similar to: UNSUBSCRIBE

https://bugzilla.netfilter.org/show_bug.cgi?id=1061 Bug ID: 1061 Summary: net-firewall/nftables-0.5-r2: limit rate: burst parameter doesn't work Product: nftables Version: unspecified Hardware: x86_64 URL: http://wiki.nftables.org/wiki-nftables/index.php/Rate_ limiting_matchings

Can someone show me how to modify one (R.oo) class's object inside another (R.oo) class's method? Is that possible with the R.oo package? A quick example or reference to an example would be outstanding... Thanks, Ben [[alternative HTML version deleted]]

Dear R Helpers, I am missing something very elementary here, and I don't seem to get it from the help pages of the ave, seq and seq_along functions, so I wonder if you could offer a quick help. To use an example from an earlier post on this list, I have a dataframe of this kind: dat = data.frame(name = rep(c("Mary", "Sam", "John"), c(3,2,4))) dat$freq =

Hi, there, I have a simple data manipulation question for you. Thank you for your help! Suppose that I have this data about people appearing in a class Mary Mary Mary Sam Sam John John John John Then I want to find out what exact time(s) the student appears at the moment such as Mary 1 Mary 2 Mary 3 Sam 1 Sam 2 John 1 John 2 John 3 John 4 the fifth row shows tha Sam show the second times

Hallo, I have got an existing data frame and want to add a new column. The existing data frame was created like this: > df <- rbind( c("Fred", "Mary", 4), c("Fred", "Mary", 7), + c("Fred", "Mary", 9), c("Barney", "Liz", 3), + c("Barney", "Liz", 5) ) > df

Hi, R function aggregate can only take summary stats functions, can I aggregate text columns? For example, for the dataframe below, > a <- rbind(data.frame(id=1, name='Tom', hobby='fishing'),data.frame(id=1, name='Tom', hobby='reading'),data.frame(id=2, name='Mary', hobby='reading'),data.frame(id=3, name='John',

Hello! I have a matrix a with 2 variables (see below) that contain character strings. I need to create a 3rd variable that contains True if the value in column x is equal to the value in column y. The code below does it. a<-data.frame(x=c("john", "mary", "mary", "john"),y=c("mary","mary","john","john"))

Hello, I have an unbalanced panel data set that looks like: ID,YEAR,HEIGHT Tom,2007,65 Tom,2008,66 Mary,2007,45 Mary,2008,50 Harry,2007,62 Harry,2008,62 James,2007,68 Jack,2007,70 Jordan,2008,72 That is, James, Jack, and Jordan are missing a YEAR. Is there any command that will "fill in" the missing YEAR such that the end result will be balanced and look like: ID,YEAR,HEIGHT

Dear people from the R help list, I have a question that I can't get my head around to start answering, that is why I am writing to the list. I have data in a format like this (tabs might look weird): John A1 1 0 1 John A2 1 1 1 John A3 1 0 0 Mary A1 1 0 1 Mary A2 0 0 1 Mary A3 1 1 0 Peter A1 1

Hi I am having trouble using read.ssd. Can someone help? The code that I have written is *sashome<-"C:/Mary/Datasets"* *read.ssd(read.ssd(file.path(sashome, "core", "sashelp"), "surv_1v",* *sascmd = file.path(sashome, "sas.exe"))* Here the path that I have given is correct, where the dataset surv_1v.sas7bdat is kept. The message that R gives

Hi, All I am currently learning on the early userspace(EUS) code. And my goal is accessing initrd while in the EUS. For example: mount an initrd with the EUS mount command, Fsck it, chroot to it, run linuxrc in it and so on... But depend on the code, One can not preserve a initrd image when loading into EUS. So I must modify the EUS code to reach the goal. Can you give me some advices about

Hi, I have a data frame with 3 columns: ID, year and score. How can I select for each unique ID, the year that has the max score? For example, for data frame ID, year, score tom, 1995, 88 rick, 1994, 90 mary, 2000, 97 tom, 1998, 60 mary, 1998,100 I shall have ID, year, score tom, 1995, 88 rick, 1994, 90 mary, 1998,100 Thanks, Richard [[alternative HTML version deleted]]

Hi I have a character vector with thousands of names which looks like this: > V=c("Fred", "Mary", "SAM") > V [1] "Fred" "Mary" "SAM" > class(V) [1] "character" I would like to change it to a list: > L=as.list(V) > L [[1]] [1] "Fred" [[2]] [1] "Mary" [[3]] [1] "SAM" but I need to

Gracias Carlos. Yo también he visto el ejemplo que te pone chatGPT, pero la salida que te da no soy capaz de interpretarla. Os paso las ordenes y las respuestas de R de la propuesta de chatGPT Ejemplo.aov<- aov(P~TRAT+CORTE+REP) > summary (Ejemplo.aov) Df Sum Sq Mean Sq F value Pr(>F) TRAT 6 0.0028 0.00046 0.777 0.590 CORTE 2 0.5022 0.25110 424.542 <2e-16

Buenas, En R, como en la mayoría del software estadístico, no se utiliza ningún nivel de confianza sino que lo que se calcula es el p-valor asociado al contraste. De forma que cuanto más cerca de 0 esté el p-valor "menos credibilidad le damos a la hipótesis nula". Dicho mejor, debemos rechazar la hipótesis nula si el p-valor está por debajo de nuestro nivel de confianza. Por ejemplo,

Dear R experts, I'm reading data from an online database via API and it gets delivered in this messy comma separated structure, > RAW.API <-

Wow ! so many thanks Arun and Rui works like a charm problem solved 2013/4/13 arun <smartpink111@yahoo.com> > Hi, > Try this; > library(reshape2) > res<-dcast(Input,people~place,value.var="time") > res[is.na(res)]<-0 > res > # people beach home school sport > #1 Joe 5 3 0 1 > #2 Marc 0 4 2 0 > #3 Mary

Hi UserRs, I know that there has to be an easy way to do this in R (probably easy enough that once someone clues me in I'll smack myself on the forehead for not figuring it out myself), but my searches on my own have not yielded any hints. I have many fields in my dataset that participants entered as "free lists" - i.e., the field constitutes a varying number of names each

Hi all, let's say I have matrix People Desc Value Mary Height 50 Mary Weight 100 Fanny Height 60 Fanny Height 200 Is there a quick way to form the following matrix? People Height Weight Mary 50 100 Fanny 60 200 (Assuming I don't know the length of people/desc and let's say these are characters matrix.. I tried

I'm learning ggplot and am a little confused. Sometimes discrete scales work like I'd expect, and sometimes they don't. For example... This works exactly like one would expect: df<-data.frame(names=c("Bob","Mary","Joe","Bob","Bob")) ggplot(df,aes(names))+geom_histogram() But this yields an error: