop into the debugger when the error occurs.

Sir, I want to solve the equation Q(u)=mean, where Q(u) represents the quantile function. Here my Q(u)=(c*u^lamda1)/((1-u)^lamda2), which is the quantile function of Davies (Power-pareto) distribution. Hence I want to solve , *(c*u^lamda1)/((1-u)^lamda2)=28353.7....(Eq.1)* where lamda1=0.03399381, lamda2=0.1074444 and c=26104.50. When I used the package 'Davies' and solved Eq 1, I got the

Hello, I've been enjoying using the "Mixture and Hidden Markov Models in R" by Visser & Speekenbrink to learn how to apply these analyses to my own data using depmixS4. I currently have a fitted 4-state mixture model with three emissions variables and one binomial covariate (HS). I am trying to compute confidence intervals using the following code, where fmms4s is the model:

Well, this would seem to work: e <- data.frame(Score = Score , Country = factor(Country) , Time = Time) ncountry <- nlevels(e$Country) func= function(dat,idx) { if(length(unique(dat[idx,'Country'])) < ncountry) NA else coef(lm(Score~ Time + Country,data = dat[idx,])) } B <- boot(e, func, R=1000) boot.ci(B, index=2, type="perc")

On Tue, 30 May 2023 17:43:31 +0000 Heather Lucas <hlucas2 at lsu.edu> wrote: > Hello, > > I've been enjoying using the "Mixture and Hidden Markov Models in R" > by Visser & Speekenbrink to learn how to apply these analyses to my > own data using depmixS4. > > I currently have a fitted 4-state mixture model with three emissions > variables and one

Hello, I am drawing contour lines for a function of 2 variables at one level of the value of the function and want to include a small arrow in any direction of increase of the function. Is there some way to do that? Below is an example that creates the contour lines. How do I add one small arrow on each line in the direction of increase of the function (at some central point of the contour

Dear All, I want to create a vector p and extract first 20 observations using subset function based on logical condition. My code is below p <- 0:100 I know i can extract the first 20 observations using the following command. q <- p[1:20] But I want to extract the first 20 observations using subset function which requires a logical condition. I am not able to frame the logical