similar to: text display using expression or not

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

? 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>

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)

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,

? 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

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

Folks: I consider my reply below rather clumsy: One has to keep track of index numbers other than that which is inserted and must separately change column names. Is there as "essentially better" way to do this, either via base R or via an R package. I leave it to you to define "essentially better." Thanks. Cheers, Bert On Mon, Aug 1, 2011 at 10:17 AM, Bert Gunter

Hello to all, I still have this doubt. I'd like to colour the different labels of my dendrogram each one with a different colour. How can I do? I guess I could do using *edgetext* and then *t.col* or* lab.col* but I don't know how to add edgetext to my dendrogram. Can you help me please? Example: require(graphics); require(utils) hc <- hclust(dist(USArrests), "ave") (dend1

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,

Sarah's solutions are good, and here's another, even more basic: tmp1 <- unique(dat1$B) tmp2 <- seq_along(tmp1) dat1$C <- tmp2[ match( dat1$B, tmp1) ] > dat1 N B C 1 1 29_log 1 2 2 29_log 1 3 3 29_log 1 4 4 27_cat 2 5 5 27_cat 2 6 6 1_log 3 7 7 1_log 3 8 8 1_log 3 9 9 1_log 3 10 10 1_log 3 11 11 3_cat 4 12 12 3_cat 4 As a single line

Hi Scott, Where did you get this function originally? I can't find anything about it. What OS are you using? What says, "not available for the version"? Where are you getting that error? What are you trying to accomplish? What does that function actually do? It's impossible to suggest a work-around for a function of unknown purpose and origin. (The posting guide for this

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

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

Hello, I was messing around with graphics, and noted an odd behavior of par("din"). If the x11 device is empty, par("din") does not return the correct size if the device has been resized manually. dev.size() works correctly. R version 2.7.1; Fedora 8 # case 1 - empty device > x11() > dev.size() [1] 6.995263 6.994187 > par("din") [1] 6.995263 6.994187 #

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; >

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

This is pretty badly mangled (please don't post in html), but ?aggregate is probably what you want. Sarah On Sun, Jun 25, 2017 at 5:27 PM, Jackson Rodrigues <jacksonmrodrigues at gmail.com> wrote: > Hi everybody, > > My name is Jackson and come here ask for a help to organize data. > I really need help from you. I have several data sets to organize. > > I have to

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

Quoting from today's PhD Comics, available at: http://www.phdcomics.com/comics.php?f=1476 What the methodology section says: "Analysis was performed using a commercially available software package." What it really means: "I put the numbers into this magic box and out came my thesis!" -- Sarah Goslee http://www.functionaldiversity.org

Hi all, I'm working with summary.gam() and noticed that the options()$width argument is ignored by some components of that function, in particular the formula, which is printed at an arbitrary length regardless of the desired width. I've tracked the problem back to print.formula(), so at a lower level than summary.gam(). Is there a way around this that I'm missing? library(mgcv) #