Suppose I have classifiers C_1 ... C_n that are disjoint in the sense that no two will return true on the same input (e.g. the nodes in a decision tree). I want to build a new classifier that is the union of some subset of these (e.g. I want to decide on which leaves of a decision tree to give a positive classification). Of course, in doing so there will be a trade off between sensitivity and positive predictive value. So I would like to see a ROC curve. In principle I could do this by enumerating all subsets of the classifiers and computing the resulting sensitivity and PPV. However, this is prohibitively expensive if n is more than 30 or so. On the other hand, there are almost certainly some combinations that are not Pareto optimal, so there might be some branch and bound strategy, or something, that avoids most of the computation in many cases.

I would like advice about whether this approach is likely to be fruitful and whether there is any work or if you have any ideas about efficiently computing the ROC curve in the situation above.

  • $\begingroup$ Are you classifying each input case to be either true or false ? $\endgroup$ Commented Aug 26, 2015 at 7:05
  • $\begingroup$ @image_doctor : yes $\endgroup$ Commented Aug 27, 2015 at 20:27
  • $\begingroup$ I"m not clear on , "... that are disjoint in the sense that no two will return true on the same input..." and you are classifying to a binary output, how you can have more than two classifiers in your ensemble, I'm probably missing something? $\endgroup$ Commented Sep 9, 2015 at 8:11
  • $\begingroup$ @image_doctor : You might be thinking that I am saying that no two classifiers return the same output on the same input. I am saying no two will return true. They can both return false. $\endgroup$ Commented Sep 12, 2015 at 14:35
  • 1
    $\begingroup$ Maybe this paper on a theoretically optimal way of combining classifiers for ROC (or papers that cite it) can help you to understand the state of art: M. Barreno, A. Cardenas, J.D. Tygar, Optimal ROC Curve for a Combination of Classifiers, Advances in Neural Information Processing Systems, 2008. $\endgroup$
    – Valentas
    Commented Oct 29, 2015 at 9:23

2 Answers 2


If I understood the question correctly, you have trained an algorithm that splits your data into $N$ disjoint clusters. Now you want to assign prediction $1$ to some subset of the clusters, and $0$ to the rest of them. And amont those subsets, you want to find the pareto-optimal ones, i.e. those who maximize true positive rate given fixed number of positive predictions (this is equivalent to fixing PPV). Is it correct?

This sounds very much like knapsack problem! Cluster sizes are "weights" and number of positive samples in a cluster are "values", and you want to fill your knapsack of fixed capacity with as much value as possible.

The knapsack problem has several algorihms for finding exact solutions (e.g. by dynamic programming). But a useful greedy solution is to sort your clusters in decreasing order of $\frac{value}{weight}$ (that is, share of positive samples), and take the first $k$. If you take $k$ from $0$ to $N$, you can very cheaply sketch your ROC curve.

And if you assign $1$ to the first $k-1$ clusters and to the random fraction $p\in[0,1]$ of samples in the $k$th cluster, you get the upper bound to the knapsack problem. With this, you can draw the upper bound for your ROC curve.

Here goes a python example:

import numpy as np
from itertools import combinations, chain
import matplotlib.pyplot as plt
n_obs = 1000
n = 10

# generate clusters as indices of tree leaves
from sklearn.tree import DecisionTreeClassifier
from sklearn.datasets import make_classification
from sklearn.model_selection import cross_val_predict
X, target = make_classification(n_samples=n_obs)
raw_clusters = DecisionTreeClassifier(max_leaf_nodes=n).fit(X, target).apply(X)
recoding = {x:i for i, x in enumerate(np.unique(raw_clusters))}
clusters = np.array([recoding[x] for x in raw_clusters])

def powerset(xs):
    """ Get set of all subsets """
    return chain.from_iterable(combinations(xs,n) for n in range(len(xs)+1))

def subset_to_metrics(subset, clusters, target):
    """ Calculate TPR and FPR for a subset of clusters """
    prediction = np.zeros(n_obs)
    prediction[np.isin(clusters, subset)] = 1
    tpr = sum(target*prediction) / sum(target) if sum(target) > 0 else 1
    fpr = sum((1-target)*prediction) / sum(1-target) if sum(1-target) > 0 else 1
    return fpr, tpr

# evaluate all subsets
all_tpr = []
all_fpr = []
for subset in powerset(range(n)):
    tpr, fpr = subset_to_metrics(subset, clusters, target)

# evaluate only the upper bound, using knapsack greedy solution
ratios = [target[clusters==i].mean() for i in range(n)]
order = np.argsort(ratios)[::-1]
new_tpr = []
new_fpr = []
for i in range(n):
    subset = order[0:(i+1)]
    tpr, fpr = subset_to_metrics(subset, clusters, target)

plt.scatter(all_tpr, all_fpr, s=3)
plt.plot(new_tpr, new_fpr, c='red', lw=1)
plt.title('All and Pareto-optimal subsets')

This code will draw a nice picture for you:

TPR, FPR, and optimal curve

The blue dots are (FPR, TPR) tuples for all $2^{10}$ subsets, and the red line connects (FPR, TPR) for the pareto-optimal subsets.

And now the bit of salt: you did not have to bother about subsets at all! What I did is sorted tree leaves by the fraction of positive samples in each. But what I got is exactly the ROC curve for the probabilistic prediction of the tree. This means, you cannot outperform the tree by hand-picking its leaves based on the target frequencies in the training set.

You can relax and keep using ordinary probabilistic prediction :)

  • $\begingroup$ Great idea. In theory there could still be an exponentially many possible numbers of "positive calls", but in practice it is probably not a problem. $\endgroup$
    – Valentas
    Commented Nov 9, 2017 at 15:47
  • $\begingroup$ Why exponential number of calls? I calculate value/weight for each cluster (takes linear time), sort them (N*log(N)), and evaluate TPR and FPR for each first K clusters (can be also made linear). $\endgroup$
    – David Dale
    Commented Nov 9, 2017 at 15:57
  • $\begingroup$ @Valentas, OK, I see your point. But still, if you give random prediction in some leaves, you can get to any point in the convex hull. So in this case the hull is the ROC itself. $\endgroup$
    – David Dale
    Commented Nov 9, 2017 at 21:09
  • $\begingroup$ @DavidDale, to summarize : 1) Every strategy that is pareto optimal with respect to (sensitivity, PPV) maximizes the number of true positives among strategies with that number of positive predictions. 2) This is the knapsack problem. 3) Choosing the nodes in order by number of positive examples/number of examples is known to be a good approximate solution to the knapsack problem. 4) But that's just the same as picking a threshold on the probabilities. $\endgroup$ Commented Nov 17, 2017 at 18:24
  • $\begingroup$ I will add that this is THE optimal solution to the knapsack problem when allowing fractional objects. And why not allow a mixed strategy in the original problem? $\endgroup$ Commented Nov 17, 2017 at 18:26

I might suggest that you use a greedy methods. Give a classifier to start, you will include the classifier that make the ensemble get the best performance improvement. If no improvement could be get through include more classifiers, then stop. You will start with every classifiers. The complexity will be at most N*N.

I have one more question, What do you mean by "Pareto optimal", especially in your context? I found from wiki this explanation, https://en.wikipedia.org/wiki/Pareto_efficiency

through reallocation, improvements can be made to at least one participant's well-being without reducing any other participant's well-being.

The improvement for the Pareto efficiency is for each participant, which might correspond to each classifier. How do you define the improvement over one classifier?

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    $\begingroup$ What I mean is this: if I have ensembles 1 and 2, with (sensitivity, positive predictive value) = (.90, .80) and (.97, .93) respectively, then 1 is not Pareto optimal, because there is another ensemble, namely 2, that beats it in every way. Regarding your proposed algorithm : there is a tradeoff between sensitivity and PPV, so "the ensemble get the best performance improvement" is not well defined. $\endgroup$ Commented Aug 30, 2015 at 14:49

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