I am using the h2o.glm module (in R). I tried to find the 'weights_column' specification value in the outputting h2o GLM model object but I can not find it. I looked into model#allparameters and model#parameters, none of these two objects contain the weights column information. Is the weight information saved anywhere in the model object?
If you specify weights_column to GLM or any of the H2O algos, it will store the column name (not the actual column data) in the model object. In R, it stores it in both model#parameters and model#allparameters. Here's an example:
library(h2o)
model <- h2o.glm(x = 1:3, y = 5,
training_frame = as.h2o(iris),
weights_column = names(iris)[4],
family = "multinomial")
You can see the relevant info here:
> model#parameters$weights_column
$`__meta`
$`__meta`$schema_version
[1] 3
$`__meta`$schema_name
[1] "ColSpecifierV3"
$`__meta`$schema_type
[1] "VecSpecifier"
$column_name
[1] "Petal.Width"
$is_member_of_frames
NULL
Related
I am trying to input matrix data into the brm() function to run a signal regression. brm is from the brms package, which provides an interface to fit Bayesian models using Stan. Signal regression is when you model one covariate using another within the bigger model, and you use the by parameter like this: model <- brm(response ~ s(matrix1, by = matrix2) + ..., data = Data). The problem is, I cannot input my matrices using the 'data' parameter because it only allows one data.frame object to be inputted.
Here are my code and the errors I obtained from trying to get around that constraint...
First off, my reproducible code leading up to the model-building:
library(brms)
#100 rows, 4 columns. Each cell contains a number between 1 and 10
Data <- data.frame(runif(100,1,10),runif(100,1,10),runif(100,1,10),runif(100,1,10))
#Assign names to the columns
names(Data) <- c("d0_10","d0_100","d0_1000","d0_10000")
Data$Density <- as.matrix(Data)%*%c(-1,10,5,1)
#the coefficients we are modelling
d <- c(-1,10,5,1)
#Made a matrix with 4 columns with values 10, 100, 1000, 10000 which are evaluation points. Rows are repeats of the same column numbers
Bins <- 10^matrix(rep(1:4,times = dim(Data)[1]),ncol = 4,byrow =T)
Bins
As mentioned above, since 'data' only allows one data.frame object to be inputted, I've tried other ways of inputting my matrix data. These methods include:
1) making the matrix within the brm() function using as.matrix()
signalregression.brms <- brm(Density ~ s(Bins,by=as.matrix(Data[,c(c("d0_10","d0_100","d0_1000","d0_10000"))])),data = Data)
#Error in is(sexpr, "try-error") :
argument "sexpr" is missing, with no default
2) making the matrix outside the formula, storing it in a variable, then calling that variable inside the brm() function
Donuts <- as.matrix(Data[,c(c("d0_10","d0_100","d0_1000","d0_10000"))])
signalregression.brms <- brm(Density ~ s(Bins,by=Donuts),data = Data)
#Error: The following variables can neither be found in 'data' nor in 'data2':
'Bins', 'Donuts'
3) inputting a list containing the matrix using the 'data2' parameter
signalregression.brms <- brm(Density ~ s(Bins,by=donuts),data = Data,data2=list(Bins = 10^matrix(rep(1:4,times = dim(Data)[1]),ncol = 4,byrow =T),donuts=as.matrix(Data[,c(c("d0_10","d0_100","d0_1000","d0_10000"))])))
#Error in names(dat) <- object$term :
'names' attribute [1] must be the same length as the vector [0]
None of the above worked; each had their own errors and it was difficult troubleshooting them because I couldn't find answers or examples online that were of a similar nature in the context of brms.
I was able to use the above techniques just fine for gam(), in the mgcv package - you don't have to define a data.frame using 'data', you can call on variables defined outside of the gam() formula, and you can make matrices inside the gam() function itself. See below:
library(mgcv)
signalregression2 <- gam(Data$Density ~ s(Bins,by = as.matrix(Data[,c("d0_10","d0_100","d0_1000","d0_10000")]),k=3))
#Works!
It seems like brms is less flexible... :(
My question: does anyone have any suggestions on how to make my brm() function run?
Thank you very much!
My understanding of signal regression is limited enough that I'm not convinced this is correct, but I think it's at least a step in the right direction. The problem seems to be that brm() expects everything in its formula to be a column in data. So we can get the model to compile by ensuring all the things we want are present in data:
library(tidyverse)
signalregression.brms = brm(Density ~
s(cbind(d0_10_bin, d0_100_bin, d0_1000_bin, d0_10000_bin),
by = cbind(d0_10, d0_100, d0_1000, d0_10000),
k = 3),
data = Data %>%
mutate(d0_10_bin = 10,
d0_100_bin = 100,
d0_1000_bin = 1000,
d0_10000_bin = 10000))
Writing out each column by hand is a little annoying; I'm sure there are more general solutions.
For reference, here are my installed package versions:
map_chr(unname(unlist(pacman::p_depends(brms)[c("Depends", "Imports")])), ~ paste(., ": ", pacman::p_version(.), sep = ""))
[1] "Rcpp: 1.0.6" "methods: 4.0.3" "rstan: 2.21.2" "ggplot2: 3.3.3"
[5] "loo: 2.4.1" "Matrix: 1.2.18" "mgcv: 1.8.33" "rstantools: 2.1.1"
[9] "bayesplot: 1.8.0" "shinystan: 2.5.0" "projpred: 2.0.2" "bridgesampling: 1.1.2"
[13] "glue: 1.4.2" "future: 1.21.0" "matrixStats: 0.58.0" "nleqslv: 3.3.2"
[17] "nlme: 3.1.149" "coda: 0.19.4" "abind: 1.4.5" "stats: 4.0.3"
[21] "utils: 4.0.3" "parallel: 4.0.3" "grDevices: 4.0.3" "backports: 1.2.1"
I'm having a problem converting rxGlm models to normal glm models. Every time I try and covert my models I get the same error:
Error in qr.lm(object) : lm object does not have a proper 'qr' component.
Rank zero or should not have used lm(.., qr=FALSE).
Here's a simple example:
cols <- colnames(iris)
vars <- cols[!cols %in% "Sepal.Length"]
form1 <- as.formula(paste("Sepal.Length ~", paste(vars, collapse = "+")))
rx_version <- rxGlm(formula = form1,
data = iris,
family = gaussian(link = 'log'),
computeAIC = TRUE)
# here is the equivalent model with base R
R_version <- glm(formula = form1,
data = iris,
family = gaussian(link = 'log'))
summary(as.glm(rx_version)) #this always gives the above error
I cant seem to find this "qr" component (I'm assuming this is related to matrix decomposition) to specify in rxGlm formula.
Anyone else dealt with this?
rxGlm objects don't have a qr component, and converting to a glm object won't create one. This is intentional, as computing the QR decomposition of the model matrix requires the full dataset to be in memory which would defeat the purpose of using the rx* functions.
as.glm is really meant more for supporting model import/export via PMML. Most of the things that you'd want to do can be done with the rxGlm object, without converting. Eg rxGlm computes the coefficient std errors as part of the fit, without requiring a qr component afterwards.
I am new to R and trying to save my svm model in R and have read the documentation but still do not understand what is wrong.
I am getting the error "object is not a matrix" which would seem to mean that my data is not a matrix, but it is... so something is missing.
My data is defined as:
data = read.table("data.csv")
trainSet = as.data.frame(data[,1:(ncol(data)-1)])
Where the last line is my label
I am trying to define my model as:
svm.model <- svm(type ~ ., data=trainSet, type='C-classification', kernel='polynomial',scale=FALSE)
This seems like it should be correct but I am having trouble finding other examples.
Here is my code so far:
# load libraries
require(e1071)
require(pracma)
require(kernlab)
options(warn=-1)
# load dataset
SVMtimes = 1
KERNEL="polynomial"
DEGREE = 2
data = read.table("head.csv")
results10foldAll=c()
# Cross Fold for training and validation datasets
for(timesRun in 1:SVMtimes) {
cat("Running SVM = ",timesRun," result = ")
trainSet = as.data.frame(data[,1:(ncol(data)-1)])
trainClasses = as.factor(data[,ncol(data)])
model = svm(trainSet, trainClasses, type="C-classification",
kernel = KERNEL, degree = DEGREE, coef0=1, cost=1,
cachesize = 10000, cross = 10)
accAll = model$accuracies
cat(mean(accAll), "/", sd(accAll),"\n")
results10foldAll = rbind(results10foldAll, c(mean(accAll),sd(accAll)))
}
# create model
svm.model <- svm(type ~ ., data = trainSet, type='C-classification', kernel='polynomial',scale=FALSE)
An example of one of my samples would be:
10.135338 7.214543 5.758917 6.361316 0.000000 18.455875 14.082668 31
Here, trainSet is a data frame but in the svm.model function it expects data to be a matrix(where you are assigning trainSet to data). Hence, set data = as.matrix(trainSet). This should work fine.
Indeed as pointed out by #user5196900 you need a matrix to run the svm(). However beware that matrix object means all columns have same datatypes, all numeric or all categorical/factors. If this is true for your data as.matrix() may be fine.
In practice more than often people want to model.matrix() or sparse.model.matrix() (from package Matrix) which gives dummy columns for categorical variables, while having single column for numerical variables. But a matrix indeed.
I have two R objects as below.
matrix "datamatrix" - 200 rows and 494 columns: these are my x variables
dataframe Y. Y$V1 is my Y variable. I have converted column V1 to a factor I am building a classification model.
I want to build a neural network and I ran below command.
model <- train(Y$V1 ~ datamatrix, method='nnet', linout=TRUE, trace = FALSE,
#Grid of tuning parameters to try:
tuneGrid=expand.grid(.size=c(1,5,10),.decay=c(0,0.001,0.1)))
I got an error - " argument "data" is missing, with no default"
Is there a way for caret package to understand that I have my X variables in one R object and Y variable in other? I dont want to combined two data objects and then write a formula as the formula will be too long
Y~x1+x2+x3.................x199+x200....x493+x494
The argument "data" is missing error is addressed by adding a data = datamatrix argument to the train call. The way I would do it would be something like:
datafr <- as.data.frame(datamatrix)
# V1 is the first column name if dimnames aren't specified
datafr$V1 <- as.factor(datafr$V1)
model <- train(V1 ~ ., data = datafr, method='nnet',
linout=TRUE, trace = FALSE,
tuneGrid=expand.grid(.size=c(1,5,10),.decay=c(0,0.001,0.1)))
Now you don't have to pull your response variable out separately.
The . identifier allows inclusion of all variables from datafr (see here for details).
If x is a Random Forest in R, for example,
x <- cforest (y~ a+b+c, data = football),
what does x[[9]] mean?
You can't subset this object, so in some sense, x[[9]] is nothing, it is not accessible as such.
x is an object of S4 class "RandomForest-class". This class is documented on help page ?'RandomForest-class'. The slots of this object are named and described there. You can also get the slot names via slotNames()
library("party")
foo <- cforest(ME ~ ., data = mammoexp, control = cforest_unbiased(ntree = 50))
> slotNames(foo)
[1] "ensemble" "where" "weights"
[4] "initweights" "data" "responses"
[7] "cond_distr_response" "predict_response" "prediction_weights"
[10] "get_where" "update"
If by x[[9]] you meant the 9th slot, then that is predict_weights and ?'RandomForest-class' tells us that this is
‘prediction_weights’: a function for extracting weights from
terminal nodes.