similar to: Question about subset

Sent from my iPhone > On Apr 8, 2018, at 9:06 PM, Sebastien Bihorel <sebastien.bihorel at cognigencorp.com> wrote: > > Hi, > > The help page for subset states "subset: logical expression indicating elements or rows to keep: missing values are taken as false." > > Before I try to re-invent the wheel, I would like to know if one of the base or recommended

Dear R-users, I have got the following problem. I need to create 4x2 arrays of xyplot's on several pages. The plots are created within a loop and plotted using the print function. It seems that I cannot find the proper grid syntax with my viewports, and the more/newpage arguments. The following script is a simplification but hopefully will suffice to illustrate my problem. Any suggestion

Dear R-users, I am trying to understand what the different padding arguments in trellis.par.set are exactly controlling the space around lattice plots. I have used the following code as a basis for testing but it did not really help me to visualize how the value of each argument changes the margins and the plotting area. I guess a better way to visualize the effects of these padding items

Dear R-users, I have 3 plant populations (fixed). Within each population there is the same number of “families” (random) – the seed progeny of the same plant. These families were exposed to 2 treatments (fixed) and their response was measured (mean values for 25 seedlings per family per treatment are presented in data table). I would like to know if there is a significant difference in the

Dear list, I have data like this: dat1 <- data.frame(subject=rep(1:10,2), cond1=rep(c("A","B"),each=5), cond2=rep(c("C","D"),each=10), choice=sample(0:1,10,replace=TRUE)) I would like to compare subjects' "choice" for (cond1=="A" & cond2=="C") vs

Dear R-Users, I would like to know whether it is possible to draw several stacked barplots (i.e. side by side on the same sheet)... my data look like : Cond1 Cond1' Cond2 Cond2' Compartment 1 11,81 2,05 12,49 0,70 Compartment 2 10,51 1,98 13,56 0,85 Compartment 3 1,95 0,63 2,81 0,22 Compartment 4 2,08 0,17

Dear UseRs, I wrote following function in order to solve Data Envelopment Analysis. Reason for posting is that the function is slow when nrow(dat) is large. I wonder if other functions could substitute the for() loop in the code, such as mapply(). Can anybody help to rewrite the dea() function as efficiently as possible? The code is as follows:

split any mixed columns into letter and number columns and then order can be used on that: DF <- data.frame(x = c("a10", "a2", "a1")) o <- do.call("order", transform(DF, let = gsub("\\d", "", x), no = as.numeric(gsub("\\D", "", x)),

1- mixedorder does not work in a "do.call(mixedorder, mydataframe)" call like the order function does This is tangential, but do.call(order, mydataframe) is not safe to use in a general purpose function either - you need to remove the names from the second argument: > d <- data.frame(method=c("New","New","Old","Old","Old"),

I am evaluating a switch from Stata to R. I don't need to extensive Statistical methods, but the main reason I am exploring the switch is the coding flexibility in R (e.g. Stata does not support linear/quadratic programming). I have been going over the R syntax and I had a quick question: In Stata, one has a very useful construction called "by", e.g. by month signal: gen xxx =

Community, (Windows Xp , 1.7.1, too little memory) This may have the flavor of an answer in search of a question, but I'll pose this question: Is their a better way to do the following than what I'll show below? Problem: In Sweave I have multiple figures from lattice, some of which have more that one page. I can easily figure out the total number of figures I have created, but its

Hi all, suppose there is a vector y with rownames: > y cond1 cond2 cond3 cond4 78.952 87.308 86.490 74.040 how can I easily plot this vector using the rownames? plot(y) gives me a plot with a x-axis from 1 to 4 in 0.5 steps, also plot(rownames(y), y) and plot(y ~ rownames(y) don't work. I know I can build a x-axis with axis(1, ...), but in this case I need a character string like

Hi, I have a general question on LoopUnswtich pass. Consider the following IR snippet: define i32 @test(i1 %cond) { br label %loop_begin loop_begin: br i1 %cond, label %loop_body, label %loop_exit loop_body: br label %do_something do_something: call void @some_func() noreturn nounwind br label %loop_begin loop_exit: ret i32 0 } declare void @some_func() noreturn After running

Typo: should be NULL not NUL of course An alternative approach closer to your original attempt is to use do.call() to explicitly evaluate the expr argument: w <- "1 + x^2" do.call(curve, list(expr = parse(text = w), ylab ="y")) Cheers, Bert Bert Gunter "The trouble with having an open mind is that people keep coming along and sticking things into it." -- Opus

Hello everyone, I am working for a few days already on a basic algorithm, very common in applied agronomy, that aims to determine the degree-days necessary for a given individual to reach a given growth stade. The algorithm (and context) is explained here: http://www.oardc.ohio-state.edu/gdd/glossary.htm , and so I implemented my function in R as follows: DD <- function(Tmin, Tmax, Tseuil,

Sebastian: This is somewhat arcane, perhaps even a bug (correction on this welcomed). The problem is that the "expr" argument to curve() must be an actual expression, not a call to parse that evaluates to an expression. If you look at the code of curve() you'll see why (substitute() does not evaluate expr in the code). Another simple workaround other than sticking in the eval()

??? > y <- sort( c("a1","a2","a10","a12","a100")) > y [1] "a1" "a10" "a100" "a12" "a2" > mixedsort(y) [1] "a1" "a2" "a10" "a12" "a100" **Please read the docs!** They say that mixedsort() and mixedorder() both take a **single

Hi: I have a large dataset mydata, of 1000 rows and 1000 columns. The rows have gene names and columns have condition names (cond1, cond2, cond3, etc). mydata<- read.table(file="c:/file1.mtx", header=TRUE, sep="") I applied PCA as follows: data_after_pca<- prcomp(mydata, retx=TRUE, center=TRUE, scale.=TRUE); Now i get 1000 PCs and i choose first three PCs and make a

x <- c( "a1", "a10", "a2" ) y <- c( "b10", "b2", "a12", "ca1" ) DF <- expand.grid( x = x, y = y ) # randomize set.seed( 42 ) DF <- DF[ sample( nrow( DF ) ), ] # missing from gtools mixedrank <- function( x ) { seq.int( length( x ) )[ gtools::mixedorder(x) ] } o <- do.call( order, lapply( DF, mixedrank ) )

I worked this out over the weekend. I appreciate that using temporary variables would be simpler but I think this makes for quite readable code: # in RProfile.site inplace <- function (f, arg=1) eval.parent(call("<-",substitute(f)[[arg+1]], f),2) # examples in code inplace(foo[bar,baz] *2) # or inplace(paste(foo[bar,baz], 1:10)) # or inplace(sub("blah",