similar to: package docs: examples format

Hi, You're mixing base plot and ggplot2 grid graphics, which as you've discovered doesn't work. Here's av strategy that does: https://cran.r-project.org/web/packages/egg/vignettes/Ecosystem.html This vignette has a good overview, well as info specific to that package. Sarah On Wed, May 30, 2018 at 4:43 AM greg holly <mak.hholly at gmail.com> wrote: > Hi all; >

I thought some of you might be amused by this. In my non-work time, I'm an avid weaver and teacher of weaving. I'm working on a project involving creating many detailed weaving patterns, so I wrote R code to automate it. Details here: http://stringpage.com/blog/?p=822 If the overlap between R users and avid tablet weavers turns out to be >> 1, I'll polish it up and turn it

I need the output to have groups and the probability any given record in that group then has of being in the response class. Just like my email in the beginning i need the output that looks like if A and if B and if C then %77 it will be D. The examples you provided are just simply not similar. They are different and would take interpretation to get what i need. On Apr 14, 2016 1:26 AM,

Since you only have 3 predictors, each categorical with a small number of categories, you can use expand.grid to make a data.frame containing all possible combinations and give that the predict method for your model to get all possible predictions. Something like the following untested code. newdata <- expand.grid( Humidity = levels(Humidity), #(High, Medium,Low)

That does clarify for me that you're missing a step: I didn't clearly follow your description at first. corrplot expects a correlation matrix, not your original data. You need to use cor() first. That's pretty clear in the documentation. See for instance the examples: data(mtcars) M <- cor(mtcars) corrplot(M) Sarah On Sat, Mar 17, 2018 at 12:00 PM Shivi Bhatia <shivipmp82 at

I still need the output to match my requiremnt in my original post. With decision rules "clusters" and probability attached to them. The examples are sort of similar. You just provided links to general info about trees. Sent from my Verizon, Samsung Galaxy smartphone<div> </div><div> </div><!-- originalMessage --><div>-------- Original message

? data=read.table("http://statcourse.com/research/boston.csv", , sep=",", header = TRUE) ? library(rpart) ? fit=rpart (MV~ CRIM+ZN+INDUS+CHAS+NOX+RM+AGE+DIS+RAD+TAX+ PT+B+LSTAT) Please: Show me the tree. Mark -------- Original Message -------- Subject: Re: [R] help with rpart From: "Stephen Milborrow" <[1]milbo at sonic.net>

Hi Sarah, Thank you for your help. I tried using CR1<-as.matrix(CR1) but gives error Error in corrplot(CR1, method = "circle") : The matrix is not in [-1, 1]!. I am using a corrplot library. Please find the reproducible example: dput(head(CR1,10)) structure(c(26L, 46L, 39L, 38L, 47L, 59L, 56L, 61L, 43L, 60L, 78L, 63L, 2L, 58L, 8L, 1L, 1L, 9L, 11L, 2L, 1037500L, 46747L, 346300L,

? data=read.table("http://statcourse.com/research/boston.csv", , sep=",", header = TRUE) ? library(rpart) ? fit=rpart (MV~ CRIM+ZN+INDUS+CHAS+NOX+RM+AGE+DIS+RAD+TAX+ PT+B+LSTAT) predict(fit,data[4,]) plot only reveals part of the tree in contrast to the results on obtains with CART or C5 -------- Original Message -------- Subject: Re: [R] help with rpart From: Sarah

Hi all; I need to put 12 different plot2 into the same matrix. So my array for the matrix will be par(mfrow=c(3,4)). I am running ggplot2 to produce my 12 plots. For some reason, par(mfrow=c(3,4)) did not turn out 3*4 matrix. my basic R codes for each plot is par(mfrow=c(3,4)) library(ggplot2) p <- ggplot(a, aes(x=Genotypes, y=Plant_hight, size=Plant_hight, color=Showing_rate)) + . . Best

Tjats great that you are familiar and thanks for responding. Have you ever done what I am referring to? I have alteady spent time going through links and tutorials about decision trees and random forrests and have even used them both before. Mike On Apr 13, 2016 5:32 PM, "Sarah Goslee" <sarah.goslee at gmail.com> wrote: It sounds like you want classification or regression trees.

Hi again; I have placed many examples in the CTFS Rd pages because my audience really really likes lots of examples in order to understand the options available in a functions. At this time I have set them all to \dontrun{}. however, this isn't because the example command line with its function and options won't work, but rather that CTFS specific datasets must exist to have the

Hi, Can R read in .xyz files? If so, what is the package, thanks -- Shane [[alternative HTML version deleted]]

Here is my problem, 100 decision trees were built(similar to random forest) and I want to replace some of them by new trees. How can I define a tree array including 100 trees, i.e. t[100], and every t[n] is an "C5.0" object, such that when a new tree comes, i can do n<-10 t[n]<-C5.0(...) -- View this message in context:

hi R-users: I want to draw three plot into one figure by layout and the script has been shown below. But I find R does not show the xlab and ylab information completely as shown the figure attached. How can I midify the script.? thank you . xxlab<-paste(cpmd," (",ro,"%)",sep=" ") yylab<-paste(rfmd," (",co,"%)",sep=" ")

Sorry for the confusion, here is the edited question. The data= Stand_Height (attached) is recorded from 12/1/2009 to 12/31/2015 (25 observations) and the other dataset (leafbiom) is recorded from 10/7/2009 to 12/29/2016 (daily observations). I want to use the 25 observations of stand height to predict the daily stand height from 10/7/2009 to 12/29/2016. The daily stand height will be multiplied

Hi, I would like to access a local mysql database. In python using the sqlite3 library it is quite straightforward. I just have to pass the path to the sqlite-file; sqlite3.connect('.../xy.sqlite') Is there something similar in R (or specifically in the RMySQL package)? It seems like I have to use the MySQL function in the RMySQL package in order to ... "initializes a MySQL

Once again, you are over-writing your variable. This time, you are overwriting the entirety of Stand_Height with the timeseries of height. Perhaps you should spend some time with one of the good introductory R resources out there, and think a bit more about your procedure. Sarah On Fri, Jun 30, 2017 at 11:23 AM, Ahmed Attia <ahmedatia80 at gmail.com> wrote: > Sorry for the confusion,

Hi Sarah, Thank you so much!! I got your good ideas. Ding -----Original Message----- From: Sarah Goslee [mailto:sarah.goslee at gmail.com] Sent: Friday, May 11, 2018 11:40 AM To: Ding, Yuan Chun Cc: r-help mailing list Subject: Re: [R] add one variable to a data frame [Attention: This email came from an external source. Do not open attachments or click on links from unknown senders or

Sarah et. al.: As a matter of aesthetics (i.e. my personal ocd-ness) I prefer using the public API of an object, i.e. *not* to makes use of the representation of a factor as essentially an integer vector with labels, but rather to use its documented behavior. (Feel free to ignore this remark!) Anyway, >cumsum(!duplicated(dat1$B)) [1] 1 1 1 2 2 3 3 3 3 3 4 4 will do it. This is very