Compare 2d numpy array to itself to form 3d array

Question

I've got a simple 2d numpy array like this

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

What I'm trying to do is multiply each row of the array by itself to form a 3d array, (so that 1*1 = 1 and anything else will be 0, essentailly an and function):

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

SCB · Accepted Answer · 2015-08-20 00:44:24Z

1

Turns out the answer is pretty simple. If the array is x:

x[np.newaxis,:,:]*x[:,np.newaxis,:]

And this works with things such as logical expressions, if you want to just compare on identical elements in each row, you could do something like:

np.logical_not(np.logical_xor(x[np.newaxis,:,:], x[:,np.newaxis,:]))

Which would return:

array([[[ True,  True,  True],
        [ True, False, False],
        [False, False,  True],
        [False,  True, False],
        [ True,  True, False]],

       [[ True, False, False],
        [ True,  True,  True],
        [False,  True, False],
        [False, False,  True],
        [ True, False,  True]],

       ...,
       ...,
       ...], dtype=bool)

answered Aug 20, 2015 at 0:44

SCB

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