similar to: Dummy quesion about environment

Hi, I am trying to apply two different functions on on a vector as follow: a<-c(NA,1,2,3,-3,-4,-6) if a>0 I would like to raise it by the power of 2: 2^a and if the a<0 I would like to have the inverse value, i.e., -1/2^a. so I thought of doing it two steps: a[a>0]<-2^[a>0] a[a<0]<-(-1)/2^a[a<0] I got the following error Error: NAs are not allowed in subscripted

Dear group members, I am trying to plot two adjacent plots using layout function. One plot is scatter plot and the other is density plot. However, I would like that the density plot would be present horizonaly in respect to the scatter plot. | | | | | | |

Dear all, I am trying to run cloud function in lattice library under r-1.7, and I wlways get gray background. I've tried to chang it by set "...par.box=list(col="white")" or by gpar(background="white") but no result. How do i do it? Thanks in advance, Ron

dear group members, I am looking for a function that assess the stability of cluster. The result of hclust function is an hclust object which can be plot as a dendrogram. However to have confidence in the tree topology usualy bootstap is applied. I understand that I can apply bootstarp on the original data and then run hclust(dist() ) as much as I resampled but how to comapre the topologies the I

Dear supporter, Can you tell me please how can I mirroring R at our site? One of the reasons I would like to do this is the fact that it is impossible to update R behind our firewalls. Thank in advance Ron Ophir, Ph.D. Bioinformatician, Biological Services Weizmann Institute of Science POB 26 Rehovot 76100 Israel e-mail: Ron.Ophir at weizmann.ac.il Phone: 972-8-9342614 Fax:972-8-9344113

Dear All: Could anybody help me figure out why I get the Error message below while I running the example code of bugs() function in R2OpenBUGS packages? I have tried the code both in Win 7 and Ubuntu 12.04, but they show the same message. My R version is 2.15.1 in win7 and 2.15.0 in Ubuntu. Many thanks ! Best, Yilong > schools.sim <- bugs(data, inits, parameters, model.file,+

Hi all: I have a quesion about ANOVA: Is SS(Sum of Square) of a specific factor constant with the number of factors changing? dat1 includes one factor g1,and g1's SS is called SS_g1_dat1. dat2 includes two factors g1,g2,and g1's SS is called SS_g1_dat2. My quesion is: Is SS_g1_dat1 equals to SS_g1_dat2? I have both "yes" and "no" reasons for the quesion,but

Hello, How can I tell RODBC to scan all the records of an xls file to determine the data type? If the first n records happen to be empty Rodbc assumes a character, and any numbers are made <NA>. And if, for instance, the first n records contain numbers, and later they also contain characters, those characters become NA. Cheers!! Albert-Jan

Hi all: My data is in the attachment. I want to analysis the mean difference of y between 2 sex. My code: result_lm<-lm(y~factor(sex) + x1 + x2) summary(result_lm) The result of "factor(sex)m" 136.83, is the mean difference of y between 2 sex,and the corresponding p value is 0.07618. My question is: how to get the mean y of sex(m) and sex(f) respectively via lm function? Many

I ran into an interesting issue with `evalq` (and also `eval(quote(...))`): ???? f <- function() { ?????? list( ???????? sys.parent(1), ???????? evalq(sys.parent(1)), ???????? evalq((function() sys.parent(2))()),? # add an anon fun layer ???????? evalq((function() sys.parent(1))()) ?????? ) ???? } ???? res <- f() ???? str(res) ???? ## List of 4 ???? ##? $ : int 0???????? # sys.parent(1)

In 3.1.2 eval does not store the result of the bquote-generated call in the given environment. Interestingly, in 3.2.1 eval does store the result of the bquote-generated call in the given environment. In other words if I run the given example with eval rather than evalq, on 3.1.2 "x" is never stored in "fenv," but it is when I run the same code on 3.2.1. However, the given

The help page of eval says: The 'evalq' form is equivalent to 'eval(quote(expr), ...)'. But the following is not equivalent. Can anyone give me some explaination? Thanks very much. > f1 <- function(x,digits=5) lapply(x, f2) > f2 <- function(x) eval(quote(print(x+1,digits=digits)),list(x=x),parent.frame(2)) > f1(list(x1=1)) [1] 2 $x1 [1] 2 > > f1 <-

On Jul 15, 2015, at 12:51 PM, William Dunlap wrote: > I think rapply() was changed to act like lapply() in this respect. > When I looked at the source of the difference, it was that typeof() returned 'language' in 3.2.1, while it returned 'list' in the earlier version of R. The first check in rapply's code in both version was: if (typeof(object) != "list")

Hello, I upgraded from 3.1.2 to 3.2.1 and am receiving errors on code that worked as I intended previously. Briefly, I am using bquote to generate expressions to modify data.table objects within a function, so I need the changes to actually be stored in the given environment. Previously, I used code like the following: test <- list(bquote(x <- 10)) fenv <- environment() rapply(test,

David, If you are referring to the solution that would be: rapply(list(test), eval, envir = fenv) I thought I explained in the question that the above code does not work. It does not throw an error, but the behavior is no different (at least in the output or result). Using the above code still results in the x object not being stored in fenv on 3.1.2. Dayne On Wed, Jul 15, 2015 at 4:40 PM,

Hi, In order to avoid deep copies by passing large arguments to functions or returning values, I'm trying to do the assignment of variables in a given environment. The problem is when I try to assign a structure: a list for example. If I have: ind <- c("a","b") my idea is doing something like l <- alist() l[ind] <- as.list(c(20,40)) in a given

Hi First, sorry if I get the terminology wrong, I am still quite new to the concept of using environments and workspaces. Say I have a statement in a package SIM like sim <- TYPE where the variable TYPE is initialized in the package to e.g. "exponential" (SIM::TYPE == "exponential"). Now, I want to give the user the option of specifying the variable TYPE, but to the

Dear mailing list, I have found some strange behaviour which I think relates to parent frames and eval. Can anyone explain what's going on here? First example: > test.parent.funcs_ function() { outer.var_ 5 subfunc1_ function() substitute( outer.var, envir=parent.frame()) print( subfunc1()) subfunc2b_ function() eval( quote( outer.var), envir=parent.frame()) print(

On 01/04/2015 1:35 PM, Gabriel Becker wrote: > Joris, > > > The second argument to evalq is envir, so that line says, roughly, "call > environment() to generate me a new environment within the environment > defined by data". I think that's not quite right. environment() returns the current environment, it doesn't create a new one. It is evalq() that created

I'm still going over old emails and trying to get my head around evaluation so I'm persistent if nothing else. A while back , an expert sent me below as an exercise in understanding and I only got around to it tonight. I understand some of the output but not all of it and I put "Why not Zero ?" next to the ones that I don't understand based on my reading of the various