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我正在尝试使用 cv2 LUT 在 Python 中进行图像传输。LUT 需要与图像具有相同数量的通道。但我无法解决一个错误：

image1Transfered = cv2.LUT(image1, lut) cv2.error: /build/buildd/opencv-2.3.1/modules/core/src/convert.cpp:1037: error: (-215) (lutcn == cn || lutcn == 1) && lut.total() == 256 && lut.isContinuous() && (src.depth() == CV_8U || src.depth() == CV_8S) 在函数 LUT

这是python代码，我相信我可以将图像拆分为多个单通道并分别应用LUT。但这是对资源的浪费。

    #!/usr/bin/python
    import sys
    import cv2
    import numpy as np

    image1 = cv2.imread("../pic1.jpg", 1)
    # apply look up table
    lut = np.arange(255, -1, -1, dtype = image1.dtype )
    lut = np.column_stack((lut, lut, lut))
    image1Converted = cv2.LUT(image1, lut)  # <-- this is where it fails

感谢您的时间。

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2 回答 2

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您正在使用np.column_stack()创建 3 通道图像，但这不是正确的功能。您必须使用np.dstack()或cv2.merge()。然后它工作正常。

例如：

In [3]: x
array([[0, 1, 2],
       [3, 4, 5],
       [6, 7, 8]])

In [5]: np.column_stack((x,x,x))
array([[0, 1, 2, 0, 1, 2, 0, 1, 2],
       [3, 4, 5, 3, 4, 5, 3, 4, 5],
       [6, 7, 8, 6, 7, 8, 6, 7, 8]])

In [6]: np.dstack((x,x,x))
array([[[0, 0, 0],
        [1, 1, 1],
        [2, 2, 2]],

       [[3, 3, 3],
        [4, 4, 4],
        [5, 5, 5]],

       [[6, 6, 6],
        [7, 7, 7],
        [8, 8, 8]]])

In [11]: cv2.merge((x,x,x))
array([[[0, 0, 0],
        [1, 1, 1],
        [2, 2, 2]],

       [[3, 3, 3],
        [4, 4, 4],
        [5, 5, 5]],

       [[6, 6, 6],
        [7, 7, 7],
        [8, 8, 8]]], dtype=int32)
于 2013-05-16T09:06:45.707 回答
0

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: 

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.

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: 

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]]])

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.)

于 2013-06-06T04:34:03.477 回答