The problem that you have described is not quite a multi-label classification problem. Multi-label classification allows an object to have any combination of labels, including no labels at all. So in your case where there are 2 labels, it would allow 4 possible outcomes. That being said, you may be able to adapt a multi-label classifier to exclude to "no label" case.
What you describe sounds more like an ordinal regression problem. That is, there are 3 labels, where one label is logically "in between" the other two. See the link for details.
The Multi-label algorithm accepts a binary mask over multiple labels. So, for example, you could do something like this:
data = [
[[0.1 , 0.6, 0.0, 0.3], 1, 10, 0, 0, 0],
[[0.7 , 0.3, 0.0, 0.0], 0, 7, 22, 0, 0],
[[0.0 , 0.0, 0.6, 0.4], 0, 0, 6, 0, 20],
#...
]
X = np.array([d[1:] for d in data])
yvalues = np.array([d[0] for d in data])
# Create a binary array marking values as True or False
from sklearn.preprocessing import MultiLabelBinarizer
Y = MultiLabelBinarizer().fit_transform(yvalues)
clf = OneVsRestClassifier(SVC(kernel='poly'))
clf.fit(X, Y)
clf.predict(X) # predict on a new X
The result for each prediction will be an array of 0s and 1s marking which class labels apply to each row input sample.
Given your data, though, I'm not sure this is what you want to do. For example, the third point has zero listed twice, which makes me think that you're not predicting multiple labels in an unordered OneVsRest manner, but actually predicting multiple ordered columns of labels: in that case, it might make sense to do a separate classification for each, e.g.
X = np.array([d[1:] for d in data])
Y = np.array([d[0] for d in data])
clfs = [SVC().fit(X, Y[:, i]) for i in range(Y.shape[1])]
Ypred = np.array([clf.predict(X) for clf in clfs]).T
With other classifiers, such as RandomForestClassifier, you can do this column-by-column prediction in one operation: e.g.
X = np.array([d[1:] for d in data])
Y = np.array([d[0] for d in data])
RandomForestClassifier().fit(X, Y).predict(X)
Of course, the array passed to predict should be on something different than the array passed to fit, but hopefully this makes the distinction clear.
Best Answer
Predict each label independently.
Because objects may have more than one label.
Thus, this isn't really a multi-class problem, because the classes aren't disjoint.