similar to: How to display data content on a only row?

Dear R experts I have the follwoing data V1 V2 1 -5.8000000 0 2 -4.8000000 0 3 -2.8666667 0 4 -0.8666667 0 5 -0.7333333 0 6 -1.6666667 0 7 -0.1333333 1 8 1.2000000 1 9 1.3333333 1 and I want to know, whether V1 can predict V2: of course it can, since there is a perfect separation between cases 1..6 and 7..9 How can I test, whether this conclusion (being able to assign an

Hello, I am having some trouble with the 'censoring-adjusted C-index' by Uno et al, in the package survAUC. The relevant function is UnoC. The question has to do with what happens when I specify a time point t for the upper limit of the time range under consideration (we want to avoid using the right-end tail of the KM curve). Copying from the example in the help file: TR <-

Hi all, I have an issue with the lm() function regarding the listing of the coefficients. My data are below, showing a list of hours (HR) relating to the time spent resting (R) by an individual animal. Simply i want to run a lm() to run in an anova() to see if there is a significant difference in resting between hours. HR R 1 2 0.6666667 2 2 0.4666667 3 2 0.8000000 4 2

Dear R People: I have a question about a "sorting" problem, please. I have a vector xx: > xx [1] -2.0 1.4 -1.2 -2.2 0.4 1.5 -2.2 0.2 -0.4 -0.9 and a vector of breaks: > xx.y [1] -2.2000000 -0.9666667 0.2666667 1.5000000 I want to produce another vector z which contains the number of the class that each data point is in. for instance, xx[1] is between xx.y[1] and

Hi there, I am trying to get the sum of rain per day. That is what the data-set looks like: Timestamp Rain_mm_Tot 2017-05-29 23:40:00 4.7999980 2017-05-29 23:50:00 1.2000000 2017-05-30 00:10:00 2.5800000 2017-05-30 00:20:00 1.2009600 2017-05-30 00:30:00 1.2000006 2017-05-30 00:40:00 2.5002480 First I tried to define the

Hello, i have following problem and I hope you can help me a little bit My dataframe looks like: df a m d typ value 1950 1 1 5 -4.1 1950 1 2 9 2.7 1950 1 3 3 -1.3 1950 1 4 5 -1.9 1950 1 5 2 0.2 1950 1 6 8 0.5 1951 1 1 4 1.3 .... It consists by daily observations from 1950- 2009. Now, I get with.... for (i

Dear R People: Suppose I have the following character string: > f1 [1] "(1/30)*(20-x)" My goal is to end up with y <- (1/30)*(20-x) How would I do this, please? I've been experimenting with eval, but no good so far. As usual, I have the feeling that this is something really simple, but I can't quite get it. Thank you in advance for any help. Sincerely, Erin --

Hi, I'm trying to get the p-value from the 'lm' regression function as a list object. For example, I can get r^2 from the following code by entering summary(fm)$r.squared. Is there a way to get the p-value? If not, is there a function where I can enter the f-value and degrees of freedom to get the p-value? Thanks. x <- c(1,2,3,4,5,6,7,8,9,10) y <- c(1,2,3,4,4,5,6,8,1,9) fm

Hello all. I have two data frames. Group Start End G1 200 700 G2 500 1000 G3 2000 3000 G4 4000 6000 G5 7000 8000 and Pos C0 C1 200 0.9 0.6 500 0.8 0.8 800 0.9 0.7 1000 0.7 0.6 2000 0.6 0.4 2500 1.2 0.8