similar to: Supressing printing from a function: ecdf

Dear R Users, Does R have an "inverse empirical cumulative distribution" function, something one can use to invert ecdf ? Thanks in advance, Tolga Generally, this communication is for informational purposes only and it is not intended as an offer or solicitation for the purchase or sale of any financial instrument or as an official confirmation of any transaction. In the event you

Dear R Users, Are there any packages in R which carries out a normalisation to variables as follows: - find the empirical distribution function, using perhaps ecdf - use the empirical distribution function to transform the variables into a series between 0 and 1 - use this series to map the variables into the normal distribution function, using qnorm - perform a regression on the transformed

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'm bit confused about ecdf (read the help files but still not sure about this). I have an analytical expression for the pdf, but want to get the empirical cdf. How do I use this analytical expression with ecdf? If this helps make it concrete, the pdf is: f(u) = \sum_{t = 1}^T 1/n_t \sum_{i = 1}^{n_t} 1/w K((u - u_{it})/w) where K = kernel density estimator, w = weights, and u_{it} =

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

Dear Group, I am using the ecdf function as follows: cawa.cdp <- ecdf(cawaocc$LEFF80) summary(cawa.cdp) Empirical CDF: 223 unique values with summary Min. 1st Qu. Median Mean 3rd Qu. Max. 0.07918 1.35700 1.68600 1.61000 1.91200 2.70000 I can see by the summary that the y-value for the 3rd quartile is 1.912. How can I obtain the x-value for a specified y-value (e.g., 0.8)?

Dear R-Users, I've written a number of functions in a .R/script file. I would like to call those functions from another script file. How can I execute all the code in a script file so that the functions are available for use in my other script file ? I have tried help.search("script") and various Google alternatives but couldn't come up with anything. Thanks in advance,

Dear R Users, I am trying to get plot.zoo to place monthy tickmarks/labels for a time series which spans daily data going back a bit over a year. Right now, I am getting only one tick mark on the x-axis for the beginning of 2008. How can I force plot.zoo to place more regular x-axis tick marks on a monthly basis ? Thanks in advance, Tolga Generally, this communication is for informational

Dear R Users, A bit of an elementary question, but somehow, I haven't been able to figure it out. I'd like to changes the column names of a data frame, so I am looking for something like col.names (as in row.names). Could someone please show me how to change the column names of a data frame ? Thanks, Tolga Generally, this communication is for informational purposes only and it is

Dear R-Users, I am looking for a way to get legends placed automagically in an empty spot on a graph. Additional complication comes through my useage of multiple graphs on the same plot through mfrow. Is there a way to achieve this in R ? I have legends for each of the sub-plots. Many thanks in advance, Tolga Generally, this communication is for informational purposes only and it is not

Dear R Users, I would like to get the pdf() command to create graphics in landscape mode, as I am plotting 8 graphs in 2 rows and 4 columns. This is being generated for an a4 printer (so I also use the paper="a4" command). Is there a way I can ask pdf() to generate the document in landscape mode ? Thanks, Tolga Generally, this communication is for informational purposes only and it

Dear R Users, Can anyone point me to a package for R vrsion 2.7.1 which implements some Hurst exponent estimation methods ? Thanks in advance, Tolga Generally, this communication is for informational purposes only and it is not intended as an offer or solicitation for the purchase or sale of any financial instrument or as an official confirmation of any transaction. In the event you are

Dear R Users, I am trying to do something quite simple: replace the elements of a zoo object. For some reason, the following code does not seem to work. How can I replace the value for the 14th of Dec of 2008 in the zoo object x below with 1 (it is currently NA). > x 2008-12-11 2008-12-12 2008-12-13 2008-12-14 2008-12-15 2008-12-16 361.667 389.875 NA NA 397.822

Dear R-Users, I have the CDF of a discrete probability distribution. I now observe a change in this CDF at one point. I would like to find a new CDF such that it has the shortest Kullback-Leibler Distance to the original CDF and respects my new observation. Is there an existing package in R which will let me do this ? Google searches based on entropy revealed nothing. Kind regards, Tolga

Dear R Users, A bit of a simple question, but I could not find the answer thru google ("sorting lists r cran","how can i sort a list r cran") or RSiteSearch... how can I sort a list ? One imagines something like sort.list would do the job, but that's not the one. > foo<-list(a=10,b=2,c=3) > sort.list(foo) Error in sort.list(foo) : 'x' must be atomic

Dear R Users, The granger.test command in the MSBVAR package estimates all possible bivariate Granger causality tests for m variables. If one passes a data frame with 3 rows, it returns 6 granger tests in two rows, one for the F-statistic and another for the p-value. For example: > a<-rnorm(1:10) > b<-c(lag(a),rnorm(1)) > c<-c(lag(b),rnorm(1)) >

Dear R Users, I am on Windows XP SP2 platform, using R version 2.7.2 . I was wondering if there is a way to find out, within R, the number of CPU's on my machine ? I would use this information to set the number of nodes in a cluster, depending on the machine. Sys.info() and .Platform do not carry this information. Thanks in advance, Tolga Uzuner Generally, this communication is for

Dear R Users, I would like to plot three time series on the same graph, two axis on the left and one axis on the right. The time series that I am graphing on the left do not share a similar scale: one has a range of 1:100 and the other a range of 25000:70000. How can I display the tick marks for both on the left hand side without superimposing them, i.e. one set of tick marks in black right

Dear R Users, Is anyone aware of a package which calculates the Yule Kendall resistant (to errors,outliers) measure of skewness ? An easy calculation to perform, but was just wondering if a package exists (as the contents of that package would probably include other cool things I would also be interested). Thanks, Tolga Generally, this communication is for informational purposes only and it

Dear Achim, R Users, What am I doing wrong in this example ? a<-zoo(rnorm(100),order.by=1:100) b<-lag(a) regr<-na.exclude(merge(a,b)) plot(regr) grangertest(regr[,1],regr[,2],3) > a<-zoo(rnorm(100),order.by=1:100) > b<-lag(a) > regr<-na.exclude(merge(a,b)) > plot(regr) > grangertest(regr[,1],regr[,2],3) Error in solve(vc[ovar, ovar]) : subscript out of bounds