similar to: Lowest values in a vector of numbers

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

> On Jun 15, 2017, at 12:37 PM, Andras Farkas via R-help <r-help at r-project.org> wrote: > > Dear All, > > we have: > > t<-seq(0,24,1) > a<-10*exp(-0.05*t) > b<-10*exp(-0.07*t) > c<-10*exp(-0.1*t) > d<-10*exp(-0.03*t) > z<-data.frame(a,b,c,d) > > res<-t(apply(z, 1, quantile, probs=c(0.3))) > > > > my

David, thanks for the response. In your response the quantile function (if I see correctly) runs on the columns versus I need to run it on the rows, which is an easy fix, but that is not exactly what I had in mind... essentially we can remove t() from my original code to make "res" look like this: res<-apply(z, 1, quantile, probs=c(0.3)) but after all maybe I did not explain

Hi Duncan I tried Ecdf and/or wtd.quantile from Hmisc and it is working (probably). Ecdf(x, q=.5) Ecdf(x, weights=xw,col=2, add=T, q=.5) wtd.quantile(x) 0% 25% 50% 75% 100% 10 10 10 100 300 wtd.quantile(x, weights=xw, type="i/n") 0% 25% 50% 75% 100% 10.0000 138.8667 192.5778 246.2889 300.0000 But could you please be more specific in this? >

It would depend on which one of the 9 quantile definitions you are using. The discontinuous ones aren't invertible, and the continuous ones won't be either, if there are ties in the data. This said, it should just be a matter of setting up the inverse of a piecewise linear function. To set ideas, try x <- rnorm(5) curve(quantile(x,p), xname="p") The breakpoints for the

Dear All, we have: t<-seq(0,24,1) a<-10*exp(-0.05*t) b<-10*exp(-0.07*t) c<-10*exp(-0.1*t) d<-10*exp(-0.03*t) z<-data.frame(a,b,c,d) res<-t(apply(z, 1, quantile, probs=c(0.3))) my goal is to do a 'reverse" of the function here that produces "res" on a data frame, ie: to get the answer 0.3 back for the percentile location when I have

Hi Petr, I think that Duncan suggests something like this: x<- c(rep(10,20), rep(300,5), rep(100, 10)) tx <- table(x) prop.x <- tx / sum(tx) vx <- as.integer(names(tx)) prop.wx <- tx * vx / sum(tx * vx) plot(ecdf(x)) plot(vx, cumsum(prop.x), ylim = 0:1) plot(vx, cumsum(prop.wx), ylim = 0:1) Best regards, Thierry ir. Thierry Onkelinx Statisticus / Statistician Vlaamse

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)?

I am trying to return the p value for a stat from the ECDF. That is the index of the first occurrence, on an ordered vector, of a value either greater than or equal to a given value. Ideally I would not have to order the vector beforehand. Currently I use: PValue<-function(stat, ECDF){ ###Get the length of the ECFD L<-length(ECDF) ###Loop through the ECDF until the p value is

Dear List, I have some discrete data and want to calculate the percentiles and the percentile ranks for each of the unique scores. I can calculate the percentiles with quantile(). I know that "ecdf" can be used to calculate the empirical cumulative distribution. However, I don't know how to exact the cumulative probabilities for each unique element. The requirement is similar

On 24/11/2017 6:27 AM, PIKAL Petr wrote: > Dear all > > Strictly speaking it is not R question but as you are the most capable persons I know I give it a try. > > I am strugling with recalculation of number weighted to volume weighted distribution. > > Suppose I have objects (cubes) with size > > x<- c(rep(10,20), rep(100, 10), rep(300,5)) > I can get >

Dear R users, I'm using Ecdf (Hmisc library) to plot four cdf in a same graphic. In this graphic I also plot the 0.99 quantile for these cdf. I successfully plot cdfs using different types of line to distinguish them, but I can't determine the type of lines showing 0.99 quantile. Is there a way to assign different line types for quantile lines in Ecdf plot? Best regards, -- Mateus da

Hello all, I have two columns of numbers. I would like to do the following: (1) Plot both cdfs, F1 and F2 on the same graph. (2) Find smoothed approximations of F1 and F2 lets call them F1hat and F2hat (3) Find values for F1hat when we substitue a value of x in it. (4) Find the corresponding densities of the cdfs. Any ideas? -- Thanks, Jim. [[alternative HTML version deleted]]

I have a problem where I need to calculate the corresponding cohort percentile ranks for each of several variables. Essentially, what I need is a function that will calculate the distribution-free percentiles from each variable's data vector, returning a corresponding vector of percentiles: e.g.: percentile.my.data<-/function/(my.data) I tried to make ecdf() perform this task but

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

Dear all Strictly speaking it is not R question but as you are the most capable persons I know I give it a try. I am strugling with recalculation of number weighted to volume weighted distribution. Suppose I have objects (cubes) with size x<- c(rep(10,20), rep(100, 10), rep(300,5)) I can get plot(ecdf(x)) or the number weighted average mean(x) [1] 77.14286 or volume weighted average

x = c(runif(1000, 0,2.5), runif(100, 2.5, 4)) plot(ecdf(x)) You will notice a sharp turn around x = 2.5 How do I get that value of x using R? -- Thanks, Jim. [[alternative HTML version deleted]]

Is there an easy way to "thin" a lattice plot? I often create plots from large data sets, and use the "pdf" command to save them to a file, but the resulting files can be huge, because every point in the underlying dataset is rendered in the plot, even though it isn't possible to see that much detail. For example: require(Hmisc) x <- rnorm(1e6)

I have a dataframe called "data" with 5 records (in rows) each of which has been scored on each of many variables (in columns). Five of the variables are named var1, var2, var3, var4, var5 using headers. The other variables are named using other conventions. I can create a new variable called var6 with the value 15 for each record with this code: > var6=var1+var2+var3+var4+var5

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