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I have a small imbalanced sample of 18 and 12 records in 0-labeled and 1-labeled groups respectively with 6 features, so the size of my feature matrix X is $30 \times 6$ and y is my target vector $30 \times 1$.

The data in CSV format is shared via the dropbox link:

THE PROBLEM

I would like to assess my classification accuracy and to do so, I perform the stratified nested cross validation: the outer loop iterates over different train-test samples, and the inner loop selects the best model. The sklearn (0.18.1) implementation of the code:

sss_outer = StratifiedShuffleSplit(n_splits=10, test_size=0.333, random_state=None)
sss_inner = StratifiedShuffleSplit(n_splits=10, test_size=0.3, random_state=None)

parameters = {'clf__C': logspace(-4,3,150)}
pipe_logistic = Pipeline([('scl', StandardScaler()),('clf', LogisticRegression(penalty='l1'))])
grid_search = GridSearchCV(estimator=pipe_logistic, param_grid=parameters, verbose=1, scoring=make_scorer(accuracy_score), cv=sss_inner)
cross_val_score(grid_search, X, y, cv=sss_outer, scoring='accuracy')

Also, in order to see how many instances are in each partitioned group, I created a simple function "nested_partitioning_calculator()", which quickly computes the number of instances in each split.

nested_partitioning_calculator(X, y, sss_inner, sss_outer)
----------------- Outer Split: Train+Validate and Test -----------------
[8, 12] [4, 6]
[8, 12] [4, 6]
[8, 12] [4, 6]
[8, 12] [4, 6]
[8, 12] [4, 6]
[8, 12] [4, 6]
[8, 12] [4, 6]
[8, 12] [4, 6]
[8, 12] [4, 6]
[8, 12] [4, 6]
----------------- Inner Split: Train and Validate -----------------
[6, 8] [2, 4]
[6, 8] [2, 4]
[6, 8] [2, 4]
[6, 8] [2, 4]
[6, 8] [2, 4]
[6, 8] [2, 4]
[6, 8] [2, 4]
[6, 8] [2, 4]
[6, 8] [2, 4]
[6, 8] [2, 4]

From this example, it's clear, that the outer CV loop splits [12, 18] data (18-in 0 class, and 12 in 1 class) into [8, 12] for the training sample and [4, 6] and then, the inner loop splits again [8, 12] into training set [6, 8] and validation set [2, 4]. So, basically, the training is performed on a small set of 6 instances from the 1 class and 8 instances from the 0 class.

THE QUESTION

Obviously, different values in the test_size parameter will yield different CV accuracy results. What is the proper way to split the small data set in the case of the nested CV? Should I be aiming at smaller test and validation samples? (aka LOOCV?).

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@John is right that sampling variability is your problem. In particular, the variance on the performance estimates.

In contrast to his advise, I'd strongly recommend not to do LOO. The main reason for that (apart from the possible complication of strong pessimistic bias due to inherent lack of stratification) is that with LOO you cannot distinguish two different sources of variance:

  • variance due to the limited number of cases tested and
  • variance due to model instability (i.e. due to the training sample size being so limited that exchanging a few training cases does make a difference). Model instability is one symptom of unsuccessful optimization.

Doing e.g. repeated k-fold cross valiation (or out-of-bootstrap, ...), you can separate these influences as you can check whether predictions for the same case by different surrogate models are the same or not (= model instability). The more aggressively you optimize in the inner loop, the more important it is to make sure the optimization yields stable results (across the surrogate models of the outer loop).

Now one consequence of your limited number of cases is that the estimates of model performance will have high variance due to the low number of test cases. If you work with 0/1 loss and e.g. accuracy*, you can do some back-of-the-envelope calculations what uncertainty to expect.

  • outer loop has 30 cases. At the end, all those have been tested. The best possible case is that all were correctly - outer loop has 30 cases. At the end, all those have been tested. The best possible case is that all were correctly predicted. A binomial 95% confidence interval for 30 out of 30 cases yields roughly 90 - 100% accuracy.

  • say you do 6-fold in the outer loop (which you can do nicely stratified for your application). Then the optimization has 25 cases, and a correspondingly higher confidence interval for its performance estimates.

Without going into calculations, I think it unlikely that the expected differences across the models compared in the optimization step differ enough to reliably measure this difference with only 25 or 30 cases available.

Thus I recommend considering not to do any optimization but restrict yourself to a model where you can fix the hyperparameters by external knowledge (if there are any). We wrote a paper on a closely related topic that may be of interest:
Beleites, C. and Neugebauer, U. and Bocklitz, T. and Krafft, C. and Popp, J.: Sample size planning for classification models. Anal Chim Acta, 2013, 760, 25-33.
DOI: 10.1016/j.aca.2012.11.007

accepted manuscript on arXiv: 1211.1323

* there are other figures of merit, e.g. proper scoring rules, that are much better behaved from a statistical point of view. Nevertheless, they usually don't provide miracles, neither.


update: Plausibility check whether doing an optimization is worth while:

  • take an unoptimized model that doesn't need hyperparameters or that is calculated with manually set plausible hyperparameters (e.g. logistic regression without regularization, or random forest with manually set hyperparameters) and cross validate this (total = 30 tested cases).
    Let's assume you get 31 correct = 70% accuracy.
  • Check e.g. by simulated McNemar's tests how much better the optimized model would need to be in order to recognize the superiority.
    In the example, McNemar's test would be significant if the optimized model had 90% accuracy in the paired test without making any error that the reference model didn't make. Or it may make one new error at accuracy > 93%.
    It is then up to you to judge how realistic it is to expect such an improvement from the optimization and whether it is worth trying.

  • similarly, you can check with a proportion test simulation what performance you'd need to observe in order to have performance significantly better than, say, random guessing of the class label.

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  • $\begingroup$ Thank you for your reply and for the detailed answer! I am a bit novice to ML and may still miss important statistical subtleties and important points, please bear with me. 1) As for the fixing hyperparameters: the only hyperparameter I have is the \lambda parameter of the LASSO penalization. I use the logistic regression with L1 penalty term and to fix the parameter of penalization, I do CV. This is due to a (relatively) large number of features (15-25) which not all of them are useful. I tried to use Random Forest to find feature importance (it worked nicely as well) $\endgroup$ – Arnold Klein Nov 25 '16 at 17:53
  • $\begingroup$ but, I decided to stick to a simple L1 penalization.** Are there any better solutions to do feature selection besides the L1?** (and other sparse methods, elastic net, or L2). I tried univariate feature selection, ANOVA-SVM and similar. $\endgroup$ – Arnold Klein Nov 25 '16 at 17:57
  • $\begingroup$ As for the binomial confidence interval and the estimation metrics errors. Please correct me, if I'm wrong: the error of the classifier accuracy (or any other metric, such as f1 score for example) at first approximation might be computed using: 1.96\sqrt(p(p-1)/n), where p is the metric (ratio of successes). For example, my classification accuracy is 0.7, thus, according to this formula the 95% CI will be -+0.16 -> 0.7-+0.16. Due to the imbalance of classes, my baseline classifier, which always predicts the most prevalent class, will have accuracy 0.6 (which lies within the CI interval). $\endgroup$ – Arnold Klein Nov 25 '16 at 18:06
  • $\begingroup$ So basically it means, that all my classification methods do no do any better (from accuracy metric point of view) that a dummy classifier. But (!), f1 score of my classifier is approx. 0.7 +-0.16 from the same binomial CI), while f1 score of the dummy classifier is only 0.4. 1) Can we conclude something based on these two results? 2) Is there any flaw in my logic? P.S. I am delving into your paper, great reading! $\endgroup$ – Arnold Klein Nov 25 '16 at 18:09
  • $\begingroup$ Arnold, a) I'd need to dig into f1 in detail to answer your questions about it. I haven't used it so far because I did not yet understand the benefit of a mean between two figures of merit with the denominator having opposite meanings. If I have prior probabilities/class prevalence I'd go with pos + neg predictive values; otherwise I usually have to restrict myself to sensitivity + specificity, i.e. I basically look at the ROC. I tend to think that one should stick rather long to at least 2 figures of merit, and then have a good argumentation how to combine them into one. IMHO an unweighted... $\endgroup$ – cbeleites Dec 1 '16 at 14:08
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When you're looking at small samples it's easy to see variable results by changing the folds in your CV. That's just sampling variability. With modern computers there's rarely a reason to do anything but LOOCV.

Consider the most single split train v. test with half of your data. You might get a great result or a terrible one. How much do you trust it? The result can depend a lot on what you happened to hold out and didn't. I think that's what you're seeing and where your indecision about what to do is coming from. You don't select LOOCV because it gives you the result that looks the nicest. You select it because on average, and at the limit, it gives you the highest quality estimate. You might be select folds that genuinely are more accurate with for a particular sample but you'd never know that you really hit that without knowing the population model.

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  • $\begingroup$ thanks for your reply. I'm wondering if there is a rule of thumb, how to select the splitting ratios for small datasets? I'm following the work of Kohavi, where he argue that working with stratified splitting is far better than LOOVC. Any thoughts? $\endgroup$ – Arnold Klein Nov 18 '16 at 17:02

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