similar to: Splitting a set of vectors in a list

Whats going on here? df<-data.frame(x=1:10,y=1:10) ggplot()+geom_point(data=df,aes(x=x,y=y)) ## this is the normal usage right? ggplot()+geom_point(data=df,aes(x=df[,1],y=df[,2])) ## but I can also feed it column indices ggplot()+geom_point(aes(x=df[,'x'],y=df[,'y'])) ## or column names. ## but if i wrap it in a function... plot.func.one<-function(dff,x.var,y.var){

Hello With due respect, have a nice time. I would like to ask some command in R. It is regarding variable selection in linear regression. In R, there is one rebuild function called "step" which selecting variables according to AIC. let say i have data [y, x1,x2,x3,x4] we start with y~b0 i compute the partial F test and choose the variable with maximum partial F to enter the

Hi all, I'm trying to use lapply on a list with the following command: out<-lapply(mylist,myfun,par1=p,par2=d) (1) where myfun<-function(x,par1,par1) {.....} (2) now this function is in fact a wrapper for some Fortran code I have written so I think this might be the problem. When I call lapply() as in (1) I get the following message: Error in get(x,

Hi All I have a dataframe: myframe<-data.frame(ID=c("first","second"),x=c(1,2),y=c(3,4)) And I have a function myfun: myfun<-function(x,y) x+y I would like to write a function myfun2 that takes myframe and myfun as parameters and returns a list as below: mylist $first [1] 4 $second [2] 6 Could you please help me with this? Doesn't seem like the

Hi, I am moving from MATLAB, where one can easily assign a number of output values from a function like this: [x,y] = myfun(a,b) Then variables x and y can be directly used in the caller workspace. I understand that R functions return a single argument, which could be a list. This in a way makes it possible to return multiple values with a single function call, but accessing the list variables

Can the value of a list element be referenced from a sibling list element during list creation without the use of a temporary variable? The following doesn't work but it's the general idea. > list(value = 2, plusplus = $value+1) such that the following would be the output from str() List of 2 $ value : num 2 $ plusplus: num 3

Hi useR's, I want to overlay an image plot over a world map and I can do it, but just not the way I need to do it. Here is the code I am using (with data file attached) to create my baseline map: library(sp) load("TM_WORLD_BORDERS_SIMPL-0.2.RData") par(bty="l") plot(wrld_simpl, axes = TRUE, ylim = c(-90, 90), xlim=c(-180, 180), asp=1.5) lim <- par("usr")

I have a question regarding accessing the names of list components when applying a function with lapply. Here is an example that demonstrates what I'd like to do. I have a list like this one: mylist <- list(a=letters[1:10], b=letters[10:1], c=letters[1:3]) Now I would like to append the names of the list components to their corresponding vectors with the c() function. I thought this

Dear R helpers Suppose I have a vector as vect1 = as.character(c("ABC", "XYZ", "LMN", "DEF")) > vect1 [1] "ABC" "XYZ" "LMN" "DEF" I want to reverse the order of this vector as vect2 = c("DEF", "LMN", "XYZ", "ABC") Kindly guide Regards Vincy [[alternative HTML

Hi, I wonder if anyone knows how to generate a list of objects, e.g. ten vectors with names: vect1, vect2, ... , vect10. My own idea was to use something like: for (i in 1:10) print(paste("vect", i,"<-NULL",sep="")) but the result is: "vect1<-NULL" ... "vect10<-NULL" and not vect1<-NULL ... vect10<-NULL as I would

Hi everybody, I am currently working on mixtures of two densities ( f(xi,teta)= (1-teta)*f1(xi) + teta*f2(xi) ), particularly on the behavior of the variance for teta=0 (so sample only comes from the first distribution). To determine the maximum likelihood estimator I use the Newton-Rapdon Iteration. But when computing the first derivative I get a none linear function (with several asymptotes)

Hello, i would like to ask you another question. Is exist anymethod to vectors that tells me the last element?That is to say,I have a vector, I want to return the position of last element. I hope having explained. A greeting, Ignacio. -- View this message in context: http://old.nabble.com/Method-tp26493442p26493442.html Sent from the R help mailing list archive at Nabble.com.

Re:_: how to replace values in x by means in subgroups created in ...(not loops) Thanks, below some code and reply: #_-_-_-_-_-_-_-_-_-_-_-_-_-_-_-_-_# #------------------------------------------------------- # my slow function with loops: # replace_x_by_locallyMean_x_4_0s_in_y.f(x,y) #-------------------------------------------------------

Hi R users: I have de following data frame (called "Sx") Descripcion Nitratos Cont85g 72.40 Cont85g 100.50 Cont85g 138.30 Cont80g 178.33 Cont80g 79.01 Cont80g 74.16 Cont75g 23.70 Cont75g 15.80 Cont75g 16.20 Patron80g

Dear R-bug-people I have encountered a problem with "unsplit", which I believe may be caused by a bug in the function. However, unexpericend with bug-reports I apologise if this is barely a user problem rather than a problem within R. The problem occurs if an object is split by several grouping factors with levels not occuring in the data, and using drop = TRUE. This may appear as

Dear group, I am trying to write functions, but as a beginner, everything is not so obvious. Let's say I want the results in a list of elemts like this : tot1, tot2, etc Here is a function: toto <- function(x,y) { for(i in x:y){ paste(c("tot",i),collapse="")<-(i*2) } } If I type this : >toto(1,5) I get this message error: Error in paste(c("tot",

The R translation teams have done a great job in making R usable for people who do not have English as their mother tongue. However, even within English speaking countries, there are groups which have trouble with the language, and it may be valuable to support the Sociolects of these groups too. Thanks to a generous contribution from Lars Polifo, these features will be made available in an

Dear group, Here is my data frame: position100415 <- structure(list(DESCRIPTION = structure(1:9, .Label = c(" SUGAR NO.11 Jul/10 ", " SUGAR NO.11 May/10 ", "CORN May/10 ", "COTTON NO.2 Jul/10 ", "CRUDE OIL miNY May/10 ", "ROBUSTA COFFEE (10) Jul/10 ", "SILVER May/10 ", "SOYBEANS Jul/10 ", "WHEAT May/10

> From: Weiwei Shi > > it works. > thanks, > > but: (just curious) > why i tried previously and i got > > > is.vector(sample.size) > [1] TRUE Because a list is also a vector: > a <- c(list(1), list(2)) > a [[1]] [1] 1 [[2]] [1] 2 > is.vector(a) [1] TRUE > is.numeric(a) [1] FALSE Actually, the way I initialize a list of known length is by

I don't know what I have changed, NT does not list the shares anymore!! I can access them by directly connecting to \\SERVER\share but cannot browse them!! Win98 does just fine! Which option is the problem?? Here's a snipped of my config: [global] keep alive = 30 log file = /var/log/smbd.log lock directory = /var/lock/samba bind interfaces only = yes interfaces = eth0 interfaces =