How to do crossvalidation on a two-part regression model? - r

I am currently working on regression analysis to model EQ-5D-5L health data. This data is inflated at the upper bound i.e. 1, and one of the approaches I use to model is with two-part models. By doing that, I combine a logistic model with binary data (1 or not 1), and continuous data.
The issue comes when trying to cross-validate (K-fold) the two-part models: I cannot find a way to include both "parts" of the model in the caret package in R, and I have not been able to find anybody that has solved the problem for me.
When I generate the predictions from the two-part model, it is essentially the coefficients from the two separate models that are multiplied together. So the models are developed separately, as they model different things from the same variable (binary and continuous outcome), but joined together when used to predict values.
Could it be possible to somehow cross-validate each part of the model separately, and get some kind of useful answer out of it?
Hope you guys can help.

Related

Imputation missing data for MLM in R

Maybe anyone can help me with this question. I conducted a follow-up study and obviously now have to face missing data. Now I am considering how to impute the missing data at best using MLM in R (f.e. participants concluded the follow up 2 survey, but not the follow up 1 survey, therefore I am missing L1 predictors for my longitudinal analysis).
I read about Multiple Imputation of multilevel data using the pan package (Schafer & Yucel, 2002) and came across the following code:
imp <- panImpute(data, formula = fml, n.burn = 1000, n.iter = 100, m = 5)
Yet, I have troubles understanding it completely. Is there maybe another way to impute missing data in R? Or maybe somebody could illustrate the process of the imputation method a bit more detailed, that would be so great! Do I have to conduct the imputation for every model I built in my MLM? (f.e. when I compared, whether a random intercept versus a random intercept and random slope model fits better for my data, do I have to use the imputation code for every model, or do I use it at the beginning of all my calculations?)
Thank you in advance
Is there maybe another way to impute missing data in R?
There are other packages. mice is the one that I normally use, and it does support multilevel data.
Do I have to conduct the imputation for every model I built in my MLM? (f.e. when I compared, whether a random intercept versus a random intercept and random slope model fits better for my data, do I have to use the imputation code for every model, or do I use it at the beginning of all my calculations?)
You have to specify the imputation model. Basically that means you have to tell the software which variables are predicted by which other variables. Since you are comparing models with the same fixed effect, and only changing the random effects (in particular comparing models with and without random slopes), the imputation model should be the same in both cases. So the workflow is:
perform the imputations;
run the model on all the imputed datasets,
pool the results (typically using Rubin's rules)
So you will need to do this twice, to end up with 2 sets of pooled results - one for each model. The software should provide functionality for doing all of this.
Having said all of that, I would advise against choosing your model based on fit statistics and instead use expert knowledge. If you have strong theoretical reasons for expecting slopes to vary by group, then include random slopes. If not, then don't include them.

How to run a truncated and inflated Poisson model in R?

My data doesn't contain any zeros. The minimum value for my outcome, y, is 1 and that is the value that is inflated. My objective is to run a truncated and inflated Poisson regression model using R.
I already know how to separate way each regression zero truncated and zero inflated. I want to know how to combine the two conditions into one model.
Thanks for you help.
For zero inflated models or zero-hurdle models, the standard approach is to use pscl package. I also wrote a package fitting that kind of models here but it is not yet mature and fully tested. Unless you have voluminous data, I still recommend you to use pscl that is more flexible, robust and documented.
For zero-truncated models, you can have a look at the VGML::vglm function. You might find useful information here.
Note that you are not doing the same distributional assumption so you won't need the same estimation data. Given the description of your dataset, I think you are looking for a zero-truncated model (since you do not observe zeros). With zero-inflated models, you decompose your observed pattern into zeros generated by a selection model and others generated by a count data model. This doesn't look to be a pattern consistent with your dataset.

How to train a multiple linear regression model to find the best combination of variables?

I want to run a linear regression model with a large number of variables and I want an R function to iterate on good combinations of these variables and give me the best combination.
The glmulti package will do this fairly effectively:
Automated model selection and model-averaging. Provides a wrapper for glm and other functions, automatically generating all possible models (under constraints set by the user) with the specified response and explanatory variables, and finding the best models in terms of some Information Criterion (AIC, AICc or BIC). Can handle very large numbers of candidate models. Features a Genetic Algorithm to find the best models when an exhaustive screening of the candidates is not feasible.
Unsolicited advice follows:
HOWEVER. Please be aware that while this approach can find the model that minimizes within-sample error (the goodness of fit on your actual data), it has two major problems that should make you think twice about using it.
this type of data-driven model selection will almost always destroy your ability to make reliable inferences (compute p-values, confidence intervals, etc.). See this CrossValidated question.
it may overfit your data (although using the information criteria listed in the package description will help with this)
There are a number of different ways to characterize a "best" model, but AIC is a common one, and base R offers step(), and package MASS offers stepAIC().
summary(lm1 <- lm(Fertility ~ ., data = swiss))
slm1 <- step(lm1)
summary(slm1)
slm1$anova

sLDA for predicting categorical response instead of continuous in R

I have a collection of documents, that might have latent topics associated with them. It is likely that each document might relate to one or more topics. I have a master file of all possible "topics"/categories and descriptions to these topics. I am seeking to create a model that predicts the topics for each document.
I could potentially use Supervised text classification using RTextTools, but that would only help me categorize documents to belong to one category or another. I am seeking to find a solution that would not only help me determine the topic proportions to the document, but also give the term-topic/category distributions.
sLDA seems like a good fit, but it seems to only predict continuous variable outcomes instead of categorical.
LDA is more of a classification method, predicting classes. other methods can be multinational logistic regression. LDA could be harder to train compared to Multinational, given a possible little improved fit it can provide.
update: LDA is a classification method where unlike logistic regression that you directly predict Pr(Y = k|X = x) using the logit link, LDA uses the Bayes theorem for prediction. It is normally a more popular compared to logistic regression (and its extension for multi-class prediction, namely multinational logistic regression) for multi-class problems.
LDA assumes that the observations are drawn from a Gaussian distribution with a common covariance matrix in each class, and so can provide some improvements over logistic regression when this assumption approximately holds. in contrast,it is suggested that logistic regression can outperform LDA if these Gaussian assumptions are not hold. To sum up, While both are appropriate for the development of linear classification models, linear discriminant analysis makes more assumptions about the underlying data as opposed to logistic regression, which makes logistic regression a more flexible and robust method when these assumptions are not hold. So what I meant was, it is important to understand your data well, and see which might fit your data better. There are good sources on read you can read and comparison of classification methods:
http://www-bcf.usc.edu/~gareth/ISL/ISLR%20Seventh%20Printing.pdf
I suggest Introduction to statistical learning, on classification chapter. Hope this helps

can we get probabilities the same way that we get them in logistic regression through random forest?

I have a data structure with binary 0-1 variable (click & Purchase; click & not-purchase) against a vector of the attributes. I used logistic regression to get the probabilities of the purchase. How can I use Random Forest to get the same probabilities? Is it by using Random Forest regression? or is it Random Forest classification with type='prob' in R which gives the probability of categorical variable?
It won't give you the same result since the structure of the two method are different. Logistic regression is given by a definitive linear specification, where RF is a collective vote from multiple independent/random trees. If specification and input feature are properly tuned for both, they can produce comparable results. Here is the major difference between the two:
RF will give more robust fit against noise, outliers, overfitting or multicollinearity etc which are common pitfalls in regression type of solution. Basically if you don't know or don't want to know much about whats going in with the input data, RF is a good start.
logistic regression will be good if you know expertly about the data and how to properly specify the equation. Or somehow want to engineer how the fit/prediction works. The explicit form of GLM specification will allow you to do that.

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