I have build a binary logistic regression for churn prediction in Rstudio. Due to the unbalanced data used for this model, I also included weights. Then I tried to find the optimum cutoff by try and error, however To complete my research I have to incorporate ROC curves to find the optimum cutoff. Below I provided the script I used to build the model (fit2). The weight is stored in 'W'. This states that the costs of wrongly identifying a churner is 14 times as large as the costs of wrongly identifying a non-churner.
#CH1 logistic regression
library(caret)
W = 14
lvl = levels(trainingset$CH1)
print(lvl)
#if positive we give it the defined weight, otherwise set it to 1
fit_wts = ifelse(trainingset$CH1==lvl[2],W,1)
fit2 = glm(CH1 ~ RET + ORD + LVB + REVA + OPEN + REV2KF + CAL + PSIZEF + COM_P_C + PEN + SHOP, data = trainingset, weight=fit_wts, family=binomial(link='logit'))
# we test it on the test set
predlog1 = ifelse(predict(fit2,testset,type="response")>0.5,lvl[2],lvl[1])
predlog1 = factor(predlog1,levels=lvl)
predlog1
confusionMatrix(pred,testset$CH1,positive=lvl[2])
For this research I have also build ROC curves for decision trees using the pROC package. However, of course the same script does not work the same for a logistic regression. I have created a ROC curve for the logistic regression using the script below.
prob=predict(fit2, testset, type=c("response"))
testset$prob=prob
library(pROC)
g <- roc(CH1 ~ prob, data = testset, )
g
plot(g)
Which resulted in the ROC curve below.
How do I get the optimum cut off from this ROC curve?
Getting the "optimal" cutoff is totally independent of the type of model, so you can get it like you would for any other type of model with pROC. With the coords function:
coords(g, "best", transpose = FALSE)
Or directly on a plot:
plot(g, print.thres=TRUE)
Now the above simply maximizes the sum of sensitivity and specificity. This is often too simplistic and you probably need a clear definition of "optimal" that is adapted to your use case. That's mostly beyond the scope of this question, but as a starting point you should a look at Best Thresholds section of the documentation of the coords function for some basic options.
Related
I am currently trying to generate a general additive model in R using a response variable and three predictor variables. One of the predictors is linear, and the dataset consists of 298 observations.
I have run the following code to generate a basic GAM:
GAM <- gam(response~ linearpredictor+ s(predictor2) + s(predictor3), data = data[2:5])
This produces a model with 18 degrees of freedom and seems to substantially overfit the data. I'm wondering how I might generate a GAM that maximizes smoothness and predictive error. I realize that each of these features is going to come at the expense of the other, but is there good a way to find the optimal model that doesn't overfit?
Additionally, I need to perform leave one out cross validation (LOOCV), and I am not sure how to make sure that gam() does this in the MGCV package. Any help on either of these problems uld be greatly appreciated. Thank you.
I've run this to generate a GAM, but it overfits the data.
GAM <- gam(response~ linearpredictor+ s(predictor2) + s(predictor3), data = data[2:5])
I have also generated 1,000,000 GAMs with varying combinations of smoothing parameters and ranged the maximum degrees of freedom allowed from 10 (as shown in the code below) to 19. The variable "combinations2" is a list of all 1,000,000 combinations of smoothers I selected. This code is designed to try and balance degrees of freedom and AIC score. It does function, but I'm not sure that I'm actually going to be able to find the optimal model from this. I also cannot tell how to make sure that it uses LOOCV.
BestGAM <- gam(response~ linearpredictor+ predictor2+ predictor3, data = data[2:5])
for(i in 1:100000){
PotentialGAM <- gam(response~ linearpredictor+ s(predictor2) + s(predictor3), data = data[2:5], sp=c(combinations2[i,]$Var1,combinations2[i,]$Var2))
if (AIC(PotentialGAM,BestGAM)$df[1] <= 10 & AIC(PotentialGAM,BestGAM)$AIC[1] < AIC(PotentialGAM,BestGAM)$AIC[2]){
BestGAM <<- PotentialGAM
listNumber <- i
}
}
You are fitting your GAM using generalised cross validation (GCV) smoothness selection. GCV is a way to get around the invariance problem of ordinary cross validation (OCV; what you also call LOOCV) when estimating GAMs. Note that GCV is the same as OCV on a rotated version of the fitting problem (rotating y - Xβ by Q, any orthogonal matrix), and while when fitting with GCV {mgcv} doesn't actually need to do the rotation and the expected GCV score isn't affected by the rotation, GCV is just OCV (wood 2017, p. 260)
It has been shown that GCV can undersmooth (resulting in more wiggly models) as the objective function (GCV profile) can become flat around the optimum. Instead it is preferred to estimate GAMs (with penalized smooths) using REML or ML smoothness selection; add method = "REML" (or "ML") to your gam() call.
If the REML or ML fit is as wiggly as the GCV one with your data, then I'd be likely to presume gam() is not overfitting, but that there is something about your response data that hasn't been explained here (are the data ordered in time, for example?)
As to your question
how I might generate a GAM that maximizes smoothness and [minimize?] predictive error,
you are already doing that using GCV smoothness selection and for a particular definition of "smoothness" (in this case it is squared second derivatives of the estimated smooths, integrated over the range of the covariates, and summed over smooths).
If you want GCV but smoother models, you can increase the gamma argument above 1; gamma 1.4 is often used for example, which means that each EDF costs 40% more in the GCV criterion.
FWIW, you can get the LOOCV (OCV) score for your model without actually fitting 288 GAMs through the use of the influence matrix A. Here's a reproducible example using my {gratia} package:
library("gratia")
library("mgcv")
df <- data_sim("eg1", seed = 1)
m <- gam(y ~ s(x0) + s(x1) + s(x2) + s(x3), data = df, method = "REML")
A <- influence(m)
r <- residuals(m, type = "response")
ocv_score <- mean(r^2 / (1 - A))
I am trying to create a GLMM in R. I want to find out how the emergence time of bats depends on different factors. Here I take the time difference between the departure of the respective bat and the sunset of the day as dependent variable (metric). As fixed factors I would like to include different weather data (metric) as well as the reproductive state (categorical) of the bats. Additionally, there is the transponder number (individual identification code) as a random factor to exclude inter-individual differences between the bats.
I first worked in R with a linear mixed model (package lme4), but the QQ plot of the residuals deviates very strongly from the normal distribution. Also a histogram of the data rather indicates a gamma distribution. As a result, I implemented a GLMM with a gamma distribution. Here is an example with one weather parameter:
model <- glmer(formula = difference_in_min ~ repro + precipitation + (1+repro|transponder number), data = trip, control=ctrl, family=gamma(link = log))
However, since there was no change in the QQ plot this way, I looked at the residual diagnostics of the DHARMa package. But the distribution assumption still doesn't seem to be correct, because the data in the QQ plot deviates very much here, too.
Residual diagnostics from DHARMa
But if the data also do not correspond to a gamma distribution, what alternative is there? Or maybe the problem lies somewhere else entirely.
Does anyone have an idea where the error might lie?
But if the data also do not correspond to a gamma distribution, what alternative is there?
One alternative is called the lognormal distribution (https://en.wikipedia.org/wiki/Log-normal_distribution)
Gaussian (or normal) distributions are typically used for data that are normally distributed around zero, which sounds like you do not have. But the lognormal distribution does not have the same requirements. Following your previous code, you would fit it like this:
model <- glmer(formula = log(difference_in_min) ~ repro + precipitation + (1+repro|transponder number), data = trip, control=ctrl, family=gaussian(link = identity))
or instead of glmer you can just call lmer directly where you don't need to specify the distribution (which it may tell you to do in a warning message anyway:
model <- lmer(formula = log(difference_in_min) ~ repro + precipitation + (1+repro|transponder number), data = trip, control=ctrl)
I have fit a robust random-effects meta-regression model using metafor package in R.
My full data, as well as reproducible R code appear below.
Questions:
(1) What are the meaning and interpretation of grey-colored diamonds appearing over CIs?
(2) I won't get an overall mean effect when I have moderators, correct?
library(metafor)
d <- read.csv("https://raw.githubusercontent.com/izeh/m/master/d.csv", h = T) ## DATA
res <- robust(rma.uni(yi = dint, sei = SD, mods = ~es.type, data = d, slab = d$study.name),
cluster = d$id)
forest(res)
1) Quoting from help(forest.rma): "For models involving moderators, the fitted value for each study is added as a polygon to the plot." So, the grey-colored diamonds (or polygons) are the fitted values and the width of the diamonds/polygons reflects the width of the CI for the fitted values.
2) No, since there is no longer a single overall effect when your model includes moderators.
I am trying to fit a copula for two variables which have extreme value distribution. for "mvdc" class, I need to define margins and parammargins. Since GEV is not included in default distribution functions of Rcopula, I got these two values by using "evd" package, by these two functions:
# pgev gives the Generalized Extreme Value distribution function
GEVmarginU1<-pgev(U1, loc=0, scale=1, shape=0, lower.tail = TRUE)
GEVmarginV2<-pgev(V2, loc=0, scale=1, shape=0, lower.tail = TRUE)
#fit a generalised extreme value distribution to my data
MU1 <- fgev(U1, scale = 1, shape = 0)
MV2 <- fgev(V2, scale = 1, shape = 0)
but when I give these values to "mvdc" function, I get an error
myMvd <- mvdc(copula = ellipCopula(family = "Frank", param = 0), margins = c(pgev, pgev),
paramMargins = list(list(MU1), list(MV2))
Most importantly, I want to be sure whether I am in a right track. Since two variables are obtained from discrete choice model, I have extreme value distribution. Also the marginal have GEV distribution, right? So I need to define GEV for "mvdc" otherwise my fitted copula will not wok well.
(1) Ui = β1Xi1 + β2Xi2 + β3Xi3 + εi
(2) Vi = γ1Yj1 + γ2Yj2 + γ3Yj3 + ηi
in summary:
(1) Ui = β'Xi' + εi
(2) Vi = γ'Yj' + ηi
Since these models are made from discrete choice modelling approach, the distribution function follows “extreme value” distribution. First step: I estimate coefficients of β1,β2,β3,γ1,γ2,γ3 separately for each variable of i and Vj by using multinomial logit model using Biogeme software. But intuitively I know that they are dependent variables, so I try to fit a copula and again estimate coefficients by considering dependency value. So, the joint probability that Ui and Vi is chosen by decision-maker n is:
These marginals are transformed to continuous, but still have extreme value distribution, am I right?!???
1) How can I define GEV when using “mvdc” copula class in Rcopula?
Second, assume I used “fitcopula” instead of “mvdc”, and got param(dependency parameter of copula), if I understood correctly, “fitcopula” is for parametric and in my case, it’s non-parametric, am I right?
2) Now, how should I update coefficients by using a joint distribution and dependency parameter???
For the first question, I found out that my marginals are logistic randomly distributed, since they are the difference between two error terms in the utility model and we know that error terms follow type 1 extreme value or Gumbel distribution, and the difference between two Gumbel distribution follow logistics distribution, according to the Wikipedia.
I am using random-forest for a regression problem to predict the label values of Test-Y for a given set of Test-X (new values of features). The model has been trained over a given Train-X (features) and Train-Y (labels). "randomForest" of R serves me very well in predicting the numerical values of Test-Y. But this is not all I want.
Instead of only a number, I want to use random-forest to produce a probability density function. I searched for a solution for several days and here is I found so far:
"randomForest" doesn't produce probabilities for regression, but only in classification. (via "predict" and setting type=prob).
Using "quantregForest" provides a nice way to make and visualize prediction intervals. But still not the probability density function!
Any other thought on this?
Please see the predict.all parameter of the predict.randomForest function.
library("ggplot2")
library("randomForest")
data(mpg)
rf = randomForest(cty ~ displ + cyl + trans, data = mpg)
# Predict the first car in the dataset
pred = predict(rf, newdata = mpg[1, ], predict.all = TRUE)
hist(pred$individual)
The histogram of 500 "elementary" predictions looks like this:
You can also use quantregForest with a very fine grid of quantiles, convert them into a "cumulative distribution function (cdf)" with R-function ecdf and convert this cdf into a density estimation with a kernel density estimator.