similar to: Re: evidence/timelines that show linux is "more secure"

I''m looking for some evidence, backup up with dates and references, that shows that the Linux community responds to security problems more quickly than other OS vendors, and thus might be considered "more secure". A number of fairly high profile corporations are starting to look for such information as they consider Linux as an alternative solution to other UNIXes. Something

# read.table is NOT part of the data.table package #library(data.table) DFM <- read.table( text= 'obs start end 1 2/1/2015 1/1/2017 2 4/11/2010 1/1/2011 3 1/4/2006 5/3/2007 4 10/1/2007 1/1/2008 5 6/1/2011 1/1/2012 6 10/5/2004 12/1/2004 ',header = TRUE, stringsAsFactors = FALSE) # cleaner way to compute D DFM$start <- as.Date( DFM$start, format="%m/%d/%Y" ) DFM$end

Since the number of choices is small (6), how about this? Starting with Jeff's initial DFM: DFM <- structure(list(obs = 1:6, start = structure(c(16467, 14710, 13152, 13787, 15126, 12696), class = "Date"), end = structure(c(17167, 14975, 13636, 13879, 15340, 12753), class = "Date"), D = c(700, 265, 484, 92, 214, 57), bin = structure(c(6L, 3L, 5L, 1L, 3L, 1L), .Label

Thank you Jeff and All, Within a given time period (say 700 days, from the start day), I am expecting measurements taken at each time interval;. In this case "0" means measurement taken, "1" not taken (stopped or opted out and " -1" don't consider that time period for that individual. This will be compared with the actual measurements taken (Observed-

I'm not sure whether or not this is a bug, but I did isolate the line where the error is thrown: src/library/base/R/dataframe.R:1395. https://github.com/wch/r-source/blob/01374c3c367fa12f555fd354f735a6e16e5bd98e/src/library/base/R/dataframe.R#L1395 The error is thrown because the line attempts to set a subset of the rownames to NULL, which fails. R> options(error = recover) R>

In my main R program, I have source("retaanalysis/Functions/doAirport.R") .... stuff to read data and calculate ads sapply(ads, function(x) {doAirport(x, base)} ) And doAirport has # analyze the flights for a given airport doAirport = function(df, base) { # Get rid of unused runway factor levels (from other airports) df$lrw <- drop.levels(df$lrw) # In gdata package #

Hi lily, As I have no idea of what the "true record" is, I can only guess. Maybe this will help: # get some fairly distinct colors rainbow_colors<-rainbow(9) # this should sort the numbers in dfm$A dfm$Acolor<-factor(dfm$A) plot(dfm$B,dfm$C,pch=ifelse(dfm$DF==1,1,19), col=rainbow_colors[as.numeric(dfm$Acolor)]) legend("bottom",legend=sort(unique(dfm$A)),

I suspect there may be a bug in base::rbind.data.frame Below there is minimal example of the problem: m <- matrix (1:12, 3) dfm <- data.frame (c = 1 : 3, m = I (m)) str (dfm) m.names <- m rownames (m.names) <- letters [1:3] dfm.names <- data.frame (c = 1 : 3, m = I (m.names)) str (dfm.names) rbind (m, m.names) rbind (m.names, m) rbind (dfm, dfm.names) #not working rbind

For more than 10 records, how to reformat the colors? Also, how to show the first legend only, but at the bottom, while the second legend in your code is not necessary? In all, the same A values have the same color, but different symbols in DF==1 and DF==2. Thanks for your help. On Sun, Jul 16, 2017 at 9:28 AM, lily li <chocold12 at gmail.com> wrote: > Hi Jim, > > For true color,

Hi Jim, For true color, I meant that the points in the figure do not correspond to the values from the dataframe. Also, why to use rainbow(9) here? And the legend is straight in the middle, is it possible to reformat it to the very bottom? Thanks again. On Sun, Jul 16, 2017 at 2:50 AM, Jim Lemon <drjimlemon at gmail.com> wrote: > Hi lily, > As I have no idea of what the "true

Hi , I am trying to create a dataframe in C and sebd it back to R. Can anyone point me to the part of the source code where it is doing , let me explain the problem I am having . -------------------------------------------------------------------- My simple implementation is like this SEXP formDF() { SEXP dfm ,df , dfint , dfStr,lsnm; char *ab[3] =

anyone ever got this error ? [Wed Jul 26 17:36:25 2006] [error] [client 192.168.11.100] (eval):21:in `seek'', referer: http://rf.mathieu.mv.sha/diary.cgi/edit/1?someting=true [Wed Jul 26 17:36:25 2006] [error] [client 192.168.11.100] : , referer: http://rf.mathieu.mv.sha/diary.cgi/edit/1?someting=true [Wed Jul 26 17:36:25 2006] [error] [client 192.168.11.100] Illegal seek, referer:

I think the following may provide a clearer explanation: subs <- c(1,3) DFA <- data.frame(id = 1:3) ar <- array(1:12, c(3,2,2)) ## yielding > ar , , 1 [,1] [,2] [1,] 1 4 [2,] 2 5 [3,] 3 6 , , 2 [,1] [,2] [1,] 7 10 [2,] 8 11 [3,] 9 12 ## array subscripting gives > ar[subs,,] , , 1 [,1] [,2] [1,] 1 4 [2,] 3 6 , , 2

So is this an expected behavior or is it a bug which should be reported somewhere else? Thanks! Georg? ? ? Gesendet:?Dienstag, 09. Mai 2023 um 19:28 Uhr Von:?"Bert Gunter" <bgunter.4567 at gmail.com> An:?"Georg Kindermann" <Georg.Kindermann at gmx.at> Cc:?"Rui Barradas" <ruipbarradas at sapo.pt>, r-help at r-project.org Betreff:?Re: [R] data.frame

Hello, Has anyone out there successfully installed NetApp's "DataFabric Manager" (DFM) on CentOS? If so, what version of DFM, CentOS, and what architecture? I am trying to install DFM 4.0.2, and have tried on both CentOS 4.8 i386 and CentOS 5.5 x86_64. I have edited my /etc/redhat-release file to be equal to RHEL's, as the DFM installer immediately aborts if that isn't

I really don't know. I would call it a request for extended capabilities of [.data.frame, rather than a feature or bug. But maybe wiser heads than mine who monitor this list can sort it out. -- Bert On Wed, May 24, 2023 at 8:52?PM Georg Kindermann <Georg.Kindermann at gmx.at> wrote: > So is this an expected behavior or is it a bug which should be reported > somewhere else? >

Hi, I ran your code, but the results were not as expected. After I ran the code by "source", it return No id variables; using all as measure variables > p2 and no line or legend is on the graph (as attached) Am I doing anything wrong? John library(ggplot2) library(reshape2) dfm<-melt(df) dfm$variable<-factor(dfm$variable, levels=levels(dfm$variable)[c(2,1,3,4)])

Hi R users, I still have the problem about plotting. I wanted to put the datasets on one figure, x-axis represents values B, y-axis represents values C, while different colors label column A. Each record uses a circle on the figure, while hollow circles represent DF=1 and solid circles represent DF=2. I put my code below, but the A labels do not correspond to the true record, so I don't know

I WROTE THIS FUNCTION BELOW test <- function(x, ...) UseMethod('test', x) test.data.frame = function(x, model, which, error, ...) { av <- aov(formula(model), data = x) res <- test.aovlist(av, which = which, error = error) return(res) } test.aovlist <- function(x, which, error, ...) { mm <- model.tables(x, "means") tabs <- mm$tables[-1]

Hi, ( As a total newbie to R ) I get this error when attempting to use read.table to read a data frame: "Error in type.convert(x, na.strings, as.is) : null string encountered" (Is there help documentation for the type.convert function ( I could not find it in the usual places) This does not always happen, but seems to occur when I am attempting to read data frames with more than a