I have taken plenty of time to try and help myself, but I keep reaching dead ends.

I have a dataset consisting of body measurements collected from a bird species, and the sex of each bird (known by molecular means). I built a logistic regression model (using the AIC information criterion) to assess which measurements explain better the sex of the birds. My ultimate goal is to have an equation which could be used by others under field conditions to predict reliably the sex of the birds by taking as few body measurements as possible.

My final model includes four independent variables, namely "Culmen", "Head-bill", "Tarsus length", and "Wing length" (all continuous). I wish my model was a little more parsimonious, but all the variables seem to be important according to AIC criterion. Because the model produced should be used as prediction tool, I decided validate it using a leave-one-out cross validation approach. In my learning process, I first tried to complete the analyses (cross-validation and plotting) by including only one explanatory variable, namely "Culmen".

The output of the cross validation (package "boot" in R) yields two values (deltas), which are the cross-validated prediction errors where the first number is the raw leave-one-out, or lieu cross-validation result, and the second one is a bias-corrected version of it.

model.full <- glm(Sex ~ Culmen, data = my.data, family = binomial)

cv.glm(my.data, model.full, K=114)

cv.glm(data = my.data, glmfit = model.full, K = 114)

[1] 114

[1] 0.05941851 0.05937288

Q1. Could anyone expalin what do these two values represent and how to interpret them?

Following is the code as presented by Dr. Markus Müller (Calimo) in a similar, albeit not identical, post (https://stackoverflow.com/questions/20346568/feature-selection-cross-validation-but-how-to-make-roc-curves-in-r) which I tried to tweak to meet my data:

k <- 114    # Number of observations or rows in dataset
n <- dim(my.data)[1]
indices <- sample(rep(1:k, ceiling(n/k))[1:n])

all.response <- all.predictor <- aucs <- c()
for (i in 1:k) {
test = my.data[indices==i,]
learn = my.data[indices!=i,]
model <- glm(Sex ~ Culmen, data = learn, family=binomial)
model.pred <- predict(model, newdata=test)
aucs <- c(aucs, roc(test$Sex, model.pred)$auc)
all.response <- c(all.response, test$outcome)
all.predictor <- c(all.predictor, model.pred)

Error in roc.default(test$Sex, model.pred) : No case observation.

roc(all.response, all.predictor)

Error in roc.default(all.response, all.predictor) : No valid data provided.


Q2. What's the reason for the first error message? I guess the second error is associated with the first one, and that it will be solved once I find a solution to the first one.

I will appreciate very much any help!!


  • $\begingroup$ (1) for the ROC curve the example you used used 10 folds. You might want to reduce K (2) for model validation you will want to include the whole model selection process - looks like you've done variable selection as well (3) do you want to use "prediction error" for validation of your logistic regression model? this looks research-like: you might want to use your research domain accepted method of model validation $\endgroup$
    – charles
    Jan 22 '16 at 15:34
  • $\begingroup$ I strongly second @charles ' comment - the model selection must be included in the CV, see e.g. kaggle.com/c/the-analytics-edge-mit-15-071x/forums/t/7837/… $\endgroup$ Jan 14 '17 at 16:27

Include more data into your test set.

Sometimes algorithms like Neural Network give classification output such that all the output labels are the same.

For example, suppose your actual labels are like this: c(1,0,0,1,1,0,0,1,1,1).
You might end up training your neural network (I am mentioning neural network specifically because I have faced this problem while applying the algorithm) in such a way that the output labels come out to be: c(1,1,1,1,1,1,1,1,1,1).

In such a case, your auc/roc functions would show the above mentioned error as these are no 0 labels in predicted data.

Hope this might help!


The first delta value outputted by cv.glm is the mean squared error between your dependent variable Sex (dummy coded as 0 and 1 values) and the predicted probability that Sex$=1$, averaged over the validation set. Since your goal is to use the model to make decisions on the field, you may want to change the cost function to the prediction error rate (that is the average number of error you expect to make if you were to classify each observation according to the output of your model), which may be more straightforward to interpret. This can be done by setting the cost argument of the cv.glm function (taken from the examples in the cv.glm manual page):

cost <- function(r, pi = 0) mean(abs(r-pi) > 0.5)
cv.glm(my.data, model.full, cost, K=114)

The second delta value is the same error after applying a correction for the bias that may be introduced if the K partitions are chosen randomly. The adjusted, bias-corrected error $\Delta _{aCV}$ is computed as $${\Delta _{aCV}} = {\Delta _{CV}} + \Delta _n^* - \sum\limits_{k = 1}^K {{p_k}} \Delta _k^*$$ where $\Delta _{CV}$ is the cross-validated error (first delta value), $\Delta _n^*$ is the training error (I used $^*$ to denote that the error is evaluated on the same observations used to fit the model, e.g. the error is measured on the training set) when the model is fit to all the observations, $\Delta _k^*$ is the training error of the $k$-th partition, and $p_k$ is the proportion of observations included in the $k$-th partition (taken from Davison and Hinkley, 1997, Bootstrap Methods and Their Application). But if you are doing leave-one-out cross-validation (K is equal to the number of observations in your dataset) then there should be no need to look at the adjusted error.

About your second question, the error message seems due to the fact that in at least one of the K splits in the dataset, due to random sampling, there were no "cases", that is observations where Sex$=1$, in the test set. You could try to extend the code by doing a stratified sampling where each of the K splits contains approximately the same proportion of observations where Sex$=1$ and Sex$=0$.


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