similar to: Empirical density estimation

But for my reporting purpose, I need to generate a bell curve like plot based on empirical PDF, that also contains original points. Any idea would be helpful. Thanks, On Mon, Mar 12, 2018 at 3:49 AM, Bert Gunter <bgunter.4567 at gmail.com> wrote: > You need to re-read ?density and perhaps think again -- or do some study -- > about how a (kernel) density estimate works. The points at

You need to re-read ?density and perhaps think again -- or do some study -- about how a (kernel) density estimate works. The points at which the estimate is calculated are *not* the values given, nor should they be! Cheers, Bert Bert Gunter "The trouble with having an open mind is that people keep coming along and sticking things into it." -- Opus (aka Berkeley Breathed in his

On 3/11/2018 3:35 PM, Christofer Bogaso wrote: > But for my reporting purpose, I need to generate a bell curve like > plot based on empirical PDF, that also contains original points. > > Any idea would be helpful. Thanks, > Christofer, something like the following may get you what you want: ## get the kernel density estimate dens <- density(Dat) ## estimate the density at

Dear all, I am attempting to export my results (data.frame) created with the help of a number of you to Excel. In the procedure my column structure is however lost and all results are placed together into the first Excel column. I have tried: write(), write.table(), write.matrix(), export() and have the same results. I Have checked the import/export FAQ and did a Google search to no avail. Any

Hello, I have a data frame that looks like Position, Relevance 0, 0.151531117072265 0, 0.245532564696541 0, 0.285207130323724 0, 0.302718099889669 0, 0.308213700400889 0, 0.341562706518953 0.0151515151515152, 0.368991215730364 0.0263157894736842, 0.256406702156839 0.0263157894736842, 0.344003157058329 0.0303030303030303, 0.200307950418176 0.0303030303030303, 0.558093143666938 0.04,

Hallo, I am totally confused why Excel and R give me totally different regression results for the data below. If you know the solution, please enlighten me. In Excel I used LINEST() and Data>Data Analysis>Regression and both (fortunately) gave the same result. The coefficients were: /b0=1.16, b1=0.957, b2=0.024, R2=0.0027, adjusted R2=-0.017/ In R I used

I have the data structure below and I'm attempting to send it into barchart using the R code below it. I don't get an error but I don't get any output either. Deepyan's new Lattice book is amazing and there are some examples sort of similar to what i'm doing but I couldn't see a way of using the formula interface to condition on what I wanted to condition on so I

This is my reproducible example tv.ms<-structure(list(inq = structure(4:17, .Label = c("D4", "D5", "D6a", "D6b", "D6c", "D7", "D8", "F4", "F5a", "F5b", "F6a", "F6b", "F6c", "F6d", "F7a", "F7b", "F8"), class =

Hello! I am getting the following errors when running optim() [I tried optim() with 3 different methods as you can see]: Error in optim(c(0.66, 0.999, 0.064), pe, NULL, method = "L-BFGS-B") : objective function in optim evaluates to length 6 not 1 > out <- optim( c(0.66, 0.999, 0.064), pe, NULL, method = "Nelder-Mead") Error in optim(c(0.66, 0.999, 0.064),

Hi, I have a matrix that I am displaying via image. I would like to obtain an image where the columns are 'grouped' (I have a grouping variable). But I am not sure how how I can draw the lines to indicate the grouping in the margin. I realize this is essentially waht heatmap does but as I have a set of custom breaks and colors I wanted to generate this via image. I can see that it is a

I am attempting to graph 12 months of temperatures, delineate the months with a vline and place the names of the months at the top of the graph. So far I have gotten everything to work except the names, despite getting a similar graph to work yesterday the day before yesterday with Baptise A's help. Can anyone suggest what I am doing wrong. Data set is below code. Thanks. Code

I had a problem annotating a graph last year ( see http://n4.nabble.com/Putting-names-on-a-ggplot-td907158.html#a907158 for the discussion) Stefan (smu) provided a solution using annotate(). However I apparently did not update the graph file and,now, when I go back to the thread and try to use Stefan's solution it does not seem to work although I am sure that it did then. The problem

Dear R users, I have a follow-up question on sqlSave(). Since most of the output from the tests I use are lists, I would like to loop to export each element of the list and append it to the sheet. Here is what I do: > library(RODBC) > test <- structure(list(m = structure(c(0.090909090909091, 0.181818181818182, 0.272727272727273, 0.363636363636364, 0.454545454545455,

I''m using the ruby gem "Color" and I''m having some troubles with to_hex. I''ve installed the gem, and when I try and run the test program I get this error: >> red = Color.new(''ff0000'') => #<Color:0x3685990 @saturation=1.0, @hue=0.0, @lightness=0.5> >> yellow = Color.new(''ffff00'') =>

Putting names on a ggplot Can anyone suggest what I am doing wrong here. I am plotting daily temperatures at Ottawa Ontario for 2008 broken down by months, I seperate them by lines and want to put the names of the months at the top of the chart ( with in the graphing area) Everything is working as I want until I try to add the names of the months. I did something similar a few days ago and I