similar to: two ecdf in the same figure

Hi list I loaded the content of a file dureetasks.txt to variable a. This file contains an interger per line. when I print a vector, it displays correctly. however, when I try to print the histogram, I get this error message > a=read.table("dureetasks.txt") > hist(a) Error in hist.default(a) : 'x' must be numeric Can you help please? regards -- PhD candidate in

Hello, How can I make a monte carlo simulation on R? Regards Adel -- PhD candidate in Computer Science Address 3 avenue lamine, cité ezzahra, Sousse 4000 Tunisia tel: +216 97 246 706 (+33640302046 jusqu'au 15/6) fax: +216 71 391 166 [[alternative HTML version deleted]]

Hello > orderulcount Group.1 Group.2 Group.3 x V5 7 C L 0.0 30 C / L 19 C L 0.2 27 C / L 31 C L 0.4 15 C / L 43 C L 0.6 7 C / L 54 C L 0.8 2 C / L 10 C S 0.0 27 C / S 22 C S 0.2 10 C / S 34 C S 0.4 6 C / S 46 C S 0.6 1 C / S 1 D

Hello, I have a question about how R can run automatically. Here is the story: A file called "data.csv" will be generated every couple of minutes in a folder (overwrite itself). What I want R to do is: 1) scan the folder to find the file. 2) if the file is a newly generated file, process the file 3) output some file identifier on the screen, such as the time when the file was

Hello, I have a file with very big values. I want to display the real values with classic ways (without exposant) > summary(a[,1]); Min. 1st Qu. Median Mean 3rd Qu. Max. 1.198e+09 1.199e+09 1.200e+09 1.200e+09 1.201e+09 1.202e+09 Can you help please? Regards Adel -- PhD candidate in Computer Science Address 3 avenue lamine, cité ezzahra, Sousse 4000 Tunisia tel:

dear list I use runif to generate a ramdom number between min and max runif(n, min=0, max=1) however , the syntaxe of rexp does not allow that rexp(n, rate = 1) and it generate a number with the corresponding rate. The question is: how to generate a number between min and max using rexp(). Regards -- PhD candidate in Computer Science Address 3 avenue lamine, cité ezzahra, Sousse 4000

Deal list > I want to generate a random number in an interval using exponential low > I know how to make that simply using uniform low > x1 <- runif(1, 5.0, 7.5) > this will generate a number between 5 et 7.5 but with uniform low. > Could you help pease.? > Regards > > -- > PhD candidate in Computer Science > Address > 3 avenue lamine, cité ezzahra, Sousse 4000

Hello, I have a problem with ggplot and colors I used this function to draw somes lines and I want them to be all black (just to test) however, I dont get any black line in the figure. Do you have any experience with ggplot and manual colors?? ggplot(cmax, aes(cmax[,3], cmax[,6],colour=interaction(cmax[,1],cmax[,2]))) + geom_line() + geom_point() +

Hello I am using this command to draw the figure attached to this mail. xyplot(g[,4]~g[,3],type="b",group=interaction(g[,1],g[,2]), auto.key =list( title="Evolution de la stabilit?" , points = FALSE, columns=2, lines = TRUE,space="right") ,data=g,xlab=expression(beta),ylab="stabilit?") I want to change name of the legend that were attributed

Hi list I am a new user of R. I ask for some "beginner" question I am searching if there is any function that computes the variation of some discrete values of a vector (mean() and sd() exists, but i have not find variation). Thanks in advance Adel -- PhD candidate in Computer Science Address 3 avenue lamine, cité ezzahra, Sousse 4000 Tunisia tel: +216 97 246 706 (+33640302046

Hi list I need some help for drawing some histograms I have a dataframe , say, X Y Z T I want to draw a histogram Z-T for each value of the couple (X-Y). When I use thus syntax > library(lattice) > histogram(law[,3] ~ law[,66] | law[,1] ) it draws multiple histograms but by selecting distinct values of law[,1] The deal is to make the same thing but for a couple of columns Thanks in

Hi list I am discovering R, and -- PhD candidate in Computer Science Address 3 avenue lamine, cité ezzahra, Sousse 4000 Tunisia tel: +216 97 246 706 (+33640302046 jusqu'au 15/6) fax: +216 71 391 166 [[alternative HTML version deleted]]

Hello list I have this file 0 121289479 25 0 212599129 1 0 285254098 21 0 444889848 45 0 469197123 30 0 640007403 82 0 718215617 8 0 758534043 56 0 799706577 4 0 814441385 93 0 843545059 37 0.2 121289479 6 0.2 285254098 3 0.2 444889848 6 0.2 469197123 13 0.2 640007403 24 0.2 718215617 3 0.2 758534043 2 0.2 799706577 2 0.2

Hello, I want to create 10 dataframe using a for loop. I don t know what to do to create 10 different dataframes whose name is parametrized with variable i. This syntaxe fails. It create a unique fataframe called dfn. Any input will help Thanks > for (i in 1:10){ + filename=paste("avail4",i,sep = "_") + dfn=read.table(filename) + } -- PhD candidate in Computer Science

Hi, Newbie here. I read the R for Beginners but i still don't get this. I have the following data (this is just an example) in a CSV file: courseid numstudents 101 209 141 13 246 140 263 8 321 10 361 10 364 28 365 25 366 23 367

Hi All, I have codes as follows to get the ecdf plots: > day.hos2<-c(6,4,6,6,4,6,5,4,7,5,6,6,8,6,17,9,8,4,6,3,5,8,7,12,5,10,6,4,6 ,13,7,6,6,25,4,9,96,6,6,6,6,9,4,5,5,4,10,5,7,6) > day.hos3<-c(5,6,7,6,4,5,6,6,6,6,19,7,5,9,8,8,7,5,6,20,40,5,8,7,7,5,6,13, 11,9,4,6,9,16,6,7,6) > f<-ecdf(day.hos2) > plot(f,col.p='red',col.h='red') >

Given a dataset x, the ecdf is ecdf(x). Then I can use ecdf(x)(y) to find the percentile of y. Given the ecdf is there a way to determine what is the value of y that is the boundary of let's say 95 percentile? In other words, is there a function I can call on the ecdf like: fomeFunc( ecdf( x ), 0.95 ) Which will return the highest value of y, for which ecdf( y ) < 0.95? The only solution

Dear all, I would like to calculate the complementary cumulative distribution function. As it is known, the ccdf is the 1-ecdf(X)==1-F(x). (From ?ecdf help is shown that ecdf returns a function which is the F(x) I would like to use that function inside my function and after call curve (which accepts function as input). My code looks like that onVector<-seq(1:100)

Dear UseRs, maybe is a silly question: how can I get Empirical CDF values from an object created with ecdf()?? Using print I obtain: Empirical CDF Call: ecdf(t) x[1:57] = 4.1, 4.4, 4.5, ..., 491.3, 671.27 Thanks in advance. Regards, Vito Diventare costruttori di soluzioni Became solutions' constructors "The business of the statistician is to catalyze the scientific

Hi. I have the following two ecdf plots in one graph: plot( ecdf(....), do.points=FALSE, verticals=TRUE, main=paste("Ecdf of distances ",DIM,sep=""), col="red" ); lines( ecdf(....), do.points=FALSE, verticals=TRUE ); How do I change the color of the resulting graph? Adding col="red" to either plot or lines results in an error