CODEWITHSUNDEEP
pythonnumpy
Tom
Tom

Reputation: 33

Loss of dimension when using numpy array inside masked array

I had a Masked array and a Numpy boolean array containing 3 dimensional values. However when I used indexing of numpy array inside masked array it led to loss of dimensions. I couldn't figure out the reason:

Masked_array = [[[--, 1, --],
                 [--, 1, --],
                 [--, 1, --]]]

Running this line gave me

masked_array = masked_array.mask
mm = ~np.logical_and.accumulate(masked_array)
list(masked_array[mm])

the output as [1, 1, 1] instead of [[1] [1] [1]] I couldnt understand the error and tried various methods. Could you please help me in clarifying the doubt. Thanks

Upvotes: 2

Views: 178

Answers (1)

yatu
yatu

Reputation: 88226

When indexing a 2D array with a mask with the same shape, you get a 1D array:

a = np.random.random((4,4))
a[np.random.choice([True,False], (4,4))].shape
# (7,)

The original shape is not preserved because as a result from the boolean indexing you'd probably get a jagged array, which numpy does not support. So by default, it just flattens out the result for you as in the example above.

If you know your know that as a result you'll be indexing a column, and you want to preserve the 2D shape, you can always add a new axis:

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

masked_array = np.ma.masked_array(a, a==0)

mask = masked_array.mask
mm = ~np.logical_and.accumulate(mask)

masked_array[mm,None].data
array([[1],
       [1],
       [1]])

Though as mentioned, you'll always end up with a squeezed array, which you'll have to reshape according to your needs.

Upvotes: 1

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