1d numpy array to blocks in a matrix

Question

I have a 1D array that I need to reshape() into a 2D array, but the reshape is tricky because it seems like some kind of grouping is needed.

I have tried a few different series of reshape() and transpose() combinations, but I have not landed on the correct arrangement of values.

I have provided two arrays below. I would like to reshape the 1D array a to have the same shape as the 2D array b. What makes this trick is that I want the 4 ones in the middle. I made some attempts to make 2x2 blocks for every 4 values then arrange into the 6x6 array, but I did not have success.

a = np.array([0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0])

b = np.array([
    [0, 0, 0, 0, 0, 0],
    [0, 0, 0, 0, 0, 0],
    [0, 0, 1, 1, 0, 0],
    [0, 0, 1, 1, 0, 0],
    [0, 0, 0, 0, 0, 0],
    [0, 0, 0, 0, 0, 0]])

Here are the shapes

That doesn't look like a reshape. in a the ones are contiguous and in b the ones are not contiguous. It seems like some shuffling of items will be needed but which items will get shuffled where. Do your real arrays look like that - ones and zeros? or is that an example? — wwii
– wwii, Commented Mar 13, 2021 at 20:05

hpaulj · Accepted Answer · 2021-03-13 20:07:49Z

2

This does it:

In [327]: a = np.array([0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 1, 1, 1, 1, 0, 0, 0,
     ...:  0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0])
     ...: 
In [328]: a.reshape(3,3,2,2)
Out[328]: 
array([[[[0, 0],
         [0, 0]],

        [[0, 0],
         [0, 0]],

        [[0, 0],
         [0, 0]]],


       [[[0, 0],
         [0, 0]],

        [[1, 1],
         [1, 1]],

        [[0, 0],
         [0, 0]]],


       [[[0, 0],
         [0, 0]],

        [[0, 0],
         [0, 0]],

        [[0, 0],
         [0, 0]]]])

the (2,2) block of 1s is visible, but I need to switch the middle 2 dimensions:

In [329]: a.reshape(3,3,2,2).transpose(0,2,1,3)
Out[329]: 
array([[[[0, 0],
         [0, 0],
         [0, 0]],

        [[0, 0],
         [0, 0],
         [0, 0]]],


       [[[0, 0],
         [1, 1],
         [0, 0]],

        [[0, 0],
         [1, 1],
         [0, 0]]],


       [[[0, 0],
         [0, 0],
         [0, 0]],

        [[0, 0],
         [0, 0],
         [0, 0]]]])

now back to 2d:

In [330]: a.reshape(3,3,2,2).transpose(0,2,1,3).reshape(6,6)
Out[330]: 
array([[0, 0, 0, 0, 0, 0],
       [0, 0, 0, 0, 0, 0],
       [0, 0, 1, 1, 0, 0],
       [0, 0, 1, 1, 0, 0],
       [0, 0, 0, 0, 0, 0],
       [0, 0, 0, 0, 0, 0]])

The basic steps of reshape, transpose, followed by reshape is common to this kind of problem. The transpose details very.

answered Mar 13, 2021 at 20:07

hpaulj

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2 Comments

Alex Over a year ago

This is perfect, but I don't really understand the operation in the transpose. When I think transpose I think of flipping the rows and columns of a 2d matrix. Could you explain how you knew what to do with the transpose?

hpaulj Over a year ago

I wanted to keep the first and last dimensions the same (last more obviously). That left the middle 2 for switching.

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