similar to: Plot MArginal distribution in the correct place

Hi Pedro, As a one-off, you just shove the coordinates around a bit: par(mar=c(11,0,6,6)) barplot(fhist$counts,axes=FALSE, space=0,horiz=TRUE,col="lightgray", ylim=c(0,24)) However, I don't think that this plot illustrates quite what you think it does. Jim On Wed, Jun 7, 2017 at 4:01 PM, Pedro p?ramo <percentil101 at gmail.com> wrote: > Please, I'm trying to put

Many thanks Jim. What I,m trying to show with the fhist plot is the empirical distribution of the values of the left plot simulation. You say: However, I don't think that this plot illustrates quite what you think it does. Can you give me a clue to try to illustrate better if it is not showing what I believe it shows a better way to show it? Many thanks in advance. El 7 jun. 2017

Please, I'm trying to put the right plot higher and centered on the left values but I don't achive. I would appreciate so much your help El 6 jun. 2017 22:37, "Pedro p?ramo" <percentil101 at gmail.com> escribi?: > Hi all, > > I have this code, but the marginal distribution plot doesn?t appear > aligned with the left plot. > > > I think could be

Please can you send me some orientation? Many thanks in advance. Only if posible one book o similar example to understand why it is not what I try. El 8 jun. 2017 7:50 PM, "Pedro p?ramo" <percentil101 at gmail.com> escribi?: > Many thanks Jim. > > What I,m trying to show with the fhist plot is the empirical distribution > of the values of the left plot simulation.

Hi Pedro, If you keep that same margins for the second plot: par(mar=c(10,0,6,6)) barplot(fhist$counts,axes=FALSE, space=0,horiz=TRUE,col="lightgray") it looks reasonably well aligned to me. Because you are plotting the counts of the values in Simulation, the ordinate (vertical axis) of the bar plot is in quite different units from that of the plot on the left side. Jim On Wed, Jun

Can you suggest a plot on the right side so that the right plot has the same units of left plot and reflect the counts of the number of time the values repeat, something like the empirical distribution? 2017-06-14 12:30 GMT+02:00 Jim Lemon <drjimlemon at gmail.com>: > Hi Pedro, > If you keep that same margins for the second plot: > > par(mar=c(10,0,6,6)) >

I'd like to substitute the Subversion revision number for a string in a package man page every time the package is installed or built. I can assume it is being installed or built from a Subversion working copy. I tried putting a target that depends on FORCE into Makefile or Makevars in the src directory, but it is not being built. What sort of make magic do I need, and where do I put it

Please can someone explain why I seem to get these contradictory results? choose(5,2) [1] 10 gamma(6)/(gamma(3)*gamma(4)) [1] 10 gamma(6)/(gamma(3)*gamma(4)) == choose(5,2) [1] TRUE # all's well so far. # now look what happens: gamma(21)/(gamma(6)*gamma(16)) == choose(20,5) [1] FALSE # check individual terms: gamma(21)/(gamma(6)*gamma(16)) [1] 15504 choose(20,5) [1] 15504 # so they are the

Hello, I'm trying to build a frequency table for a vector, broken down by the combination of factors. For further analyses, I need to have the result arranged in a new data frame, with the upper limit of the histogram's breaks, the per bin count, and the factors identifying each record. My problem is including the latter: ---<---------------cut

Hello All! SPEEX introduces an additional delay to the audio data, I found out by reverse enginiering (it is NOT the lookahead time): narrow band : delay = 200 - framesize + lookahead = 200 - 160 + 40 = 80 samples wide band : delay = 400 - framesize + lookahead = 400 - 320 + 143 = 223 samples uwide band : delay = 800 - framesize + lookahead = 800 - 640 + 349 = 509 samples To get the

I believe if you add the *decoder* lookahead time to the encoder lookahead time you are already referencing, you will get the numbers that you have calculated. --John On 4/11/2012 1:00 PM, speex-dev-request at xiph.org wrote: > Message: 1 > Date: 10 Apr 2012 22:07:20 +0200 > From: Thilo K?hler<koehlerthilo at gmx.de> > Subject: [Speex-dev] Speex Codec Delay Problem > To:

Dear R helpers I have following data stocks <- c("ABC", "DEF", "GHI", "JKL") prices_df <- data.frame(ABC = c(17,24,15,22,16,22,17,22,15,19),                                          DEF = c(22,28,20,20,28,26,29,18,24,21),                                           GHI = c(32,27,32,36,37,37,34,23,25,32),                                          

I have a dataset in "long" format that looks something like this: ID TEST RESULT 1 A 17 1 A 12 1 B 15 1 C 12 2 B 8 2 B 9 Now what I would like to do is transpose it like so: ID TEST A TEST B TEST C 1 17 15 12 1 12 . . 2

Sorry about the odd terminology, but I suspect that my intent might be completely missed had I used "aggregate" or "classify" (each of which appears to have some rather special meanings in statistical analysis and modeling). I have some data about software builds; one of the characteristics of each is the name of the branch. A colleague has generated some fairly interesting

Hi all I'm have a character vector and would like to suppress the blanks if there are more than one after the other. Example: Character value is: "abc def ghi" The result should be: "abc def ghi" I know that it's possible to delete the leading blanks with the command "trim". But how can I delete blanks within a character? Thank you very much in

Hi all I want to add a band of fluctuaci?n (exponential decreading) to a linear deacrecing values Imagine: I have a matrix like c(10,9,8,7,6,5,4,3,2,1) The thing is I want two new lines so that the m?ximum value of the new colum on the m?ximum is from 10% to 5% higher and the same lower for the m?nimum the final two matix will be something like c(10+0.10*10,9+0.089*9,8+0.075*8.....,1+0.05*1)

I have a set of labels arranged along a timeframe in a. Each label has a timestamp and marks a state until the next label. The dataframe a contains 5 such timestamps and 5 associated labels. This means, on a continious scale between 1-100, there are 5 markers. E.g. 'abc' marks the timestampls between 10 and 19, 'def' marks the timestamps between 20 and 32, and so on. a <-

Hello - What I want to do seems very simple - I want to make sure any H.323 traffic gets processed before anything else entering or leaving this network. The network has a videoconferencing device on the LAN at 192.168.16.4. A Linux firewall NATs an external IP Address to this internal address and I have appropriate SNAT and DNAT rules that work. The NAT and connection tracking rules all work

Hi, my data frame is x<-data.frame(ID=c("abc/def","abc/def/ghi","abc","mno/pqr/st/ab")) I want to split my column ID using "/" as the place to split. How can I do that without telling the code how many sub-columns. I could use nchar(gsub("[^/]","",x$ID)) to get how many "/" are in each row of the column, but could

I''m posting this question to both mailing lists as I''m not sure whether it''s a Mechanize problem or a Nokogiri problem. Using Nokogiri and Mechanize to load and parse a web page encoded with Shift-JIS. I have an HTML construct like: <head> <meta http-equiv="content-type" content="text/html; charset=Shift_JIS"> </head> <body>