similar to: Plot qualitative y axis

Hi Pedro, melt() is probably working. The problem is I did not finish the copy and paste.? It would have been better if I had included the ggplot() command. Try ============================================================== library(reshape2) library(ggplot2) dat1? <- structure(list(N = c("I", "II", "III", "IV", "V", "VI",

Many thanks, My goal is to make a plott like attached but the Y axis starts in XIV and end at top in I. Generally for instance in excel X axis is categories but Y axis is numbers I want the contrary plotted in lines, your last help is near what I look but barplot is not needed. Hope you can help me thanks in advance. 2018-05-22 0:58 GMT+02:00 Jim Lemon <drjimlemon at gmail.com>: > Hi

See ?barplot and set the horiz argument to TRUE. (This is in the base R plotting version. The ggplot2 and lattice systems have other ways of doing this) Note: if you search on e.g. "barplots in R" or similar, you should find numerous examples with code. Cheers, Bert Bert Gunter "The trouble with having an open mind is that people keep coming along and sticking things into

Hi all, I?m trying to plot this data N M W I 10 106 II 124 484 III 321 874 IV 777 1140 V 896 996 VI 1706 1250 VII 635 433 VIII 1437 654 IX 693 333 X 1343 624 XI 1221 611 XII 25 15 XIII 3 XIV 7 8 So that in de Y axis will be the level (qualitative data) and in the X axis will be M and W variables. So x axis will be wwith a lenght between 0 and 2000. I would like to plot a line with M and other

R-help, I have a data frame with one column containing roman numbers The data are not sorted as : I II III IV V VI VII VIII IX X XI XII XIII XIV XV Using data[order(data$Roman),] does not do the job. How can this be done? Thanks in advance.

Hello all, I've been invited to do a presentation on Asterisk for the Ohio Linuxfest in Columbus this weekend (http://www.ohiolinux.org). Rough estimates are that nearly 500 people will be attending. I've been working on an outline for a couple of weeks and I would like to have some peer review of the information presented. I am going to have to cut down the content to make it fit in

I'm a Gentoo amd64 user. This is what I get when I try to run the Autopano (it installed just fine): Code: szczerb at nomad ~ $ wine .wine/drive_c/Program\ Files/Kolor/Autopano\ Pro/Autopano_Win32.exe fixme:actctx:parse_depend_manifests Could not find dependent assembly L"Microsoft.VC80.CRT" err:module:import_dll Library MSVCP80.dll (which is needed by

Hi, Iam trying to setup Samba version 3.2.3 on Redhat (RHEL5) server to use Active Directory for authentication. I followed the instructions from article in following website: http://technet.microsoft.com/en-au/magazine/dd228986.aspx Setup Winbind + Samba + Kerberos and it seems to work fine. I can see the users in Active Directory through winbind as well as authenticate users using NTLM

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)) >

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

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.

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

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

Hi Ace, You can do it with plotrix: library(plotrix) barpos<-barp(c(1,5,38),width=0.5,col=c("white","lightgray","darkgray"),ylim=c(0,70)) ehplot(c(1,0.8,0.9,0.8,1.1,1,4,3,5,14,3,2,32,27,33,30,50,61), c(1,1,1,1,1,1,2,2,2,2,2,2,3,3,3,3,3,3),median=FALSE,add=TRUE,cex=2, pch=21,bg="white") dispersion(barpos$x,barpos$y,c(0.1,1,5),lwd=2,arrow.cap=0.03)

I explain better: I have this percentage z<-1/(n*365) where n=20 so z is 0,000136986 Imagine: a matrix with length(n*365) so that matrix result is resultc( 1, 1-z result previous row-z result previous row-z .... 0) If I plot this is a linear decreasing line I want to plot an upper line wich is 10% higher tan resultc on the first values and then decreasing so that last value will be 0.05.

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

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 "Bloom County" comic strip ) On Thu, Jun 1, 2017 at 11:20 AM, Pedro p?ramo <percentil101 at gmail.com> wrote: > Hi all > > I want to add a band of fluctuaci?n (exponential decreading) to a linear > deacrecing

Sorry, For me a vector is a matrix with mx1 dimmensions. But it is true that it is not the way I correctly must talk in R. Can you guide me in what I?m trying to do? I?m trying to find what I want in Excel using something like a parabolic function but dind?t get yet (then to try to replicate in R). 2017-06-02 7:01 GMT+02:00 Bert Gunter <bgunter.4567 at gmail.com>: > Bert Gunter >

I have experimented with the cheese data example from McCullagh&Nelder, page 175. With a proportional odds model they obtain a residual deviance of 20.31. Estimating the same model with polr(MASS) gives a residual deviance of 762.11 !, while using ordglm(gnlm) gives a deviance of 523.94. Can anybody explain these differences? The data frame with the data are: > cheese Cheese N

R-help, I have a data frame: > mydata strata nh Nh Wh fh 1 I 10 26 0.048 0.385 2 II 32 84 0.154 0.381 3 III 16 42 0.077 0.381 4 IV 4 11 0.020 0.364 5 V 10 26 0.048 0.385 7 VII 64 168 0.309 0.381 8 VIII 49 129 0.237 0.380 9 IX 22 58 0.107 0.379 91 VI 0 0 0.000 0.000 and I wish to rearrange the data are sorted according to the roman