similar to: Color of ecdf plots

Displaying 20 results from an estimated 8000 matches similar to: "Color of ecdf plots"

2001 Oct 22
3
Two questions
Greetings, I have two questions that I could not answer from the documentation. A - ecdf and confidence intervals : Is there a (simple) way to generate confidence intervals (95%) for a ecdf? B - cross-validation of rpart trees : a colleague is using S to generate decision tree and mentioned to me the use of cross-validation. Is this function enabled in R ? if so, how should one proceed to
2003 Jul 21
3
Confidence Band for empirical distribution function
Hi, I was trying to draw an empirical distribution function with uniform confidence bands. So I tried to find a way to calculate values of the Kolmogorov-Smirnov Distribution but failed. I guess it must be hidden somewhere (since the ks-test is implemented), but I was unable to find it. Is there any way to do this? Thanks Leif Boysen
2008 Nov 08
2
lines, ecdf and colors
Hi. I'm trying to plot two ecdf's on the same graph using two different colors. I can plot using the same color, but it doesn't work when I change colors? Any suggestions? Thanks in advance for your help. x <- c(0.80, 0.83, 1.89, 1.04, 1.45, 1.38, 1.91, 1.64, 0.73, 1.46) y <- c(1.15, 0.88, 0.90, 0.74, 1.21) plot(ecdf(x)) # it works without col='blue', but doesn't
2009 Sep 07
1
Plot 2 ecdf in one graph
Hi r-users,   I would like to compare the cdf between historical and predicted. My x.obs and x.pre are the frequency data in classes of 0-300. I tried: plot(ecdf(x.obs),ecdf(x.pre),type="l",col="red")   and it gives me: Error in plot.stepfun(x, ..., ylab = ylab, verticals = verticals, pch = pch) :   argument 4 matches multiple formal arguments   Thank you so much for any
2004 Jun 03
1
ecdf plots, lines, and y values
Hi, I have a question for the group, perhaps someone can help me figure this out. I've already looked in the help files and they were no help to me. I have a vector of values and I am plotting an ecdf graph. 1. How can i draw a continuous line through the ecdf points? (lines and type for the plot with an ecdf object does not work) 2. Supposing I have this line drawn. I can add a
2005 Nov 22
1
Kolmogorov-Smirnov test help
Hi I am conducting 2-sample Kolmogorov Smirnov tests for my Masters project to determine if two independant tree populations have the same size-class distribution or not. The trees have been placed into size-class categories based on their basal diameters. Once I started running the stats on my data, I got confused with the results. Just to show an example of what I was testing I ran stats
2006 Aug 25
2
plot question
Hi everyone, I have what may appear to be a newbie question, but I have looked everywhere I can think to look and I cannot find an answer. On page 35 of "An Introduction to R" the following command appears: plot(ecdf(eruptions), do.points=FALSE, verticals=TRUE). What is the do.points argument? I know what it does (suppresses printing of the points) but where can I find help on it?
2012 May 26
1
Kolmogorov-Smirnov test and the plot of max distance between two ecdf curves
Hi all, given this example #start a<-c(0,70,50,100,70,650,1300,6900,1780,4930,1120,700,190,940, 760,100,300,36270,5610,249680,1760,4040,164890,17230,75140,1870,22380,5890,2430) length(a) b<-c(0,0,10,30,50,440,1000,140,70,90,60,60,20,90,180,30,90, 3220,490,20790,290,740,5350,940,3910,0,640,850,260) length(b) out<-ks.test(log10(a+1),log10(b+1)) # max distance D
2004 Oct 17
3
ecdf with lots of ties is inefficient (PR#7292)
Full_Name: Martin Frith Version: R-2.0.0 OS: linux-gnu Submission from: (NULL) (134.160.83.73) I have large vectors containing 100,000 to 20,000,000 numbers. However, they only contain a few hundred *distinct* numbers (e.g. positive integers < 200). When I do ecdf(v), it either runs out of memory, or it succeeds, but when I plot the ecdf with postscript, the output is unnecessarily bloated
2007 Jul 10
3
ECDF, distribution of Pareto, distribution of Normal
Hello all, I would like to plot the emperical CDF, normal CDF and pareto CDF in the same graph and I amusing the following codes. "z" is a vector and I just need the part when z between 1.6 and 3. plot(ecdf(z), do.points=FALSE, verticals=TRUE, xlim=c(1.6,3),ylim=c(1-sum(z>1.6)/length(z), 1)) x <- seq(1.6, 3, 0.1) lines(x,pgpd(x, 1.544,0.4373,-0.2398), col="red") y
2006 Jun 04
2
slanted ends of horizontal lines for certain line widths
Hello, if I plot a horizontal line, e.g., plot(c(1,2),c(1,1),xlim=c(0,3),lwd=2,type="l") or plot(c(1,2),c(1,1),xlim=c(0,3),lwd=4,type="l") then the left end (1st example) or both ends (2nd example) of the lines are not rectangular but slanted on the graphical display (screen). That behavour first occurred when I was trying to plot a stepfun, e.g., y <-
2008 Aug 14
1
ecdf manipulation
I’m using the edcf function to look at a number of empirical distributions graphically for run-time analyses of stochastic optimization algorithms. When dealing with problems where the optimal solution for these problems is always found everything is fine and the graphs are very useful for comparative observations. These distributions have a vertical axis height of one i.e. a probability of one.
2012 May 04
1
Absolute cumulative curve with ecdf/stepfun?
Hi, I have two variables ranging both from 0 to 1 (n=500 each). Now I am interested in plotting them both in one plot (using ggplot2). So far I used ecdf() (from an example I found with google) to get values for the cumulatice distribution function which gives a relative curve. I also want to do the same plot but using absolute cumulative values instead of relative. Can that be done with ecdf or
2007 Jul 10
1
Fraction ECDF
Hi all, I would like to plot part of the emperical CDF. Suppose the variable is x, I just need the part when x>1,therefore, I am using the following codes. tail <- x>1 plot(ecdf(x[tail]), do.points=FALSE, verticals=TRUE) The "x" value starts from 1, but the yaxs still begins from 0, not the corresponding value when "x" is 1. How can I make it match? Could anyone
2005 Nov 02
2
help with the coordinates of the ECDF object
Hi all R users I would like to know how acess the coordinates of the ECDF object. I look for the example, in this part: ###################### print(ls.Fn12 <- ls(env= environment(Fn12))) ###################### but I do not know to extract the Y coordinate and put it in other variable. My objective is to make a plot and identify the points with labels. ############# Example by
2011 Sep 24
2
two ecdf in the same figure
Hello, is ot possible to draw two ecdf of vectors (say s1 and s2) on the same figire with R. plot function draws a new plot and there is no function like points or lines to draw a second ecdf on the figure. Regards -- *PhD candidate in Computer Science Address 3 avenue lamine, cité ezzahra, Sousse 4000 Tunisia tel: +216 97 246 706 fax: +216 71 391 166* [[alternative HTML version deleted]]
2011 Oct 16
2
ecdf
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
2007 May 30
1
Sort in ecdf
Hi! I've noticed the ecdf() R code (R ver. 2.5.0) contains two call to sort: --- [R-code] --- ecdf <- function(x) x <- sort(x) n <- length(x) if (n < 1) stop("'x' must have 1 or more non-missing values") vals <- sort(unique(x)) rval <- approxfun(vals, cumsum(tabulate(match(x, vals)))/n, method
2008 Feb 19
1
How to count from larger value to smaller value in ecdf (Empirical Cumulative Distribution Function)
Hi, all ecdf function (Empirical Cumulative Distribution Function) in "stats" package counts from smaller values to larger values. However, I want to draw it by counting from larger value to smaller values and I couldn't find options for this purpose. How can I draw ecdf or ecdf like graph by counting from larger values to smaller values. Thank you in advance. Hyunchul Kim
2005 Nov 17
3
ECDF values
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