The target is to predict the values to the test set with the package monreg model, but this si not working with the predict function, because there isn't a model object to use the prediction function.
Giving an example:
require(monreg) # Package ‘monreg’
x <- rnorm(100)
y <- x + rnorm(100)
x_train=x[0:80]
y_train=y[0:80]
x_test=x[81:100]
y_test=y[81:100]
mon1 <- monreg(x, y, hd = .5, hr = .5)
# I was expecting to get the prediction over the test partion as R usualy works
predict(mon1,h=length(y_test))
But this is not working. In the case this package doesnt have any predict function, I would accept any advice to implement the Narayada Watson regression in R in order to predict values like this example I gave.
Related
Using the dlm package in R I fit a dynamic linear model to a time series data set, consisting of 20 observations. I then use the dlmForecast function to predict future values (which I can validate against the genuine data for said period).
I use the following code to create a prediction interval;
ciTheory <- (outer(sapply(fut1$Q, FUN=function(x) sqrt(diag(x))), qnorm(c(0.05,0.95))) +
as.vector(t(fut1$f)))
However my data does not follow a normal distribution and I wondered whether it would be possible to
adapt the qnorm function for other distributions. I have tried qt, but am unable to apply qgamma.......
Just wondered if anyone knew how you would go about sorting this.....
Below is a reproduced version of my code...
library(dlm)
data <- c(20.68502, 17.28549, 12.18363, 13.53479, 15.38779, 16.14770, 20.17536, 43.39321, 42.91027, 49.41402, 59.22262, 55.42043)
mod.build <- function(par) {
dlmModPoly(1, dV = exp(par[1]), dW = exp(par[2]))
}
# Returns most likely estimate of relevant values for parameters
mle <- dlmMLE(a2, rep(0,2), mod.build); #nileMLE$conv
if(mle$convergence==0) print("converged") else print("did not converge")
mod1 <- dlmModPoly(dV = v, dW = c(0, w))
mod1Filt <- dlmFilter(a1, mod1)
fut1 <- dlmForecast(mod1Filt, n = 7)
Cheers
I'm trying to do stepwise regression for following data:
y <- c(1.2748, 1.2574, 1.5571, 1.4178, 0.8491, 1.3606, 1.4747, 1.3177, 1.2896, 0.8453)
x <- data.frame(A = c(2,3,4,5,6,2,3,4,5,6),
B = c(2,4,1,3,5,1,3,5,2,4)*100,
C = c(9,5,11,5,11,7,13,7,13,9),
D = c(6,5,3,7,6,4,3,7,5,4),
E = c(1,1,0.8,0.8,0.6,0.6,0.4,0.4,0.2,0.2))
x$A2 <- x$A^2
x$B2 <- x$B^2
x$C2 <- x$C^2
x$D2 <- x$D^2
x$E2 <- x$E^2
x$AB <- x$A*x$B
As we can see, it has 10 observations and 11 independent variables so I can't build a linear regression model for it. In fact, only a few factors is useful and in this case, I need to use stepwise regression and "forward" to add independent variables into my formula. But stats:: step function cannot be used. I wonder if there is a method to do it. I know there is a package called "StepReg" but I don't fully understand how to use it and how to read the results. Thank you!
I just run stepwise regression with the data you provided using R package StepReg
Here is the code, and hope this can help you.
library(StepReg)
df <- data.frame(y,x)
# forward method with information criterion 'AIC', you can choose other information criterion
stepwise(df, y="y", exclude=NULL, include=NULL, Class=NULL,
selection="forward", select="AIC")
# forward method with significant levels, significant level for entry = 0.15
stepwise(df, y="y", exclude=NULL, include=NULL, Class=NULL,
selection="forward", select="SL",sle=0.15)
You can use olsrr package which gives similar results as that of SPSS. Here is the solution
library(olsrr)
df <- data.frame(y, x)
model <- lm(y ~ ., data = df)
smlr <- ols_step_both_p(model, pent = 0.05, prem = 0.1) #pent p value; variables with p value less than pent will enter into the model.
#premp value; variables with p more than prem will be removed from the model.
You can get the model details by calling
smlr
smlr$model
smlr$beta_pval #regression coefficients with p-values
I have kept the values of pent and prem same as the default values used by SPSS.
Bert-toolkit is a very nice package to call R functions from Excel. See: https://bert-toolkit.com/
I have used bert-toolkit to call a fitted neuralnet (avNNnet fitted with Caret) within a wrapper function in R from Excel VBA. This runs perfect. This is the code to load the model within the wrapper function in bert-toolkit:
load("D:/my_model_avNNet.rda")
neuraln <- function(x1,x2,x3){
xx <- data.frame(x1,x2,x3)
z <- predict(my_model_avNNET, xx)
z
}
Currently I tried to do this with a fitted GAM (fitted with package mgcv). Although I do not succeed. If I call the fitted GAM from Excel VBA it gives error 2015. If I call the fitted GAM from a cell it gives #VALUE! At the same time the correct outcome of the calculation is shown in the bert-console!
This is the code to load the model in the wrapperfunction in bert-toolkit:
library(mgcv)
load("D:/gam_y_model.rda")
testfunction <- function(k1,k2){
z <- predict(gam_y, data.frame(x = k1, x2 = k2))
print (z)
}
The difference between the avNNnet-model (Caret) and the GAM-model (mgcv) is that the avNNnet-model does NOT need the Caret library to be loaded to generate a prediction, while the GAM-model DOES need the mgcv library to be loaded.
It seems to be not sufficient to load the mgvc-library in the script with the GAM-model which loads the GAM-model in a wrapper function in bert-toolkit, as I did in the code above. Although the correct outcome of the model is shown in the bert-console. It does not generate the correct outcome in Excel.
I wonder how this is possible and can be solved. It seems to me that maybe there are two instances of R running in bert-toolkit.
How can I load the the mgcv-library in such a way that it can be used by the GAM-model within the function called from Excel?
This is some example code to fit the GAM with mgcv and save to model (after running this code the model can uploaded in bert-toolkit with the code above) :
library(mgcv)
# construct some sample data:
x <- seq(0, pi * 2, 0.1)
x2 <- seq(0, pi * 20, 1)
sin_x <- sin(x)
tan_x2 <- tan(x2)
y <- sin_x + rnorm(n = length(x), mean = 0, sd = sd(sin_x / 2))
Sample_data <- data.frame(y,x,x2)
# fit gam:
gam_y <- gam(y ~ s(x) + s(x2), method = "REML")
# Make predictions with the fitted model:
x_new <- seq(0, max(x), length.out = 100)
x2_new <- seq(0, max(x2), length.out = 100)
y_pred <- predict(gam_y, data.frame(x = x_new, x2 = x2_new))
# save model, to load it later in bert-toolkit:
setwd("D:/")
save(gam_y, file = "gam_y_model.rda")
One of R's signatures is method dispatching where users call the same named method such as predict but internally a different variant is run such as predict.lm, predict.glm, or predict.gam depending on the model object passed into it. Therefore, calling predict on an avNNet model is not the same predict on a gam model. Similarly, just as the function changes due to the input, so does the output change.
According to MSDN documents regarding the Excel #Value! error exposed as Error 2015:
#VALUE is Excel's way of saying, "There's something wrong with the way your formula is typed. Or, there's something wrong with the cells you are referencing."
Fundamentally, without seeing actual results, Excel may not be able to interpret or translate into Excel range or VBA type the result R returns from gam model especially as you describe R raises no error.
For example, per docs, the return value of the standard predict.lm is:
predict.lm produces a vector of predictions or a matrix of predictions...
However, per docs, the return value of predict.gam is a bit more nuanced:
If type=="lpmatrix" then a matrix is returned which will give a vector of linear predictor values (minus any offest) at the supplied covariate values, when applied to the model coefficient vector. Otherwise, if se.fit is TRUE then a 2 item list is returned with items (both arrays) fit and se.fit containing predictions and associated standard error estimates, otherwise an array of predictions is returned. The dimensions of the returned arrays depends on whether type is "terms" or not: if it is then the array is 2 dimensional with each term in the linear predictor separate, otherwise the array is 1 dimensional and contains the linear predictor/predicted values (or corresponding s.e.s). The linear predictor returned termwise will not include the offset or the intercept.
Altogether, consider adjusting parameters of your predict call to render a numeric vector for easy Excel interpretation and not a matrix/array or some other higher dimension R type that Excel cannot render:
testfunction <- function(k1,k2){
z <- mgcv::predict.gam(gam_y, data.frame(x = k1, x2 = k2), type=="response")
return(z)
}
testfunction <- function(k1,k2){
z <- mgcv::predict.gam(gam_y, data.frame(x = k1, x2 = k2), type=="lpmatrix")
return(z)
}
testfunction <- function(k1,k2){
z <- mgcv::predict.gam(gam_y, data.frame(x = k1, x2 = k2), type=="linked")
return(z$fit) # NOTICE fit ELEMENT USED
}
...
Further diagnostics:
Check returned object of predict.glm with str(obj) and class(obj)/ typeof(obj) to see dimensions and underlying elements and compare with predict in caret;
Check if high precision of decimal numbers is the case such as Excel's limits of 15 decimal points;
Check amount of data returned (exceeds Excel's sheet row limit of 220 or cell limit of 32,767 characters?).
I am using an accelerated failure time / AFT model with a weibull distribution to predict data. I am doing this using the survival package in R. I am splitting my data in training and test, do training on the training set and afterwards try to predict the values for the test set. To do that I am passing the the test set as the newdata parameter, as stated in the references. I get an error, saying that newdata does not have the same size as the training data (obviously!). Then the function seems to evaluate predict the values for the training set.
How can I predict the values for the new data?
# get data
library(KMsurv)
library(survival)
data("kidtran")
n = nrow(kidtran)
kidtran <- kidtran[sample(n),] # shuffle row-wise
kidtran.train = kidtran[1:(n * 0.8),]
kidtran.test = kidtran[(n * 0.8):n,]
# create model
aftmodel <- survreg(kidtransurv~kidtran.train$gender+kidtran.train$race+kidtran.train$age, dist = "weibull")
predicted <- predict(aftmodel, newdata = kidtran.test)
Edit: As mentioned by Hack-R, there was this line of code missing
kidtransurv <- Surv(kidtran.train$time, kidtran.train$delta)
The problem seems to be in your specification of the dependent variable.
The data and code definition of the dependent was missing from your question, so I can't see what the specific mistake was, but it did not appear to be a proper Surv() survival object (see ?survreg).
This variation on your code fixes that, makes some minor formatting improvements, and runs fine:
require(survival)
pacman::p_load(KMsurv)
library(KMsurv)
library(survival)
data("kidtran")
n = nrow(kidtran)
kidtran <- kidtran[sample(n),]
kidtran.train <- kidtran[1:(n * 0.8),]
kidtran.test <- kidtran[(n * 0.8):n,]
# Whatever kidtransurv was supposed to be is missing from your question,
# so I will replace it with something not-missing
# and I will make it into a proper survival object with Surv()
aftmodel <- survreg(Surv(time, delta) ~ gender + race + age, dist = "weibull", data = kidtran.train)
predicted <- predict(aftmodel, newdata = kidtran.test)
head(predicted)
302 636 727 121 85 612
33190.413 79238.898 111401.546 16792.180 4601.363 17698.895
i trying do create a plot for my model create using SVM in e1071 package.
my code to build the model, predict and build confusion matrix is
ptm <- proc.time()
svm.classifier = svm(x = train.set.list[[0.999]][["0_0.1"]],
y = train.factor.list[[0.999]][["0_0.1"]],
kernel ="linear")
pred = predict(svm.classifier, test.set.list[[0.999]][["0_0.1"]], decision.values = TRUE)
time[["svm"]] = proc.time() - ptm
confmatrix = confusionMatrix(pred,test.factor.list[[0.999]][["0_0.1"]])
confmatrix
train.set.list and test.set.list contains the test and train set for several conditions. train and set factor has the true label for each set. Train.set and test.set are both documenttermmatrix.
Then i tried to see a plot of my data, i tried with
plot(svm.classifier, train.set.list[[0.999]][["0_0.1"]])
but i got the message:
"Error in plot.svm(svm.classifier, train.set.list[[0.999]][["0_0.1"]]) :
missing formula."
what i'm doing wrong? confusion matrix seems good to me even not using formula parameter in svm function
Without given code to run, it's hard to say exactly what the problem is. My guess, given
?plot.svm
which says
formula formula selecting the visualized two dimensions. Only needed if more than two input variables are used.
is that your data has more than two predictors. You should specify in your plot function:
plot(svm.classifier, train.set.list[[0.999]][["0_0.1"]], predictor1 ~ predictor2)