similar to: Boxplot with Log10 and base-exponent axis

Hello, I'm using fArma package to estimate the value of Hurst exponent using R/S method. However, for a certain set of data I get H ~ 1.8. How do I interpret this? Following are the output that I get for this set: > mean(data[,2]) [1] 400.5433 > sd(data[,2]) [1] 1139.786 > > rsFit(data[,2], levels = 64) Title: Hurst Exponent from R/S Method Call: rsFit(x = data[, 2], levels

Hi all, I want to label an axis with exponents, but can't get it done with expression. Any hints would be very welcome! # simulated data, somewhat similarly distributed to my real data: set.seed(12); d <- rbeta(1e6, 0.2,2)*150 ; d <- d[d>1e-8] hist( d? , breaks=100) # now on a logarithmically scaled axis: hist(log10(d), breaks=100, xaxt="n") abline(v=

Estimad en s, quería realizar una consulta Dado un data frame llamado aves, como puedo sustiuir las variables area, dist y distm por sus logaritmos en base10 y volver a crear un data frame? Muchas gracias Saludos Susana [[alternative HTML version deleted]]

I am having difficulty with the calculation of the power law exponent for set of nodes within a graph. Specifically, I am interested in the distribution of in-degree and out-degree among communities of web pages where the web pages are the nodes of the graph and the hyperlinks the edges. According to the literature, the distribution of incoming and outgoing links obeys a power law distribution

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

Hi all, in my search for a nice binary2decimal method, I received this nice code (thanx to Uwe Ligges): bindec <- function(b) sum(as.integer(unlist(strsplit(b, ""))) * 2^(floor(log10(b)):0)) It fails, however, with: > bindec(1000) [1] 4 Warning message: longer object length is not a multiple of shorter object length in: as.integer(unlist(strsplit(b, "")))

Dear R users: Sorry for this simple question: I am writing a function where I would need to pickup p values and make -log10 of it. The p values are from an anova output and sometime it can yield me 0. -log10 (0) [1] Inf I can not replace Inf with 0, which not case here. This is restricting me to go further in the function and out me the error. You help is highly appreciated. Thanks; --

In y <- substr(x, i, 1) your third integer needs to be the location not the number of digits, so change it to y <- substr(x, i, i) and you should get what you want. Cheers, Tim > Date: Sun, 21 Jan 2018 10:50:31 -0500 > From: Ek Esawi <esawiek at gmail.com> > To: Luigi Marongiu <marongiu.luigi at gmail.com>, r-help at r-project.org > Subject: Re: [R] substr

Dear all, Has anyone worked on coding for calculating Lyapunov Exponent for a time series data? or any package is available for computing Lyapunov? Please advice and many thanks in advance. Catherine X Wang

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

Dear All, I am a novice when it comes to time-series analysis and at the moment I am actually interested in calculating the Hurst exponent of a time series. This question has already been asked quite some time ago http://bit.ly/98dZsi and I trust some progress has been made ever since. I was able to find some functions in the packages http://cran.r-project.org/web/packages/Rwave/index.html

Hello, Does anybody know if there is somewhere in R a function to calculate the Lyapunov exponent in a time series? Thanks, Philippe Grosjean .......................................................<??}))><.... ) ) ) ) ) ( ( ( ( ( Prof. Philippe Grosjean \ ___ ) \/ECO\ ( Numerical Ecology of Aquatic Systems /\___/ ) Mons-Hainaut University, Pentagone / ___ /( 8, Av. du

Hi, I have a histogram with values logged to the base 10 hist(log10(x),breaks=60) How do I show the log values on the x-axis and a second x-axis showing the real values? Thanks -- Shane [[alternative HTML version deleted]]

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 =

I am attempting to deploy OpenSSH. The trouble is I keep getting the rsa_public_encrypt() exponent too small or not odd with the SSH 1 or 1.5 protocols. I can't get OpenSSH to communicate with itself with any protocal other than SSH 2. Platform notes: HP-UX 11.00 Dart 51 64bit OpenSSL 0.9.6 Zlib 1.1.3 Cflags: -Ae I have tried with and without optimizations. I noticed that this problem

Dear all, I am trying to apply the logistic regression to determine the limit of detection (LOD) of a molecular biology assay, the polymerase chain reaction (PCR). The aim of the procedure is to identify the value (variable "dilution") that determine a 95% probability of success, that is "positive"/"total"=0.95. The procedure I have implemented seemed to work looking

Dear R experts, I really have difficulty when I try to deal with this question. suppose X is a square symmetric matrix. The exponent of X is defined by the matrix limit as following: exp(X) = lim (I + X/n)^n, note: the limit is from n to infinite. How can I write R function for the above? Thank you very much -- View this message in context:

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:

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),