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...ly those individuals with the specified value for that proportion. Below is an example of the code that I am using. I have also attached the data file that I am using to create this dataframe. #Read in Data age.int <- read.csv('C:/intage.csv',header = TRUE) asubs112 <- subset(age.int, rs1042364 != "(2,2)") ages112 <- sort(unique(na.omit(asubs112$first_drink))) for ( i in ages112) { indce <- which(na.omit(asubs112$first_drink == i)) prop <- length(indce)/nrow(asubs112) asubs112[indce,5] <- prop asubs112[...

Hi everyone, Can I suggest the following cosmetic change to cut.default (R-1.3.1), to adjust the labels in the case that include.lowest = TRUE: arrow:dma0jcr% diff -u cut.R my.cut.R --- cut.R Tue Nov 13 12:28:35 2001 +++ my.cut.R Tue Nov 13 12:35:16 2001 @@ -25,6 +25,11 @@ ch.br[-nb], ",", ch.br[-1], if(...

Dear R developers, We have a big SGI Origin computation server with 32 cpu's and 64 Gb of RAM. In R 2.0.0 we could run large jobs, allocating 8 Gb of RAM was not a problem, for example by running: > v1 <- seq(1,2^29) > v2 <- seq(1,2^29) > v3 <- seq(1,2^29) > v4 <- seq(1,2^29) This yields an R process, consuming about 8 Gb of RAM: PID PGRP USERNAME PRI SIZE RES STATE TIME WCPU% CPU% COMMAND 177484 177484 mirjam 20 8225M 8217M sleep 1:18...

...roportion. Below is an example of the code that I am using. The data is made up for the dataframe. Should give you an idea, but the original has 'NA' in many rows. The original data is what is reported in the output below. #Read in Data age.int <- data.frame(IND_ID = seq(1, 140, 10), rs1042364 = sample( c("(1,1)","(1,2)","(2,2)"),14,replace = T), first_drink = sample(5:17,14,replace = T)) asubs112 <- subset(age.int, rs1042364 != "(2,2)") ages112 <- sort(unique(na.omit(asubs112$first_drink)))...

Hi all, Simple question which i thought i had the answer but it isnt so simple for some reason. I am sure someone can easily help. I would like to categorize the values in NP into 1 of the five values in "Per", with the last category("4") representing values >=4(hence 4:max(NP)). The problem is that R is reading max(NP) as multiple values instead of range so the lengths of the labels and the breaks are not matching. Suggestions? Per <- c("NA&q...

Hello * i have to rename a lot of variables, and, given that they have regular name constructs, I would like to use regexps. Here's a dump of my head(names(df)) varnames <- c("id.quest", "txt.1.3", "col1.1.3", "col2.1.3", "col3.1.3", "col4.1.3", "col5.1.3", "txt.2.3", "col1.2.3", "col2.2.3", "col3.2.3", "col4.2.3", "col5.2.3", "txt.3.3", "col1.3.3", &quo...

On Mon, 15 May 2006, m.vroonhoven at erasmusmc.nl wrote: > Dear R developers, > > We have a big SGI Origin computation server with 32 cpu's and 64 Gb of > RAM. In R 2.0.0 we could run large jobs, allocating 8 Gb of RAM was not > a problem, for example by running: > > v1 <- seq(1...

Hi R community, I have some data set and construct the likelihood as follows likelihood <- function(alpha,beta){ lh<-1 d<-0 p<-0 k<-NULL data<-read.table("epidemic.txt",header = TRUE) attach(data, warn.conflicts = F) k <-which(inftime==1) d <- (sqrt((x-x[k])^2+(y-y[k])^2))^(-beta) p<-1 - exp(-alpha*d) for(i in 1:100){ if(i!=k){ i...

Hi List, I'm trying to work out how to use which(), or another function, to find the top-level index of a list item based on a condition. An example will clarify my question. a <- list(c(1,2),c(3,4)) a [[1]] [1] 1 2 [[2]] [1] 3 4 I want to find the top level index of c(1,2), which should return 1 since; a[[1]] [1] 1 2 I can't seem to work out the syntax. I've tried; which(a == c(1,2)) and an error about coercing to double is returned. I can find the index of elements of...

Hello, I recently made a feature request with a patch for terser mocks. http://rubyforge.org/tracker/index.php?func=detail&aid=10412&group_id=1917&atid=7480 Here are some examples with their current equivalents: the_mock.expects.foo(1,2) {|a, b| :bar} the_mock.expects.foo(1,2).returns(:bar) the_mock.expects(:foo).with(1,2) {|a, b| :bar} the_mock.expects(:foo).with(1,2).returns(:bar) The patch is pretty simple too....

Hi I have a problem with substitutions involving named backreferences. I have a vector American.dates: > American.dates [1] "5/15/1976" "2.15.1970" "1.9.2006" which I want to change into British.dates: > British.dates [1] "15/5/1976" "15/2/1970" "9/1/2006" I know I can do it like this: British.dates<-sub("(\\d{1,2})\\D(\\d{1,2})\\D", "...

Buenos días. Hoy ando un poco (o bastante) espeso y no doy con la tecla de una cosa que seguro que es muy simple.. Pongo un ejemplo. var1 <- c(rep(0,3),rep(1,2)) var2 <- c(rep(1,2),0,0,1) var3 <- c(rep(1,2),rep(0,3)) var4 <- c(rep(1,2),rep(0,3)) datos <- data.frame(fila=1:5,var1, var2, var3, var4) datos datos fila var1 var2 var3 var4 1 1 0 1 1 1 2 2 0 1 1 1 3 3 0 0 0 0 4...

...lass 'A' should be colourised red and data points in class 'B' should be colourised 'blue'. No matter what I try, the data points are always blue. I attached a (very) minimal example to reproduce my problem. Thanks, Philippe library("Rcmdr") scatter3d(x=c(1,2), y=c(1,2), z=c(1,2), groups=as.factor(c(1,2)), point.col=c('red','blue'), surface=F)

hi i try to do a graph of a time series which shows in red the values > -0.05, in blue the values >0.05 and in white the values between -0.05 and 0.05 for un exemple :http://www.appinsys.com/globalwarming/enso.htm thanks !!!!! denisse -- View this message in context: http://r.789695.n4.nabble.com/how-to-do-a-graph-with-tree-different-colors-tp4635206.html Sent from the R help mailing

Carlos, en principio si sería algo así, sólo que en vez de quedarme con todas las columnas var1 a var4 tuviera sólo 3, ya que en mis datos no hay ningún caso que tenga el valor 1 en más de 3 variables.. Había llegado a una solución (mucho menos elegante que usando reshape), que implicaba un for sobre las filas. Jorge, creo que tu solución me vale. Muchas gracias a los dos.. Saludos E...

Dear R Users, I have 12 data frames, each of 12 rows and 2 columns. e.g. FeketeJAN MEAN SUM_ AMAZON 144.4997874 68348.4 NILE 5.4701955 1394.9 CONGO 71.3670036 21196.0 MISSISSIPPI 18.9273250 6511.0 AMUR 1.8426874 466.2 PARANA 58.3835497 13486.6 YENISEI...

Dear R-devel, When I was playing around with different kind of indices when subsetting I noticed some unexpected behaviours when using non-integer numeric indices, especially near zero. From the docs: ?Numeric values are coerced to integer as by as.integer<http://127.0.0.1:14277/help/library/base/help/as.integer> (and hence truncated towards zero).? But some behaviour differs from that, and the behaviour also differs between [ and [[ : c(1,2)[as.integer(.5)] --> numeric(0) # As expected c(1,2)[.5] --> num...

dear list, i have a problem using the par function. in one graphic device i want to have two plots so i tried to use par(mfror=c(1,2)). of course it worked out, but the height now is twice the length of the width for each single plot. what i actually wanted is something like par(mfrow=c(2,2)) where only the top (or bottom) two plots are drawn with entire length proportions of 1 to 2 ( height 1, width 2) i want to create a pdf...

Hola, ¿qué tal? Tengo el siguiente problema. Me llegan (simplificando) cadenas del tipo 1,2,"algo"; 3,"cosa"; 4,2,3,7; y tengo que partirlas por el caracter ";" para meterlas en una lista. Lo hago con strsplit y no tengo problemas... ... hasta que recibo cadenas como 1,2,"algo;todo"; 3,"cosa"; 4,2,3,7; en las que existen puntos y com...

...3 columns. The first column represents an "incident" type The second column represents a "month" The third column represents a "time" Code for a sample data.frame incidents <- rep(c('a','b','d','e'), each =25) months <- rep(c(1,2), each =10) times <-rnorm(100) # make my sample data DF <- data.frame(Incidents=as.factor(incidents),Months=as.factor(months),Time=times) # now calculate a mean for the "by" groups of incident type and month pivot <- aggregate(DF$Time,by=list(Incidents=DF$In...