numpy filter across dimensions for specific values

Question

I'm trying to filter an image with shape (224, 224, 3) in RGB which has been loaded into a numpy array, where the last axis is the channels.

The logic is simple: look for specific pixel values across the 3 channels and assign a label to a new 1-Dim/1-Channel array. This is used for Segmentation Networks, where a pixel-wise label is assigned to represent the area of the object in the image.

The current loop looks like this:

mask     = Image.open(args.input).convert('RGB')
array    = np.array(img_mask)

h, w, c = array.shape
bg = np.zeros((w, h))
fg = np.zeros((w, h))

for i in range(0, h):
    for k in range(0, w):
        if np.array_equal(array[i,k], [0, 0, 0]):
            bg[i,k] = 1
        elif np.array_equal(array[i,k], [255, 200, 0]):
            fg[i,k] = 2

label = np.stack((bg, fg), axis=0)

This produces a (2, 224, 224) where axis = 0 is the channels and then axis=1 has the pixel labels.

I have tried to replace the two for loops using np.where however, the result is still a 3D array, whereas what I want is a 1D array, created for each pixel coordinate which matches the condition:

bg    = np.where(array == [0, 0, 0],     1, 0)
fg    = np.where(array == [255, 200, 0], 2, 0)
label = np.stack((bg, fg), axis=0)

However here I get: (2, 224, 224, 3)

The current for loop implementation is painfully slow, and I suspect this can be done using built-in numpy functions, but I've failed to achieve the same result so far.

Did you try an intermediate solution : first reshape in (224x224, 3) with native numpy function (C speed), then use where on the last dimension, then reshape again.? Personnaly I'm not a big fan of too many dimensions array, it's always create a lot of problems. — Guilhem L.
– Guilhem L., Commented Feb 11, 2020 at 11:35
@GuilhemL. No I didn't because depending on the channel colors I'd have to be careful which dimensions I'm reducing? Unless I've not understood this well; if the same values are broadcast across all 3 axes, then I could just reduce/remove to only one. — Ælex
– Ælex, Commented Feb 11, 2020 at 11:44

Quang Hoang · Accepted Answer · 2020-02-11 12:19:07Z

2

I'm sure you can use select for masking, and ravel to get 1-D array:

array = np.array([[[0,0,0],[128,128,128]],
                  [[128,128,128],[255,255,0]]
                 ])

label = np.select([(array==(0,0,0)).all(-1),      # all make sure all channels are identical
                   (array==(255,255,0)).all(-1)],
                  [1,2], 0)

print(label.ravel())

Output:

array([1, 0, 0, 2])

answered Feb 11, 2020 at 12:19

Quang Hoang

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11 Comments

Ælex Over a year ago

Is there a way to do the opposite? E.g., where the condition doesn't match, to select.

Quang Hoang Over a year ago

Not sure I understand your question, wouldn't array != (0,0,0) work?

Daniel F Over a year ago

or use the default (0 here), if you just want to select everything that doesn't match the other queries.

Ælex Over a year ago

yes, so I suppose it would be label = np.select([(array==(0,0,0)).all(-1),((array!=(0,0,0)).all(-1)], [1,2], 0) ?

Quang Hoang Over a year ago

Yes, either at assignment of fg or of label.

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