similar to: have to point it out again: a distribution question

Hi, r-gurus: I happened to have a question in my work: I have a dataset, which has only one dimention, like 0.99037297527605 0.991179836732708 0.995635340631367 0.997186769599305 0.991632565640424 0.984047197106486 0.99225943762649 1.00555642128421 0.993725402926564 .... the data is saved in a file called f392.txt. I used the following codes to play around :)

Hi, When qqnorm on a vector of length 10M+ I get a huge pdf file which cannot be loaded by acroread or evince. Any suggestions? (apart from sampling the data). Thanks. -- Sam Steingold (http://sds.podval.org/) on Ubuntu 11.10 (oneiric) X 11.0.11004000 http://mideasttruth.com http://honestreporting.com http://camera.org http://openvotingconsortium.org http://pmw.org.il

Hi all, I am trying to test wheater the distribution of my samples is normal with QQ plot. I have a values of water content in clays in around few hundred samples. Is the code : qqnorm(w) #w being water content qqline(w) sufficient? How do I know when I get the plots which distribution is normal and which is not? Thanks, m [[alternative HTML version

R 1.7.1 on Windows XP The "normal Q-Q plot" produced by plot.lm() mislabels points when the model is fitted using na.action=na.exclude. Example: x <- 1:50 y <- x + rnorm(50) y[c(5,10,15)] <- NA # insert some NA's y[40] <- 50 # add an outlier plot(lm(y ~ x, na.action=na.omit)) # outlier correctly labeled in all # four plots

Hi all, I am a freshman of R,but I am interested in it! Those days,I am learning pages on NIST,with url http://www.itl.nist.gov/div898/handbook/eda/section3/probplot.htm, I am meeting a problem about probability plot and I don't know how to plot a data set with R. Could somebody tell me the answer,and a example is the best! I will look forward to your answer. Thank you very much.

Thanks a lot, Ista! I really appreciate it. How about a slightly different case as the following: set.seed(1) (tmp <- data.frame(x = 1:10, R1 = sample(LETTERS[1:5], 10, replace = TRUE), R2 = sample(LETTERS[2:6], 10, replace = TRUE))) x R1 R2 1 C B 2 B B 3 C E 4 E C 5 E B 6 D E 7 E E 8 D F 9 C D 10 A E Notice that the factor levels between

Suppose that we have the following dataframe: set.seed(1) (tmp <- data.frame(x = 1:10, R1 = sample(LETTERS[1:5], 10, replace = TRUE), R2 = sample(LETTERS[1:5], 10, replace = TRUE))) x R1 R2 1 1 B B 2 2 B A 3 3 C D 4 4 E B 5 5 B D 6 6 E C 7 7 E D 8 8 D E 9 9 D B 10 10 A D I want to do the following: if the difference between the level index of factor

I'm new to R (and to S) and am wondering about code from pages 72 and 83 of MASS (Venables+Ripley, 3rd edition), to draw an envelope on a QQ plot. Copying from the book, I've got: #... code whose gist is "a.fit <- nls(..." num.points <- length(resid(a.fit)) qqnorm(residuals(a.fit)) # illustrate data-model residuals qqline(residuals(a.fit)) samp <-

Hi, I am trying to figure out how to get the correlation coefficient for a QQ plot (residual plot). So to be more precise, I am creating the plot like this: qq.plot(rstudent(regrname), main = rformula, col=1) But want to also access (or compute) the correlation coefficient for that plot. Thanks, Tom [[alternative HTML version deleted]]

How about foo <- with(list(r1 = tmp$R1, r2 = tmp$R2, swapme = (as.numeric(tmp$R1) - as.numeric(tmp$R2)) %% 2 != 0), { tmp[swapme, "R1"] <- r2[swapme] tmp[swapme, "R2"] <- r1[swapme] tmp }) Best, Ista On Thu, Jul 6, 2017 at 4:06 PM, Gang Chen <gangchen6 at gmail.com> wrote: > Suppose that we have the following

Untested, but I expect that setting the levels to be the same across the two factors levels(tmp$R1) <- levels(tmp$R2) <- LETTERS[1:6] and proceeding as before should be fine. Best, Ista On Jul 6, 2017 6:54 PM, "Gang Chen" <gangchen6 at gmail.com> wrote: Thanks a lot, Ista! I really appreciate it. How about a slightly different case as the following: set.seed(1) (tmp

> From: Weiwei Shi > > it works. > thanks, > > but: (just curious) > why i tried previously and i got > > > is.vector(sample.size) > [1] TRUE Because a list is also a vector: > a <- c(list(1), list(2)) > a [[1]] [1] 1 [[2]] [1] 2 > is.vector(a) [1] TRUE > is.numeric(a) [1] FALSE Actually, the way I initialize a list of known length is by

Hello! I am trying with this question again: I would like to test few distributional assumptions for some behavioral response data. There are few theories about true distribution of those data, like: normal, lognormal, gamma, ex-Gaussian (exponential-Gaussian), Wald (inverse Gaussian) etc. The best way would be via qq-plot, to show to students differences. First two are trivial: qqnorm(dat$X)

Hello! I am using probplot in the e1071 package and want to do something like the following, only with the the 2nd plot overlaying the first. I can't seem to make it work. Any suggestions? *library(e1071) **x <- rnorm(100, mean=5)* *y <- rnorm(100, mean=3)* *probplot(x, line=FALSE) * *probplot(y, line=FALSE) * *Regards,* *Tom* [[alternative HTML version deleted]]

Hi, there. Is there any function in R to plot the probability-probability plot (PP plot)? Suppose I am testing some data against normal. Thanks. Yulei $$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$ Yulei He 1586 Murfin Ave. Apt 37 Ann Arbor, MI 48105-3135 yuleih at umich.edu 734-647-0305(H) 734-763-0421(O) 734-763-0427(O) 734-764-8263(fax) $$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$$

Estimados todos: Me es grato escribir a esta lista de ayuda para R, ya que comparto 100% la filosofía de Software libre en especial software en Estadística ya que es la carrera que estoy siguiendo. El motivo de este mensaje es por un par de dudas que no pude resolver: 1. He utilizado las funciones para realizar pruebas de normalidad (kolmogorov-smirnov, cuando n>50) y (Shapiro, cuando

Good Morning: I have attached a text file with one hundred thirty six observations. I would like to create a qq plot with the following features: 1. Observed values on the y-axis. 2. Normal approximation line on the plot. 3. X-axis with vertical reference lines at the following percentiles of the data: 1, 10, 20, 50, 80, 90 and 99. 4. Data appearing on the plot as distinct points. I assume that

I've looked around and I just can't find anything that will work for my needs. This is a bit of a 2 part question but pertaining to the same topic so bare with me. The first is with my qq plot. On the Y axis of my qq plot it'll have my sample quantities but because my data is log-normal it'll show numbers between 0 - 5 (depending on the data). I'd like to know how to get it,

Hi, Try either: set.seed(28) stats1<- as.data.frame(matrix(rnorm(5*10000),ncol=5)) pdf(paste("test",1,".pdf",sep="")) par(mfrow=c(2,1)) lst1<- lapply(names(stats1),function(i) {hist(stats1[,i],100,col="lightblue",main=paste0("Histogram of ",i),xlab=i );qqnorm(stats1[,i])}) dev.off() #or

Hi, I have a dataset with 2194651x135, in which all the numbers are 0,1,2, and is bar-delimited. I used the following approach which can handle 100,000 lines: t<-scan('fv', sep='|', nlines=100000) t1<-matrix(t, nrow=135, ncol=100000) t2<-t(t1) t3<-as.data.frame(t2) I changed my plan into using stratified sampling with replacement (col 2 is my class variable: 1 or 2).