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Dear friends, I hope this email finds you all well. This is the dataset I am working with: dput(random_mod12_data2) structure(list(Index = c(1L, 5L, 11L, 3L, 2L, 8L, 9L, 4L), x = c(5, 13, 25, 9, 7, 19, 21, 11), n = c(500, 500, 500, 500, 500, 500, 500, 500), r = c(100, 211, 391, 147, 122, 310, 343, 176), ratio = c(0.2, 0.422, 0.782, 0.294, 0.244, 0.62, 0.686, 0.352)), row.names = c(NA, -8L),

On Wed, Jul 01, 2015 at 12:02:50PM -0700, Linus Torvalds wrote: > On Wed, Jul 1, 2015 at 2:31 AM, Michael S. Tsirkin <mst at redhat.com> wrote: > > virtio/vhost: cross endian support > > Ugh. Does this really have to be dynamic? > > Can't virtio do the sane thing, and just use a _fixed_ endianness? > > Doing a unconditional byte swap is faster and simpler

Dear R-users, It is my understanding that cat(shQuote(a.string)) should print the origintal a.string. Is this right? I am not sure cat() correctly prints strings which are generated by triple-shQuote(): > shQuote(shQuote("a")) [1] "\"\\\"a\\\"\"" > cat(shQuote(shQuote(shQuote("a"))), '\n')

Dear R-users, It is my understanding that cat(shQuote(a.string)) should print the origintal a.string. Is this right? I am not sure cat() correctly prints strings which are generated by triple-shQuote(): > shQuote(shQuote("a")) [1] "\"\\\"a\\\"\"" > cat(shQuote(shQuote(shQuote("a"))), '\n')

On Wed, Jul 01, 2015 at 12:02:50PM -0700, Linus Torvalds wrote: > On Wed, Jul 1, 2015 at 2:31 AM, Michael S. Tsirkin <mst at redhat.com> wrote: > > virtio/vhost: cross endian support > > Ugh. Does this really have to be dynamic? > > Can't virtio do the sane thing, and just use a _fixed_ endianness? > > Doing a unconditional byte swap is faster and simpler

On Thu, Jul 02, 2015 at 11:12:56AM +0200, Greg Kurz wrote: > On Thu, 2 Jul 2015 08:01:28 +0200 > "Michael S. Tsirkin" <mst at redhat.com> wrote: > > > On Wed, Jul 01, 2015 at 12:02:50PM -0700, Linus Torvalds wrote: > > > On Wed, Jul 1, 2015 at 2:31 AM, Michael S. Tsirkin <mst at redhat.com> wrote: > > > > virtio/vhost: cross endian support

On Thu, Jul 02, 2015 at 11:12:56AM +0200, Greg Kurz wrote: > On Thu, 2 Jul 2015 08:01:28 +0200 > "Michael S. Tsirkin" <mst at redhat.com> wrote: > > > On Wed, Jul 01, 2015 at 12:02:50PM -0700, Linus Torvalds wrote: > > > On Wed, Jul 1, 2015 at 2:31 AM, Michael S. Tsirkin <mst at redhat.com> wrote: > > > > virtio/vhost: cross endian support

Dear R users, Please disregard my previous post "converting result of substitute to 'ordidnary' expression". The problem I have has nothing to do with substitute. Consider: > dat <- data.frame(x=1:10, y=1:10) > subsetexp <- expression(5<x) > ## this does work > subset(dat, eval(subsetexp)) x y 6 6 6 7 7 7 8 8 8 9 9 9 10 10 10 > ##

Dear R users, As substitute() help page points out: Substituting and quoting often causes confusion when the argument is 'expression(...)'. The result is a call to the 'expression' constructor function and needs to be evaluated with 'eval' to give the actual expression object. And indeed I am confused. Consider: > dat <- data.frame(x=1:10,

Dear R users, As substitute() help page points out: Substituting and quoting often causes confusion when the argument is 'expression(...)'. The result is a call to the 'expression' constructor function and needs to be evaluated with 'eval' to give the actual expression object. And indeed I am confused. Consider: > dat <- data.frame(x=1:10,

"Beast" is more appropriate than you know. It''s definately Frankenstein code. (read: it ain''t pretty, but it works - some of the time). Anyway, it entertains my wife; not sure if hardcore programmers will get anything out of it or not. You might find it interesting (or pointless) that in a complete bastardization of rails'' main reason for being it

Hey, Did you hear about Scour? It is the next gen search engine with Google/Yahoo/MSN results and user comments all on one page. Best of all we get paid for using it by earning points with every search, comment and vote. The points are redeemable for Visa gift cards! It's like earning credit card or airline points just for searching! Hit the link below to join for free and we will both get

Dear R-developers, I am trying to figure out a way to call browser() when an error occur, and naturally I want the browser() to be called in the environment of the error. I tried something simple in vain: > f <- function() { x <- 1; stop('ok') } > tryCatch(f(), error=browser()) Called from: tryCatch(f(), error = browser()) ## if browser() was called in the local environment

Dear R-users, Could someone please explain why the message printed by function stopifnot2, see below, is different from that of stopifnot itself? Thank you for your help, Vadim > stopifnot2 <- function(...) stopifnot(...) > stopifnot(F) Error: F is not TRUE > stopifnot2(F) Error: ..1 is not TRUE > version _ platform i386-pc-mingw32 arch i386 os

Dear R-help, When the result of a matrix subscription degenerates to a scalar the names implied by the dimnames are discarded. > x <- matrix(0, 1, 1, dimnames=list('a', 'x')) ## below I expected result to have names='x', it's not > x[1,] [1] 0 ## below I expected result to have names='a', it's not > x[,1] [1] 0 This is probably a side effect

Dear R-users, How can I detect a NULL in a recursive list? For a regular list I could use lapply: > lapply(list(x=NULL), is.null) $x [1] TRUE However that doesn't work for structures like list(list(x=NULL)). I tried rapply but it treats NULL as a list and discards them: > rapply(list(a=1, b=list(x=NULL)), is.null) a FALSE Any suggestion? Thank you for your help, Vadim Note:

Dear R-devel, When a character vector is used to subscript a list and when some of the subscripts are not present in the list names R returns NULL for those subscripts and generate NA names for each of them: > list(b=1)[c('a','b')] $<NA> <<-- generated name NULL $b [1] 1 Wouldn't it be more intuitive to use the subscript name rather than to generate an NA?

I'd like to train a decision tree on a set of weighted data points. I looked into the rpart package, which builds trees but doesn't seem to offer the capability of weighting inputs. (There is a weights parameter, but it seems to correspond to output classes rather than to input points). I'm making do for now by preprocessing my input data by adding multiple instances of each data

Dear R-users, Suppose I have an expression: expr = expression(a>0) and now I want to modify it to expression(a>0 & b>0). The following doesn't work: expr = expression(expr & b>0) What would be a good way of doing this? Thanks, Vadim ________________________________ Note: This email is for the confidential use of the named addressee(s) only and may contain

Hi, I need to run a .reg file. In windows, I think I'd just double click on it. If I try to run it in wine, it says it's not a valid executable. How do I run it?