similar to: A question on dummy variable

Hello R-users I was puzzled by some strange results of an analysis and I found out what is for me a strange behaviour (I won't dare to say a bug) in both plot and points (and I suspect lines and other kind of lower level plots). If you try the following code: > data<-data.frame(dummy1=c(1:10),dummy2=c(1:10)) > plot(data$dummy1,data$dummy3) A plot of the values of dummy1 against

I R-helpers #I have a data panel of thousands of firms, by year and industry and #one dummy variable that separates the firms in two categories: 1 if the firm have an auditor; 0 if not #and another variable the represents the firm dimension (total assets in thousand of euros) #I need to create two separated samples with the same number os firms where #one firm in the first have a corresponding

Hello, Is there a better way to use paste such as: a = paste(colnames(list.indep)[1],colnames(list.indep)[2],colnames(list.indep)[3],colnames(list.indep)[4],colnames(list.indep)[5],sep="+") > a [1] "aa+dummy1+dummy2+bb+cc" I tried a = paste(colnames(list.indep)[1:5],sep="+") > a [1] "aa" "dummy1" "dummy2"

Hello, I have a data frame > mdl.summary est.coef std.err t.stat intercept 0.0011625517 0.0002671437 4.351784 aa -0.0813727439 0.0163727943 -4.969997 dummy1 -0.0002534873 0.0001204000 -2.105376 dummy2 -0.0007784864 0.0001437537 -5.415417 bb -0.0002856727

Dear all, I have a simple question. I couldn't find a solution in the forums/R-user manual; I have also asked to my friends who use R, but couldn't get any answer from them either. I would appreciate any solutions. I want to write formatted text file like in Fortran. More specifically with the format choice of mine for any given line (more specifics are given below). The R function

Dear all, (sorry I got the wrong button for subscribing a minute ago) At the moment I'm writing on a package for random field simulation that I'd like to make publically availabe in near future. To this end I've asked Martin Maechler to have a look at my R-code. He was very surprised about how I perform the ".C"-calls, and encouraged me to make this request for comments.

tl;dr - a renamed dummy interface is not persisting across reboots on Cent6. I have a situation where I need to rename a dummy interface on my system. I have a total of 3 dummy interfaces: dummy0 dummy1 adummy0 I've been doing some puppet testing in a vbox VM to get this all sorted out to deploy to a group of boxes. Every time I reboot my VM, it comes up WITHOUT adummy0, but I notice

Hello all, I have developed a instrumented pass which insert some variables between the original variables, as well as insert some code into the original source code. just like: ============= original source code ============= int a[10]; void fun1 () { // some source code here } ========================================= ============= instrumented source code ============= int

Dear all, let say I have following data: dat <- structure(list(V1 = structure(c(1L, 4L, 5L, 3L, 3L, 5L, 6L, 6L, 4L, 3L, 5L, 6L, 5L, 5L, 4L, 4L, 6L, 2L, 3L, 4L, 3L, 3L, 2L, 5L, 3L, 6L, 3L, 3L, 6L, 3L, 6L, 1L, 6L, 5L, 2L, 2L), .Label = c("C", "G", "I", "O", "R", "T"), class = "factor"), V2 = c(0L, 0L, 0L, 1L, 1L, 1L, 1L, 0L,

On Wednesday 17 June 2015 16:19:31 Chen Hanxiao wrote: > We should not use tmp lines buffer as return value, > for lines buffer will be freed. s/tmp/temporary/ > Signed-off-by: Chen Hanxiao <chenhanxiao@cn.fujitsu.com> > --- > v4: take advantage of sscanf's '%m'. > v3: fix test case failure > > daemon/btrfs.c | 40

Hello I'm having trouble figuring out how to use the output of "segmented()" with a new set of predictor values. Using the example of the help file: ??set.seed(12) xx<-1:100 zz<-runif(100) yy<-2+1.5*pmax(xx-35,0)-1.5*pmax(xx-70,0)+15*pmax(zz-.5,0)+rnorm(100,0,2) dati<-data.frame(x=xx,y=yy,z=zz) out.lm<-lm(y~x,data=dati) o<-## S3

Hello again, Let say I have a matrix: Mat <- matrix(1:12, 4, 3) And a vector: Vec <- 5:8 Now I want to do following: Each element of row-i in 'Mat' will be divided by i-th element of Vec Is there any direct way to doing that? Thanks for your help

Frank Even wrote: > On Wed, May 27, 2015 at 1:37 AM, Frank Even > <lists+centos.org at elitists.org> wrote: >> tl;dr - a renamed dummy interface is not persisting across reboots on >> Cent6. >> >> I have a situation where I need to rename a dummy interface on my >> system. >> >> I have a total of 3 dummy interfaces: >> >> dummy0

Thank you guys for your reply. I tried but it still does not work On 10/14/2015 5:37 PM, Stéphane PURNELLE wrote: > > These parameters must be put on share section, not in global section > > hide unreadable = yes > hide unwriteable files = yes > example > > [smb_shr1] > path = /mnt/LV002/share1 > read only = no > create mode = 0777

s/ =/ = Signed-off-by: Chen Hanxiao <chenhanxiao@cn.fujitsu.com> --- daemon/btrfs.c | 4 ++-- 1 file changed, 2 insertions(+), 2 deletions(-) diff --git a/daemon/btrfs.c b/daemon/btrfs.c index 8b5779a..306cade 100644 --- a/daemon/btrfs.c +++ b/daemon/btrfs.c @@ -512,7 +512,7 @@ do_btrfs_subvolume_list (const mountable_t *fs) goto error; } - struct

Hello all, Let say I have 2-way contingency table: Tab <- matrix(c(8, 10, 12, 6), nr = 2) and the Chi-squared test could not reject the independence: > chisq.test(Tab) Pearson's Chi-squared test with Yates' continuity correction data: Tab X-squared = 1.0125, df = 1, p-value = 0.3143 However I want to get all possible contingency tables under this independence

Below is my full implementation (tried to make it simple as for demonstration) Lapply_me = function(X = X, FUN = FUN, Apply_MC = FALSE, ...) { if (Apply_MC) { return(mclapply(X, FUN, ...)) } else { if (any(names(list(...)) == 'mc.cores')) { myList = list(...)[!names(list(...)) %in% 'mc.cores'] } return(lapply(X, FUN, myList)) } } Lapply_me(as.list(1:4), function(xx) { if (xx ==

Hi all, please see my code: > library(zoo) > a <- as.yearmon("March-2010", "%B-%Y") > b <- as.yearmon("May-2010", "%B-%Y") > > nn <- (b-a)*12 # number of months in between them > nn [1] 2 > as.integer(nn) [1] 1 What is the correct way to find the number of months between "a" and "b", still

The reason that it works for Apply_MC=TRUE is that in that case you call mclapply(X,FUN,...) and the mclapply() function strips off the mc.cores argument from the "..." list before calling FUN, so FUN is being called with zero arguments, exactly as it is declared. A quick workaround is to change the line Lapply_me(as.list(1:4), function(xx) { to Lapply_me(as.list(1:4),

My modified function looks below : Lapply_me = function(X = X, FUN = FUN, Apply_MC = FALSE, ...) { if (Apply_MC) { return(mclapply(X, FUN, ...)) } else { if (any(names(list(...)) == 'mc.cores')) { myList = list(...)[!names(list(...)) %in% 'mc.cores'] } return(lapply(X, FUN, myList)) } } Here, I am not passing ... anymore rather passing myList On Sun, Mar 4, 2018 at 10:37 PM,