similar to: substr gives empty output

Dear all, I have a string, let's say "testing", and I would like to extract in sequence each letter (character) from it. But when I use substr() I only properly get the first character, the rest is empty (""). What am I getting wrong? For example, I have this code: >>> x <- "testing" k <- nchar(x) for (i in 1:k) { y <- substr(x, i, 1)

Hi Put fill outside aes p+geom_ribbon(aes(ymin = slope_1*x + intercept_1 - 1/w[2], ymax = slope_1*x + intercept_1 + 1/w[2]), fill = "blue", alpha=0.1) The "hole" is because you have two levels of data (red and blue). To get rid of this you should put new data in ribbon call. Something like newdat <- trainset newdat$z <- factor(0) p+geom_ribbon(data=newdat, aes(ymin =

Thank you, but this split the area into two and distorts the shape of the plot. (compared to ``` p + geom_abline(slope = slope_1, intercept = intercept_1 - 1/w[2], linetype = "dashed", col = "royalblue") + geom_abline(slope = slope_1, intercept = intercept_1 + 1/w[2], linetype = "dashed", col = "royalblue") ``` Why there

Hi What about something like p+geom_ribbon(aes(ymin = slope_1*x + intercept_1 - 1/w[2], ymax = slope_1*x + intercept_1 + 1/w[2], fill = "grey70", alpha=0.1)) Cheers Petr > -----Original Message----- > From: Luigi Marongiu <marongiu.luigi at gmail.com> > Sent: Friday, October 23, 2020 11:11 AM > To: PIKAL Petr <petr.pikal at precheza.cz> > Cc: r-help

As usual, a web search ("find statistical mode in R") brought up something that is possibly useful -- Did you try this before posting? If not, please do so in future and let us know what your results were if you subsequently post here. Here's what SO suggested: Mode <- function(x) { ux <- unique(x) ux[which.max(tabulate(match(x, ux)))] } # ergo:

Stop using as.factor() and start using factor()? Be explicit about what levels you want and in what order. As for extracting subsets, use split() and do your subsets on each set of rows with the same level. It will be on you to decide what statistical properties (proportions?) you want to maintain between the full data set and your subset. On October 20, 2020 6:42:27 AM PDT, Luigi Marongiu

Hi, I think you're misunderstanding which set of variables go on either side of the formula. Is this what you're looking for? > aggregate(OD ~ Time + Target + Conc, data = df, FUN = "mean") Time Target Conc OD 1 1 BACT 1 765.3333 2 1 BACT 2 745.3333 3 1 BACT 3 675.0000 > aggregate(ODnorm ~ Time + Target + Conc, data = df, FUN =

Thank you. The problem was not finding the mode but applying it the R way (I have the tendency to loop into each line of the dataframes, which I believe is NOT the R way). I'll try them. Best regards Luigi On Sat, Oct 31, 2020 at 5:40 PM Bert Gunter <bgunter.4567 at gmail.com> wrote: > > As usual, a web search ("find statistical mode in R") brought up something that is

Hi Luigi, If I understand your request: library(prettyR) apply(as.matrix(df),1,Mode) [1] "C" "B" "D" ">1 mode" ">1 mode" ">1 mode" "D" [8] "C" "B" ">1 mode" Jim On Sat, Oct 31, 2020 at 7:56 PM Luigi Marongiu <marongiu.luigi at gmail.com> wrote:

Dear all, Is there a simple way to add reference lines in background? I am trying with abline() or grid() but the lines, since they are executed after the plot function, are draw on top. How can I draw such lines beneath the main plot? Here is an example: x<-rnorm(100) boxplot(x) abline(h=c(-1,0,1)) grid(NA, 4, lwd = 2) regards, Luigi Marongiu, MSc [[alternative HTML

Dear all, I would like to ask if it possible to draw a median line in a Stripchart, either using the built-in STRIPCHART function or with another package. I was expecting stripchart() to work more or less like boxplot(), but I couldn't find an easy way to draw a mean or median line, neither I found this option in other packages. Could somebody help? Best regards, Luigi Marongiu, MSc

I don't know about the configuration. I installed R using the standard protocol for Chromebook http://blog.sellorm.com/2018/12/20/installing-r-and-rstudio-on-a-chromebook/ the rest, it was done by the system itself... On Tue, Jul 14, 2020 at 1:30 PM Dirk Eddelbuettel <edd at debian.org> wrote: > > > There is something wrong with your system / setup I did not notice first: >

also from this site: https://plotly.com/ggplot2/geom_ribbon/ I get the answer is geom_ribbon but I am still missing something ``` #! plot p = ggplot(data = trainset, aes(x=x, y=y, color=z)) + geom_point() + scale_color_manual(values = c("red", "blue")) # show support vectors df_sv = trainset[svm_model$index, ] p = p + geom_point(data = df_sv, aes(x=x, y=y),

On Tue, Oct 27, 2020 at 6:04 PM Luigi Marongiu <marongiu.luigi at gmail.com> wrote: > > Hello, > I am using e1071 to run support vector machine. I would like to plot > the data with lattice and specifically show the hyperplanes created by > the system. > I can store the hyperplane as a contour in an object, and I can plot > one object at a time. Since there will be

Also, > aggregate(cbind(OD, ODnorm) ~ Time + Target + Conc, data = df, FUN = "mean") Time Target Conc OD ODnorm 1 1 BACT 1 765.3333 108.33333 2 1 BACT 2 745.3333 88.33333 3 1 BACT 3 675.0000 18.00000 (You might wish for "cbind(OD,ODnorm) ~ . - Well", but aggregate.formula is not smart enough for that.) -pd > On 24 Oct 2023, at

Thank you, it looks like I have already libisl: ``` apt search libisl Sorting... Done Full Text Search... Done libisl-dbg/oldstable 0.18-1 amd64 manipulating sets and relations of integer points bounded by linear constraints libisl-dev/oldstable 0.18-1 amd64 manipulating sets and relations of integer points bounded by linear constraints libisl15/oldstable,now 0.18-1 amd64

Dear all, I would like to (i) produce boxplot graphs with axis in logarithm in base 10 and (ii) showing the values on the axis in 10^exponent format rather than 10E+exponent. To illustrate with an example, I have some widely spread data that I chart plot using boxplot() [figure on the left]; the log="y" option of boxplot() I obtained the natural logarithm conversion of the data and

Hi Did you try google? I got several answers using your question e.g. https://stackoverflow.com/questions/54687321/fill-area-between-lines-using-g gplot-in-r Cheers Petr > -----Original Message----- > From: R-help <r-help-bounces at r-project.org> On Behalf Of Luigi Marongiu > Sent: Friday, October 23, 2020 9:59 AM > To: r-help <r-help at r-project.org> > Subject:

Thank you, but this now sounds more difficult: what would be the point in having these ready-made functions if I have to do it manually? Anyway, How would I implement the last part? On Thu, Mar 23, 2023 at 1:23?AM Ebert,Timothy Aaron <tebert at ufl.edu> wrote: > > If you are open to other options: > The null hypothesis is that there is no difference. > If I have two equations

On 05/03/2024 7:16 a.m., Luigi Marongiu wrote: > A <- rep(c(0, 3, 6, 12, 24), 3) > B <- c(rep(0,5), rep(1,5), rep(2,5)) > V <- c(27.5, 27.01, 27.75, 27.98, 27.4, 25.69, > 26.17, 27.78, 26.08, > 24.97, 23.18, 21.78, 22.49, 21.85, 22.2) > df <- data.frame(Conc = A, Time = B, Value = V) > df$Conc = as.factor(df$Conc) >