similar to: Understanding padding in lattice

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, 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()

Good morning! I have the following data frame (df): X.outer Y.outer X.PAD1 Y.PAD1 X.PAD2 Y.PAD2 X.PAD3 Y.PAD3 X.PAD4 Y.PAD4 73 574690.0 179740.0 574690.2 179740.0 574618.3 179650 574729.2 179674 574747.1 179598 74 574680.6 179737.0 574693.4 179740.0 574719.0 179688 574831.8 179699 574724.9 179673 75 574671.0 179734.0 574696.2 179740.0 574719.0 179688 574807.8 179787 574729.2

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 there, I'm a bit confused concerning the axis tck setting in the lattice package as the ticks on left sided axis aren't drawn at all with the following setting: dados <- data.frame(varsep = factor(rep(1:2,10)), i = runif(20)) library(lattice) my.theme <- list( axis.components = list(left = list(tck = 1, pad1 = 1, pad2 = 2), top = list(tck = 0, pad1

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

Hello R Folks, I am new to R. I have been struggling to overlay a barchart with a xyplot together on one plot but did not get this worked out. Any help and idea are greatly appreciated. I attached R scripts for barchart and xyplot below and also data I used. What I am trying to do is just to put the barchart and xyplot together on one plot. Huapeng <<dispersal_infestation.csv>>

Here is an example: I can define two classes: A and Apad: class A { bool b1, b2; double d1; int i1; }; class Apad { bool b1, b2; bool pad1[6]; double d1; int i1; bool pad2[4]; }; A and Apad will have the same layout, from the LLVM IR level: %class.A = type <{ i8, i8, [6 x i8], double, i32, [4 x i8] }> %class.Apad = type { i8, i8, [6 x i8], double, i32, [4

Yes. LLVM types != C++ types. There is no mapping except that produced if you add debug info. On Wed, Dec 28, 2016 at 6:51 PM, Hong Hu via llvm-dev < llvm-dev at lists.llvm.org> wrote: > Here is an example: > > I can define two classes: A and Apad: > > class A { > bool b1, b2; > double d1; > int i1; > }; > > class Apad { > bool b1, b2;

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 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 <-