similar to: a problem in random forest

Hi, there: I am trying nnet as followed: > mg.nnet<-nnet(x=trn3[,r.v[1:100]], y=trn3[,209], size=5, decay = 5e-4, maxit = 200) # weights: 511 initial value 13822.108453 iter 10 value 7408.169201 iter 20 value 7362.201934 iter 30 value 7361.669408 iter 40 value 7361.294379 iter 50 value 7361.045190 final value 7361.038121 converged Error in y - tmp : non-numeric argument to binary operator

I've built two RF objects (RF1 and RF2) and have tried to combine them, but I get the following error: Error in rf$votes + ifelse(is.na(rflist[[i]]$votes), 0, rflist[[i]]$votes) : non-conformable arrays In addition: Warning message: In rf$oob.times + rflist[[i]]$oob.times : longer object length is not a multiple of shorter object length Both RF models use the same variables, although

Dear R helpers Suppose val1 = c(10, 20, 35, 80, 12) val2 = c(3, 8, 11, 7) I want to select either val1 or val2 depending on value of third quantity val3. val3 assumes either of the values "Monthly" or "Yearly". If val3 = "Monthly", then val = val1 and if val3 = "Yearly", then val = val2. I tried the ifelse statement as ifelse(val3 =

Hi, I am using dlda algorithm from supclust package and I am wondering if the output can be a continuous probability instead of discrete class label (zero or one) since it puts some restriction on convariance matrix, compared with lda, while the latter can. thanks, -- Weiwei Shi, Ph.D Research Scientist GeneGO, Inc. "Did you always know?" "No, I did not. But I believed..."

What is the best way to produce a barplot from my data? I would like the barplot to show each person with the values stacked val1+val2+val3, so there is one bar for each person When I use barplot(data.matrix(realdata)), it shows one bar for each value instead. To post here, I created an artificical data set, but it works fine. fakedata <- as.data.frame(list(LETTERS[1:3])) colnames(fakedata)

Hello list, I am wondering if a joining "one-to-many" can be done a little bit easier. I tried merge function but I was not able to do it, so I end up using for and if. Suppose you have a table with locations, each location repeated several times, and some attributes at that location. The second table has the same locations, but only once with a different set of attributes. I would

Dear all, I want to calculate mean values for multiple rows: structure(list(Name = structure(c(1L, 1L, 1L, 1L, 1L, 2L, 2L, 2L, 2L), .Label = c("AKT", "CKT"), class = "factor"), val1 = c(2, 3, 2, 2, 2, 5, 3, 8, 2), val2. = c(4, 5, 4, 8, 4, 8, 4, 7, 4), val3 = c(5, 6, 5, 9, 5, 9, 5, 9, 5)), .Names = c("Name", "val1", "val2.",

Hello R Gurus, I am perplexed by the different results I obtained when I ran code like this: set.seed(100) test1<-randomForest(BinaryY~., data=Xvars, trees=51, mtry=5, seed=200) predict(test1, newdata=cbind(NewBinaryY, NewXs), type="response") and this code: set.seed(100) test2<-randomForest(BinaryY~., data=Xvars, trees=51, mtry=5, seed=200, xtest=NewXs, ytest=NewBinarY) The

You need to give us more details, like how you call randomForest, versions of the package and R itself, etc. Also, see if this helps you: http://finzi.psych.upenn.edu/R/Rhelp02a/archive/32918.html Andy From: Eleni Rapsomaniki > > Dear all, > > I am trying to train a randomForest using all my control data > (12,000 cases, ~ 20 explanatory variables, 2 classes). > Because

Hi, XML doesn't use the `$` to access child nodes. Instead use either `[name]` to get a list of children of that name or `[[name]]` to get the just the first child of that name encountered in the genealogy. Thus for your example... > root$child1 NULL > root[['child1']] <child1 name1="A" name2="B" name3="C"/> On the other hand, you might

Big thanks. newXMLNode works great. Wonder why it is not included in the documentation. There is newXMLDoc and newXMLNamespace, but no mention of newXMLNode. Stephen From: Ben Tupper [mailto:btupper at bigelow.org] Sent: Wednesday, March 21, 2018 6:18 PM To: Bond, Stephen Cc: r-help Subject: Re: [R] how to add a child to a child in XML Hi, XML doesn't use the `$` to access child nodes.

Hello, I would like to be able to plot an array on a plot, something like: |arg1 | arg2 | arg3 val1| 0.9 | 1.1 | 2.4 val2| 0.33 | 0.23 | -1.4 val3| hello| stop | test I know Rwave is good to report but don't want to use it. ? Is there a package that allow quick and dirty plot of dataframes like this ? Thanks a lot -- View this message in context:

Hi, there: I think i need to re-phrase my question since last time I did not get any reply but i think the question is not that hard, probably i did not make the question clear: I want to find cases like 35, 90, 330, 330, 335 from the rest which look like 3, 3, 3, 3.2, 3.3 4, 4.4, 4.5, 4.6, 4.7 .... basically there is one (or more) big 'gap' in the case i seek. thanks, weiwei --

Hi there, I need a network builder and it can change the node size and color; I am not sure if network package in R can do this or not. The other functions I wanted have been found in that package. BTW, if there is another package in R relating to this, please suggest too. Thanks, Weiwei -- Weiwei Shi, Ph.D Research Scientist "Did you always know?" "No, I did not. But I

Hi, I have a question on read.table. I have a dataset with 273,000 lines and 195 columns. I used the read.table to load the data into R: trn<-read.table('train1.dat', header=F, sep='|', na.strings='.') I found it takes forever. then I run 1/10 of the data (test) using read.table again. And this time it finished quickly. So, there might be something wrong in my data

Dear Sir or Madam, I am trying to plot forest plot. I extracted odds ratio and their corresponding 95% confidence interval from papers, then I calculated the log(OR) and standard error using the following command OR<-metagen(logOR,selogOR,sm="OR") forest(OR,comb.fixed=TRUE,comb.random=TRUE,digits=2) However, it does not produce a forest plot. Can someone kindly help? Thank

Hi, It's a reasonable question. The answer is that it actually is included, but there are many instances across packages where multiple functions are documented on a single help page. The following brings up such a page... (for XML_3.98-1.9) > library(XML) > ?newXMLNode You can see the same on line... https://www.rdocumentation.org/packages/XML/versions/3.98-1.9/topics/newXMLDoc

Hi, I have a dataset and want to build a generalized linear model on it. Unfortunately, complete.cases(df) returns null, which means I have to find a way to "fill" those missings. One way is following my previous post to use median to replace(or use most freq. of level to replace for catergorical case), but I am wondering if there are other ways, when glm or something like it is

> From: Weiwei Shi > > it works. > thanks, > > but: (just curious) > why i tried previously and i got > > > is.vector(sample.size) > [1] TRUE Because a list is also a vector: > a <- c(list(1), list(2)) > a [[1]] [1] 1 [[2]] [1] 2 > is.vector(a) [1] TRUE > is.numeric(a) [1] FALSE Actually, the way I initialize a list of known length is by

Hi, I have a dataset which has around 138 variables and 30,000 cases. I am trying to calculate a mahalanobis distance matrix for them and my procedure is like this: Suppose my data is stored in mymatrix > S<-cov(mymatrix) # this is fine > D<-sapply(1:nrow(mymatrix), function(i) mahalanobis(mymatrix, mymatrix[i,], S)) Error in solve.default(cov, ...) : system is computationally