sklearn LabelBinarizer returns vector when there are 2 classes

this should do it

labels = ['yes', 'no', 'no', 'yes']
np.array([[1,0] if l=='yes' else [0,1] for l in labels])

After so many years, now there is a nice elegant solution. The MultiLabelBinirazer.

The difference with LabelBinarizer is that treats all features as multi-class.

In scikit-learn 0.18 it cannot handle unseen values.

In scikit-learn 0.20 it can handle unseen values and assign them [0,0,...,0,0] which fixed recently.

I think there is no direct way to do it especially if you want to have inverse_transform.

But you can use numpy to construct the label easily

In [18]: import numpy as np

In [19]: from sklearn.preprocessing import LabelBinarizer

In [20]: lb = LabelBinarizer()

In [21]: label = lb.fit_transform(['yes', 'no', 'no', 'yes'])

In [22]: label = np.hstack((label, 1 - label))

In [23]: label
Out[23]:
array([[1, 0],
       [0, 1],
       [0, 1],
       [1, 0]])

Then you can use inverse_transform by slicing the first column

In [24]: lb.inverse_transform(label[:, 0])
Out[24]:
array(['yes', 'no', 'no', 'yes'],
      dtype='<U3')

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Based on the above solution, you can write a class that inherits LabelBinarizer, which makes the operations and results consistent for both binary and multiclass case.

from sklearn.preprocessing import LabelBinarizer
import numpy as np

class MyLabelBinarizer(LabelBinarizer):
    def transform(self, y):
        Y = super().transform(y)
        if self.y_type_ == 'binary':
            return np.hstack((Y, 1-Y))
        else:
            return Y

    def inverse_transform(self, Y, threshold=None):
        if self.y_type_ == 'binary':
            return super().inverse_transform(Y[:, 0], threshold)
        else:
            return super().inverse_transform(Y, threshold)

Then

lb = MyLabelBinarizer()
label1 = lb.fit_transform(['yes', 'no', 'no', 'yes'])
print(label1)
print(lb.inverse_transform(label1))
label2 = lb.fit_transform(['yes', 'no', 'no', 'yes', 'maybe'])
print(label2)
print(lb.inverse_transform(label2))

gives

[[1 0]
 [0 1]
 [0 1]
 [1 0]]
['yes' 'no' 'no' 'yes']
[[0 0 1]
 [0 1 0]
 [0 1 0]
 [0 0 1]
 [1 0 0]]
['yes' 'no' 'no' 'yes' 'maybe']