Making Integer Vector into Binary Array with 1 at location of integer

Question

Suppose that I have a numpy array of integers like this: [0,1,2,0,0,1,3]. How can I make it into a binary array whose non-zeros entries are defined by [0,1,2,0,0,1,3], i.e.:

array([[1., 0., 0., 0.],
       [0., 1., 0., 0.],
       [0., 0., 1., 0.],
       [1., 0., 0., 0.],
       [1., 0., 0., 0.],
       [0., 1., 0., 0.],
       [0., 0., 0., 1.]])

I'm currently using the following code to do it:

dummy_indices = np.array([0,1,2,0,0,1,3])
dummy_array = np.zeros([dummy_indices.shape[0],(1+np.max(dummy_indices))])
for i in range(dummy_array.shape[0]):
    dummy_array[i,dummy_indices[i]] =1

Is there a faster loop-free way to go about it?

Bruno Mello · Accepted Answer · 2020-08-09 14:43:22Z

1

You could do something like this:

x = np.array([0,1,2,0,0,1,3]) #your init array
y = np.zeros(shape=(x.shape[0], x.max()+1))
y[np.arange(x.shape[0]), x] = 1

y
array([[1., 0., 0., 0.],
       [0., 1., 0., 0.],
       [0., 0., 1., 0.],
       [1., 0., 0., 0.],
       [1., 0., 0., 0.],
       [0., 1., 0., 0.],
       [0., 0., 0., 1.]])

answered Aug 9, 2020 at 14:43

Bruno Mello

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