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Thank you Abid, I like your blog. I am going through you Python CV post one by one. It's a great help on learning Python opencv. You did a very nice work. Here is what I ended up with: <pre><code>lut3 = np.column_stack((lut, lut, lut)) lutIdxDot = np.array( [0, 1, 2], dtype=int) lutIdx0 = np.zeros( image1.shape[0] * image1.shape[1], dtype=int) lutIdx1 = np.ones( image1.shape[0] * image1.shape[1], dtype=int) lutIdx2 = lutIdx1 * 2 lutIdx = np.column_stack((lutIdx0, lutIdx1, lutIdx2)) lutIdx.shape = image1.shape image1Rev = lut3[image1, lutIdx] # numpy indexing will generate the expected LUT result. </code></pre> I used the numpy indexing to get the result. I didn't use the cv LUT function. The performance is unknown to me. The last line of code was strange to me at first. The indexing is a very interesting feature of numpy. When the code run to the last line, the lut3 is: <pre><code>ipdb> p lut3 array([[255, 255, 255], [254, 254, 254], [253, 253, 253], [252, 252, 252], ... [ 2, 2, 2], [ 1, 1, 1], [ 0, 0, 0]], dtype=uint8) ipdb> p lutIdx array([[[0, 1, 2], [0, 1, 2], [0, 1, 2], ..., ..., [0, 1, 2], [0, 1, 2], [0, 1, 2]]]) </code></pre> lutIdx has the same shape of the image1. The lut3[image1, lutIdx] is asking for an array as the result that its shape is same with image1, and lutIdx. Its values are from lut3. For each item of image1, use lut[image1's Value of that spot, lutIdx's Value of that spot] to find that output value. (I wish I can draw a diagram.)
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