Fix Triplets predict function. Made a test to show the point.

Author: mvargas33

metric_learn/base_metric.py

@@ -602,7 +602,9 @@ def predict(self, triplets):
     prediction : `numpy.ndarray` of floats, shape=(n_constraints,)
       Predictions of the ordering of pairs, for each triplet.
     """
-    return np.sign(self.decision_function(triplets))
+    return np.array([-1 if (t <= 0) else 1 for t in
+                   self.decision_function(triplets)])
+    #return np.sign(self.decision_function(triplets))
 
   def decision_function(self, triplets):
     """Predicts differences between sample distances in input triplets.

test/test_triplets_classifiers.py

@@ -6,7 +6,7 @@
 from metric_learn.sklearn_shims import set_random_state
 from sklearn import clone
 import numpy as np
-
+from numpy.testing import assert_array_equal
 
 @pytest.mark.parametrize('with_preprocessor', [True, False])
 @pytest.mark.parametrize('estimator, build_dataset', triplets_learners,
@@ -63,3 +63,49 @@ def test_accuracy_toy_example(estimator, build_dataset):
   # we force the transformation to be identity so that we control what it does
   estimator.components_ = np.eye(X.shape[1])
   assert estimator.score(triplets_test) == 0.25
+
+
+@pytest.mark.parametrize('estimator, build_dataset', triplets_learners,
+                         ids=ids_triplets_learners)
+def test_no_zero_prediction(estimator, build_dataset):
+  """
+  Test that all predicted values are in {-1, 1}, even when the
+  distance d(x,y) and d(x,z) is the same for a triplet of the
+  form (x, y, z).
+  """
+  # Dummy fit
+  triplets, _, _, X = build_dataset(with_preprocessor=False)
+  # Force 3 dimentions only, to use cross product and get easy orthogonal vectors.
+  triplets = np.array([ [t[0][:3], t[1][:3], t[2][:3]] for t in triplets])
+  X = np.array([x[:3] for x in X])
+  # Dummy fit
+  estimator = clone(estimator)
+  set_random_state(estimator)
+  estimator.fit(triplets)
+  # we force the transformation to be identity, to force euclidean distance
+  estimator.components_ = np.eye(X.shape[1])
+
+  # Get two orthogonal vectors in respect to X[1]
+  k = X[1]/np.linalg.norm(X[1])  # Normalize first vector
+  x = X[2] - X[2].dot(k) * k  # Get random orthogonal vector
+  x /= np.linalg.norm(x)  # Normalize
+  y = np.cross(k, x)  # Get orthogonal vector to x
+  # Assert these orthogonal vectors are different
+  with pytest.raises(AssertionError):
+    assert_array_equal(X[1], x)
+  with pytest.raises(AssertionError):
+    assert_array_equal(X[1], y)
+  # Assert the distance is the same for both
+  assert estimator.get_metric()(X[1], x) == estimator.get_metric()(X[1], y)
+
+  # Form the three scenarios where predict() gives 0 with numpy.sign
+  triplets_test = np.array(  # Critical examples
+    [[X[0], X[2], X[2]],
+     [X[1], X[1], X[1]],
+     [X[1], x, y]
+    ])
+  # Predict
+  predictions = estimator.predict(triplets_test)
+  # Count non -1 or 1 values
+  not_valid = [e for e in predictions if e not in [-1, 1]]
+  assert len(not_valid) == 0