similar to: Can a data.frame column contain lists/arrays?

-----BEGIN PGP SIGNED MESSAGE----- Hash: SHA1 Hi, This may or may not be a bug. However, it is DEFINITELY NOT how I would expect and want to see sshd work! If you run lsof against sshd on a privilege separated user, it shows that sshd's CWD is /. I would hope that the CWD would be at a minimum /var/empty/sshd and I would really have thought it would be something along the lines of

Hello I am am new to R and any help with the following would be appreciated: I have a list (example attached) and I would like to create a new list which is a filtered version of this list. I.e I would like a list that only contains elements with this value: Chr 10 : 21853562 - 21855482 Any pointers/tips would be great. Thanks! Nikol -- Bioinformatician/Computer Associate Cambridge Institute

Hello dear all, I am unable to understand why when I run the following three lines: set.seed(4254) > a <- data.frame(y = rnorm(40), x=ordered(sample(1:5, 40, T))) > summary(lm(y ~ x, a)) The output I get includes factor levels which are not relevant to what I am actually using: Call: > lm(formula = y ~ x, data = a) > Residuals: > Min 1Q Median 3Q Max >

Hi, I want to test for a contrast from a regression where I am regressing the columns of a matrix. In short, the following. X <- matrix(rnorm(50),10,5) Y <- matrix(rnorm(50),10,5) lm(Y~X) Call: lm(formula = Y ~ X) Coefficients: [,1] [,2] [,3] [,4] [,5] (Intercept) 0.3350 -0.1989 -0.1932 0.7528 0.0727 X1 0.2007 -0.8505 0.0520

Hi; Through samba4 BDC server not update DNS record in PDC DNS server. please see logs 04-Jul-2012 10:22:19.800 client 192.253.8.25#1081: send 04-Jul-2012 10:22:19.800 client 192.253.8.25#1081: sendto 04-Jul-2012 10:22:19.800 client 192.253.8.25#1081: senddone 04-Jul-2012 10:22:19.800 client 192.253.8.25#1081: next 04-Jul-2012 10:22:19.800 client 192.253.8.25#1081: endrequest 04-Jul-2012

I'm working on an IVR that gives the users the option (number 5 in the main menu) to dial by extension: exten => 5,1,Set(TIMEOUT(digit)=5) ; Dial Extension exten => 5,2,Set(TIMEOUT(response)=10) exten => 5,3,Background(LCL/prompt-60) exten => 5,4,WaitExten(15) When going option 5 you can dial some extensions such as 2802, it goes to the extension (all extens start with 28 on the

hello, > subset(swiss, Agriculture > 60 & Examination != c(14,16), select = c(Agriculture,Examination,Catholic)) Agriculture Examination Catholic Broye 70.2 16 3.30 Glane 67.8 14 4.20 Aigle 62.0 21 5.16 Avenches 60.7 19 5.23 Cossonay 69.3 22 5.62

Hello you all from the R Help mailing list! I am working on a PowerBook with Mac Os X and use R 2.5.0. I used the distance function from the analogue package to perform a similarity analysis using the Gowers Index and weighted Variables. My variables are bivariate data and measurements as well as interval data transformed into minimum and maximum variables. I used this Code:

Hello, suppose I have a list with matrices: a=list(x1=matrix(rnorm(10),5,2),x2=matrix(rnorm(10),5,2),x3=matrix(rnorm(10),5,2)) I want to compute for all combination of xi and xj (x1,x2 x1,x3 and x2,x3) a value. This value is given for the pair x1,x2 by trace(x1%*%t(x1)%*%x2%*%t(x2)) / trace(x1%*%t(x1))*trace(x2%*%t(x2)) I know that product matrices t(xi)%*%xi can be obtained by:

I am trying to plot a graph but the points on the graph should be different symbols and colors. It should represent what is in the legend. I tried using the points command but this does not work. Is there another command in R that would allow me to use different symbols and colors for the points? Thank you kindly. data(mtcars) plot(mtcars$wt,mtcars$mpg,xlab= "Weight(lbs/1000)",

Hi, I would like to make a suggestion for a small syntactic modification of FUN argument in the family of functions [lsv]apply(). The idea is to allow one-liner expressions without typing "function(item) {...}" to surround them. The argument to the anonymous function is simply referred as ".". Let take an example. With this new feature, the following call

Dear list, I want the 1st, 2nd, 5th, and 6th columns of mtcars. After copying them, the columns become numeric class rather than data frame. But, when I copy rows, they data frame retains its class. Why is this? I don't see why copying rows vs columns is so different. > class(mtcars) [1] "data.frame" > head(mtcars) mpg cyl disp hp drat wt qsec vs

I'm trying to pick up a ringing SIP phone (203) across the office with exten => *9,1,Pickup(783743) where 783743 is the local part of the number that our ISDN works on. I tried all of these first: exten => *9,1,Pickup(203) exten => *9,1,Pickup(SIP/203) exten => *9,1,Pickup(203@internal) and got a "declined" message back from my phone (snom 300), so I then

First I have a matrix called stocks1: > class(stocks1) [1] "matrix" Here are the first 5 rows of the last 4 columns: > stocks1[1:5,2:5] [,1] [,2] [,3] [,4] [1,] 80.73 31.95 25.4 25.69 [2,] 83.66 31.95 27.12 25.2 [3,] 83.27 32.93 28.74 26.29 [4,] 83.9 34.07 29.77 26.6 [5,] 82.74 35.18 30.24 27.41 Now, why can't I convert this into a dataframe?

Dear R People: I have a data frame, xm1, which has 12 rows and 4 columns. If I put is xm1[,-4], I get all rows, and columns 1 - 3, which is as it should be. Now, is there a way to use the names of the columns to omit them, please? Thanks so much in advance! Sincerely, Erin -- Erin Hodgess Associate Professor Department of Computer and Mathematical Sciences University of Houston - Downtown

Hi Friends, I have a data frame X, and I want to add ?%? & ?$? in row 4 and 5 respectively. when I?m trying using below logic, I?m getting warning message. Can anyone help me out on this. X: Summary G Y R T Accts 582 644 0 1226 AcctCov 230 165 0 395 Cov% 40 26 0 32 UnCov% 60 74 0 68 EqVol11$MM8.5 10.6 0 19.1 Using this logic:

Hello, When plotting a barchart with ggplot it drops the levels of the factor for which no counts are available. For example: library(ggplot) mtcars$cyl<-factor(mtcars$cyl) ggplot(mtcars[!mtcars$cyl==4,], aes(cyl))+geom_bar() levels(mtcars[!mtcars$cyl==4,]) This shows my problem. Because no counts are available for factorlevel '4', the label 4 dissapears from the plot. However, I

Dear all Is there an easier way to retrieve the name of an object? For example, > tmp <- 1:10 > as.character(quote(tmp)) [1] "tmp" > as.character(quote(mtcars$cyl)) [1] "$" "mtcars" "cyl" > as.character(quote(mtcars$cyl))[3] [1] "cyl" The last call more than anything seems a hack. Is there a better way? Thank you Liviu

I am looking for ways to add points to three different plots in parallel. I generate three scatter plots and name them as s3d1, s3d2 and s3d3 s3d1<-scatterplot3d(mtcars[,3],mtcars[,4],mtcars[,5],main="common",pch=20) s3d2<-scatterplot3d(mtcars[,3],mtcars[,4],mtcars[,5],main="common",pch=20)

I'm sure this exists elsewhere, but, as a trade-off, could you achieve what you want with a separate helper function F(expr) that constructs the function you want to pass to [lsv]apply()? Something that would allow you to write: sapply(split(mtcars, mtcars$cyl), F(summary(lm(mpg ~ wt,.))$r.squared)) Such an F() function would apply elsewhere too. /Henrik On Thu, Apr 16, 2020 at 9:30 AM