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Haar分解的矩阵归一化

[英]Matrix Normalization for Haar Decomposition

I've implemented the Wavelet transformation described in the paper below for image retrieval. 我已经实现了以下论文中描述的小波变换,用于图像检索。

http://grail.cs.washington.edu/projects/query/ http://grail.cs.washington.edu/projects/query/

Overall, the system works great but my question doesn't concern the paper at all; 总体而言,该系统运行良好,但我的问题根本与论文无关。 rather, the proper way of normalizing the intensity matrix between values of [0, 1]. 而是在[0,1]值之间归一化强度矩阵的正确方法。

Normally, I would just loop through the matrix, find the largest value, loop through the matrix again, and then divide by the maximum value. 通常,我将遍历矩阵,找到最大值,再次遍历矩阵,然后除以最大值。 But since I am trying to compare images, shouldn't the divisor be 255 regardless of what the maximum intensity value in the matrix is? 但是由于我要比较图像,所以不管矩阵中的最大强度值是多少,除数都不应该是255吗?

For example, if the maximum intensity value is only 200, when normalizing, do I divide by 200 or 255? 例如,如果最大强度值仅为200,则在归一化时,我要除以200还是255?

I don't know much about images, but it sounds like you are trying to have two values for two operations: one for comparison across images, and the other normalized for the image itself. 我对图像了解不多,但是听起来您正在为两个操作尝试使用两个值:一个用于跨图像比较,另一个用于图像本身标准化。

Do you need both? 你们都需要吗? If so, save both. 如果是这样,请同时保存两者。 If you only need one or the other, save whichever you need. 如果只需要一个,则保存所需的任何一个。

Again, not knowing anything about images, I would guess that the cross-image comparison does not require normalized values at all, (unless getting all values between 0 and 1 is a requirement), because the absolute values themselves are fine for comparing one image to another. 同样,我对图像一无所知,我猜想跨图像比较根本不需要归一化的值(除非要求所有值都介于0和1之间),因为绝对值本身对于比较一张图像是很好的到另一个。

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