CODEWITHSUNDEEP
pythonarraysnumpyreshapetranspose
janchytry
janchytry

Reputation: 350

Reshape 3D numpy array stack of images to vectors and back to 3D, preserving image structure

I observe the change from repeated imagery in time. I have a 3D numpy array, which should represent a stack of 2 images, each 3x3 px (for simplicity - shape: 2×3×3):

x = np.array([
    [
        [1,2,3],
        [4,5,6],
        [7,8,9]
    ],
    [
        [10,11,12],
        [13,14,15],
        [16,17,18]
    ]
])

I transposed and reshaped it such, that now I have "pixel vectors in time", like this:

y = x.transpose(2,1,0).reshape(-1, x.shape[0])

>>> array([[ 1, 10],
       [ 4, 13],
       [ 7, 16],
       [ 2, 11],
       [ 5, 14],
       [ 8, 17],
       [ 3, 12],
       [ 6, 15],
       [ 9, 18]])

I made a few scientific computations and ended up with Boolean values for each pixel vector, assessing certain characteristics:

>>> array([[ True, False, True],
       [ True, True, True],
       [False, False, True],
       [False, False, False],
       [ True, False, True],
       [ True, False, True],
       [ True, False, True],
       [ True, True, True],
       [False, False, True]])

Now, how do I reshape/transpose this array to a shape of (number of characteristics)×3×3 and preserve a logical order of pixels of a single image, as in array x?

I will try to clarify further if necessary. Thanks!

Upvotes: 0

Views: 518

Answers (1)

yatu
yatu

Reputation: 88236

You can use y (not the mask) to verify that you successfully reconstruct x, but essentially, keep track of the shape of the intermediate array before reshaping, and just reshape back to that shape and transpose again:

y_t = x.transpose(2,1,0) # same as x.T
y = y_t.reshape(-1, x.shape[0])

y.reshape(y_t.shape).T
array([[[ 1,  2,  3],
        [ 4,  5,  6],
        [ 7,  8,  9]],

       [[10, 11, 12],
        [13, 14, 15],
        [16, 17, 18]]])

Otherwise you can also just use the initial array's shape as a reference:

y = x.transpose(2,1,0).reshape(-1, x.shape[0])
y.reshape(x.shape[::-1]).T
array([[[ 1,  2,  3],
        [ 4,  5,  6],
        [ 7,  8,  9]],

       [[10, 11, 12],
        [13, 14, 15],
        [16, 17, 18]]])

Upvotes: 1

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