I was referring to this link which explains the usage of ROCR package for plotting ROC curves and other related accuracy measurement metrics. The author mentions about logistic regression in the beginning, but do these functions(prediction, performance from ROCR) apply to other classification algorithms like SVM, Decision Trees, etc. ?
I tried using prediction() function with results of my SVM model, but it threw a format error despite the arguments being of same type and dimensions. Also I am not sure that if we try coming up with ROC curves for these algorithms, we would get a shape like the one we generally see with logistic regression(like this).
The prediction and performance functions are model-agnostic in the sense that they only require the user to input actual and predicted values from a binary classifier. (More precisely, this is what prediction requires, and performance takes as input an object output by prediction). Therefore, you should be able to use both functions for any classification algorithm that can output this information - including both logistic regression and SVM.
Having said this, model predictions can come in different formats (e.g., propensity scores vs. classes; results stored as numeric vs. factor), and you'll need to ensure that what you feed into prediction is appropriate. This can be quite specific, for example, while the predictions argument can represent continuous or class information, it can’t be a factor. See the “details” section of the function’s help file for more information.
Related
Unfortunately, I had convergence (and singularity) issues when calculating my GLMM analysis models in R. When I tried it in SPSS, I got no such warning message and the results are only slightly different. Does it mean I can interpret the results from SPSS without worries? Or do I have to test for singularity/convergence issues to be sure?
You have two questions. I will answer both.
First Question
Does it mean I can interpret the results from SPSS without worries?
You do not want to do this. The reason being is that mixed models have a very specific parameterization. Here is a screenshot of common lme4 syntax from the original article about lme4 from the author:
With this comes assumptions about what your model is saying. If for example you are running a model with random intercepts only, you are assuming that the slopes do not vary by any measure. If you include correlated random slopes and random intercepts, you are then assuming that there is a relationship between the slopes and intercepts that may either be positive or negative. If you present this data as-is without knowing why it produced this summary, you may fail to explain your data in an accurate way.
The reason as highlighted by one of the comments is that SPSS runs off defaults whereas R requires explicit parameters for the model. I'm not surprised that the model failed to converge in R but not SPSS given that SPSS assumes no correlation between random slopes and intercepts. This kind of model is more likely to converge compared to a correlated model because the constraints that allow data to fit a correlated model make it very difficult to converge. However, without knowing how you modeled your data, it is impossible to actually know what the differences are. Perhaps if you provide an edit to your question that can be answered more directly, but just know that SPSS and R do not calculate these models the same way.
Second Question
Or do I have to test for singularity/convergence issues to be sure?
SPSS and R both have singularity checks as a default (check this page as an example). If your model fails to converge, you should drop it and use an alternative model (usually something that has a simpler random effects structure or improved optimization).
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.
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.
Background
The reference manual for the gbm package states the interact.gbm function computes Friedman's H-statistic to assess the strength of variable interactions. the H-statistic is on the scale of [0-1].
The reference manual for the dismo package does not reference any literature for how the gbm.interactions function detects and models interactions. Instead it gives a list of general procedures used to detect and model interactions. The dismo vignette "Boosted Regression Trees for ecological modeling" states that the dismo package extends functions in the gbm package.
Question
How does dismo::gbm.interactions actually detect and model interactions?
Why
I am asking this question because gbm.interactions in the dismo package yields results >1, which the gbm package reference manual says is not possible.
I checked the tar.gz for each of the packages to see if the source code was similar. It is different enough that I cannot determine if these two packages are using the same method to detect and model interactions.
To summarize, the difference between the two approaches boils down to how the "partial dependence function" of the two predictors is estimated.
The dismo package is based on code originally given in Elith et al., 2008 and you can find the original source in the supplementary material. The paper very briefly describes the procedure. Basically the model predictions are obtained over a grid of two predictors, setting all other predictors at their means. The model predictions are then regressed onto the grid. The mean squared errors of this model are then multiplied by 1000. This statistic indicates departures of the model predictions from a linear combination of the predictors, indicating a possible interaction.
From the dismo package, we can also obtain the relevant source code for gbm.interactions. The interaction test boils down to the following commands (copied directly from source):
interaction.test.model <- lm(prediction ~ as.factor(pred.frame[,1]) + as.factor(pred.frame[,2]))
interaction.flag <- round(mean(resid(interaction.test.model)^2) * 1000,2)
pred.frame contains a grid of the two predictors in question, and prediction is the prediction from the original gbm fitted model where all but two predictors under consideration are set at their means.
This is different than Friedman's H statistic (Friedman & Popescue, 2005), which is estimated via formula (44) for any pair of predictors. This is essentially the departure from additivity for any two predictors averaging over the values of the other variables, NOT setting the other variables at their means. It is expressed as a percent of the total variance of the partial dependence function of the two variables (or model implied predictions) so will always be between 0-1.
R package randomForest reports mean squared errors for each tree in the forest. I need, however, a measure of confidence for each case in the data. Since randomForest calculates the casewise predictions by averaging the predictions of the single trees, I guess that it should also be possible to calculate a casewise standard error and thus a confidence interval. Can this be done using the output randomForest object (if so: how?) or do I have to dig into the source code?
No need to dig into the source code. You only need to read the documentation. ?predict.randomForest states that one of its arguments is called predict.all:
predict.all Should the predictions of all trees be kept?
So setting that to TRUE will keep a prediction for each case, for each tree, which you can then use to calculate standard error for each case.
I have recently been made aware of this paper by Stefan Wager, Trevor Hastie and Brad Efron which investigates more rigorously the idea of standard errors for the predictions generated by random forests (and other bagged predictors).