similar to: fidelity of generated raster images (R and perl)

Colleagues, I am trying to create a PDF document in which I use margin text with two different fonts. The resulting text might be: XXXXXyZZZ where X and Z are one font and Y is the other. My plan was to do this in the following manner: mtext("XXXXX ZZZ", cex=2, adj=0.5, family=SOMEFONT) mtext("Y", cex=2, adj=??, family=DIFFERENTFONT) My question regards how to calculate

Hi, I have a data.frame in wide format which I would like to reshape into a long format: example (nonsense) data: > dput(perus2) structure(list(id = c(30L, 38L, 21L, 12L, 22L, 28L, 31L, 44L, 8L, 47L, 23L, 20L, 41L, 42L, 29L, 50L, 5L, 33L, 4L, 17L, 11L, 1L, 18L, 6L, 9L, 32L, 16L, 14L, 39L, 48L, 37L, 43L, 25L, 27L, 35L, 40L, 45L, 49L, 2L, 34L, 10L, 15L, 24L, 26L, 46L, 7L, 13L, 3L, 19L, 36L),

Sorry, I pasted the wrong file in earlier... this is the correct one: pValCalculator(b, n=20, m=20) { ind <- 1:min(c(n,m)) prob <- (1-pnorm(b,sd=std*sqrt(ind))) prob1 <- sum((n-ind+1)*(m-ind+1)*prob) prob1 } inputData <- scan("/users/lvssso/projects/LAMA/output/pValLamaScore.tmp", list(block1 = "",block2 = "",width1 = 0,width2 = 0,alignment = 0,score

Dear developers, I wonder would you consider making a save to a graphics file format (as opposed to ps)? What prompts this is that we have just finished my wife's thesis using R heavily for stats and graphs. The combination of latex, bibtex and R generated .eps worked a treat and we were very pleased with both the final outcome and the efficiency of gernerating it, especially when all the

I've installed new fifa and written in the code of the terminal had opened up to play EA Sports intro then comb a new error message in there Standt: "In the program were difficult manager11.exe gent fault and it must be-beeendet We apologize for this unnanehmlichkeiten. The cause may be a problem in the program or may fail to run in wine sein.sie www.appdb.winehq.org tips under the

+ Evgeny We have a jump threading pass downstream for this that we would love to upstream. I believe Evgeny was working on exactly this, i.e. preparing it for upstreaming. ________________________________ From: llvm-dev <llvm-dev-bounces at lists.llvm.org> on behalf of Eli Friedman via llvm-dev <llvm-dev at lists.llvm.org> Sent: 23 September 2020 19:16 To: Phipps, Alan <a-phipps

"Or use <<- assignment I think. (I usually return, but return can only return one object and I think you want two or more" You can return any number of objects by putting them in a list and returning the list. Use of "<<-" is rarely a good idea in R. -- Bert On Fri, Aug 9, 2024 at 1:53?AM CALUM POLWART <polc1410 at gmail.com> wrote: > > OK. The fact

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Dear R users, Probably, this is quite a simpe question, but I do not find the proper way to obtain want I need. To explain the problem, I constructed a simple example. Suppose I have the following function: try1<-function(x){ y<-x[1:2] z<-x[3:4] y[1]*(z[1]+z[2])+y[2]*(z[1]+z[2]) } This function will be part of a for loop. This is what I like to obtain for every k: if k=2

Hello world, I am actually transferring a course in data management for students in biology, geography and agriculture from statistica to R - it works surprisingly well. If anyone is interested in my scratch/notepad (in German language), please see www.hydrology.uni-kiel.de/~schorsch/statistik/statistik_datenauswertung.pdf (pages 40-52) The dataset is:

Dear all, I have a character string object: > chara [1] "The name of first slot" and a list object: > class( try1) [1] "list" what I want to do is to use the chara as a slot's name of "try1". Of course I could do it like: > try1$"The name of first slot" <- matrix("", 3, 4) to create a slot of 3x4 matrix with the name

Hi Sjoerd We (at Huawei) also have a pass for this. Originally we implemented this back in 2018 and meant to upstream it, but there were some issues with the implementation that required some changes in the code. We started revising it,a few weeks ago. I thought now that there are multiple options, maybe we can discuss our approaches, and see if there is a preference in the community for one

Allow to skip testing the filesystem "foo" if the environment variable SKIP_TEST_CHARSET_FIDELITY_foo=1 is set. This way it possible to not test one or more filesystems without disabling the test altogether. --- tests/charsets/test-charset-fidelity.c | 14 +++++++++++++- 1 file changed, 13 insertions(+), 1 deletion(-) diff --git a/tests/charsets/test-charset-fidelity.c

Thanks. Hmm. The loop is doing what it is supposed to do. > try1<-function(joint12=FALSE,marg1=FALSE,marg2=FALSE, +??????????????? cond12=FALSE,cond21=FALSE){ + # *************************************************** + # Testing if loop + # *************************************************** + if(joint12){ +?? {print ("joint12"); cat(joint12,"\n")} +?? {print

Hi, Sorry to bother you guys with this syntax problem, this is my first R program, and I seem to not have grasped something right. I keep getting an error message from 'parse' on the line with the *** : pValCalculator(b, n=20, m=20) { ind <- 1:min(c(n,m)) prob <- (1-pnorm(b,sd=std*sqrt(ind))) prob1 <- sum((n-ind+1)*(m-ind+1)*prob) prob1 } inputData <-

I am generating images via lattice from Frank Harrell's RMS package. These images are characterized by coloured lines and grey-scale confidence intervals. I need to port them to Openoffice/etc, and have tried both png and jpeg (at high quality), but in neither format can I subsequently see the the grey scale confidence intervals. Other than moving to LaTex, does anyone have other

I'm having a problem when using heatmap. Even though the diagonal of my matrix is all the same value, the diagonal of my heatmap is not all the same color. Any suggestions? Here is some reproducible code: ######################################### # Get data nba <- read.csv("http://datasets.flowingdata.com/ppg2008.csv", sep=",") # Reorder nba <-

OK. The fact it's in a function is making things clearer. Are you trying to update the values of an object from within the function, and have them available outside the function. I don't speak functional programming articulately enough but basically v <- 1 funA <- function() { v <- v+1 } funA() cat (v) # 1 You either return the v from the function so funB <- function() {

Is something wrong in the initialisation part that we don't see? joint12 <- marg1 <-F marg1 <-T if (joint12) { print ("joint 12") cat (joint12) } if (marg1) { print("marg 1") cat(marg1) } Would probably be my diagnostic approach On Fri, 9 Aug 2024, 04:45 Steven Yen, <styen at ntu.edu.tw> wrote: > Can someone help me with the if loop below?

I have a vector Pes with 600 elements some of which are NA's. How can I form a vector of the indices of the NA's. for(i in 1:600) if(is.na(Pes[i])) print(i) prints the indices of the NA's but I can't figure out how to put the results in a vector. Thank you very much. Peter B. -- Peter B. Mandeville mandevip at deimos.tc.uaslp.mx Jefe del Depto. de