similar to: Problem with viewports, print.trellis and more/newpage

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, The sasxport.get function (from the Hmisc package) automatically defines the class of imported variables. I have noticed that the class of theoretically numeric variables is simply "labelled", although character variables might end up been defined as "labelled" "Date" or "labelled" "factor". Is there a way to tell sasxport.get to

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"),

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

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)),

??? > 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

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

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 ) )

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()

Hi, Searching for functions that would order strings that mix characters and numbers in a "natural" way (ie, "a1 a2 a10" instead of "a1 a10 a2"), I found the mixedsort and mixedorder from the gtools package. Problems: 1- mixedorder does not work in a "do.call(mixedorder, mydataframe)" call like the order function does 2- gtools has not been updated in 2.5

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 packages would contain a variant of the subset function that would consider missing values as true. Thanks

Dear R-users, I am trying to use the grid.text and expression functions to display several character strings and plotmath text on a viewport. Some strings can include a variable portion (PI.limits in the following example), which I thought could be implemented by combining the bquote and the expression functions. Unfortunately, my expressions do not seem to be evaluated. I would greatly

Dear all, I am trying (!!!) to generate pdfs that have 8 plots on one page: df = data.frame( day = c(1,2,3,4), var1 = c(1,2,3,4), var2 = c(100,200,300,4000), var3 = c(10,20,300,40000), var4 = c(100000,20000,30000,4000), var5 = c(10,20,30,40), var6 = c(0.001,0.002,0.003,0.004), var7 = c(123,223,123,412), var8 = c(213,123,234,435), all = as.factor(c(1,1,1,1)))

Dear R-users, I am trying to produce (standard and trellis) scatterplots which use the same range of the x and y axes. In addition, I would like the plots to be physically square. Is there one or more specific argument(s) to plot and xyplot what would do that? I have looked at various combinations of asp and pin value, but could not get what I wanted.. Thank you in advance for your help

Dear R-users, The following question bothered me for the whole afternoon: how can one pass a string as the conditioning argument to subset? I tried plain mystr, eval(mystr), expression(mystr), etc... I don't to be able to find the correct syntax > foo <- data.frame(a=1:10,b=10:1,c=rep(1:2,5)) > mystr<-"c==1" > subset(foo,c==1) a b c 1 1 10 1 3 3 8 1 5 5 6

Hi, Down a cascade of function calls, I want to use the curve function with an expression that is a variable. For various reason, this variable must be a character object and cannot be an expression as required by the curve function. How do I convert my variable into a expression that is accepted by curve? Thanks in advance for your help. ## The following attempts do not work myf <-

Dear R-users, I would like to know how expressions could be passed as arguments to do.call functions. As illustrated in the short example below, concatenating lists objects and an expression creates an expression object, which is not an acceptable argument for do.call. Is there a way to avoid that? Thanks you Sebastien foo <- list(x=1:10, y=1:10) mylist <- list(pch=6, col=2) title

Thanks. That works great! > df <- data.frame(x=c(1,1,NA,NA,2), y=c('a','a','a','b','b'), z=c(TRUE,FALSE,TRUE,FALSE,TRUE)) > cond1 <- 'x==1' > cond2 <- 'x==1 & z' > df x y z 1 1 a TRUE 2 1 a FALSE 3 NA a TRUE 4 NA b FALSE 5 2 b TRUE > subset(df, subset = ifelse(is.na(eval(parse(text=cond1))), TRUE,

hello, is there a way for rsync to create non-existing parent directories when syncing files to a destination? something similar to 'mkdir -p non-existing/directory'. i'm doing 'rsync -r /path/to/source /path/to/non-existing/directory', and the 'non-existing/directory' location doesn't exist. the errors i get is... rsync: mkdir

Dear R-users, I am trying to use the grep function to test whether a particular string is of the form "n.../mydir/myfile.mytype.myext". Anything between n and mytype could vary, and anything after mytype could vary. I tried to proceed by steps to build my regular expression... but I do not really understand why the last call of the following code do not work. Any help would be