如何用八度正常化的形象呢？ [英] How to normalize an image using Octave?

问题描述

在他们的文章中描述的维奥拉 - 琼斯目标检测框架（稳健实时人脸检测由中提琴和琼斯），它是说：

In their paper describing Viola-Jones object detection framework (Robust Real-Time Face Detection by Viola and Jones), it is said:

用于培训的所有示例子窗口被方差标准化，以尽量减少不同的照明条件的影响。

All example sub-windows used for training were variance normalized to minimize the effect of different lighting conditions.

我的问题是，如何实现八度图像归？

My question is "How to implement image normalization in Octave?"

我不是在寻找特定的实现，中提琴和放大器;琼斯使用，但类似的一个，产生几乎相同的输出。我一直在关注很多哈尔培训教程（试图检测手），但尚未能够输出一个很好的检测器（XML）。

I'm NOT looking for the specific implementation that Viola & Jones used but a similar one that produces almost the same output. I've been following a lot of haar-training tutorials(trying to detect a hand) but not yet able to output a good detector(xml).

我一直试图联系作者，但还是没有反应呢。

I've tried contacting the authors, but still no response yet.

推荐答案

我已经回答了如何在一般准则，以它这个线程。

I already answered how to to it in general guidelines in this thread.

下面是如何做到的方法1倍频程（规范化为标准的正常偏差）（展示一个随机矩阵 A ，当然，也可以应用到任何矩阵，如何将图片重新presented）：

Here is how to do method 1 (normalizing to standard normal deviation) in octave (Demonstrating for a random matrix A, of course can be applied to any matrix, which is how the picture is represented):

>>A = rand(5,5)
A =

   0.078558   0.856690   0.077673   0.038482   0.125593
   0.272183   0.091885   0.495691   0.313981   0.198931
   0.287203   0.779104   0.301254   0.118286   0.252514
   0.508187   0.893055   0.797877   0.668184   0.402121
   0.319055   0.245784   0.324384   0.519099   0.352954

>>s = std(A(:))
s =  0.25628
>>u = mean(A(:))
u =  0.37275
>>A_norn = (A - u) / s
A_norn =

  -1.147939   1.888350  -1.151395  -1.304320  -0.964411
  -0.392411  -1.095939   0.479722  -0.229316  -0.678241
  -0.333804   1.585607  -0.278976  -0.992922  -0.469159
   0.528481   2.030247   1.658861   1.152795   0.114610
  -0.209517  -0.495419  -0.188723   0.571062  -0.077241

在上面的使用：

• 要获得其矩阵的标准偏差： S = STD（A（：））
• 要获得其矩阵的平均值： U =平均（A（：））
• 然后下面的公式 A'[I] [J] =（A [I] [J] - U）​​/ S 的 量化版本： A_norm =（A - U）​​/ S

• To get the standard deviation of the matrix: s = std(A(:))
• To get the mean value of the matrix: u = mean(A(:))
• And then following the formula A'[i][j] = (A[i][j] - u)/s with the vectorized version: A_norm = (A - u) / s

使用矢量归正火这也很简单：

Normalizing it with vector normalization is also simple:

>>abs = sqrt((A(:))' * (A(:)))
abs =  2.2472
>>A_norm = A / abs
A_norm =

   0.034959   0.381229   0.034565   0.017124   0.055889
   0.121122   0.040889   0.220583   0.139722   0.088525
   0.127806   0.346703   0.134059   0.052637   0.112369
   0.226144   0.397411   0.355057   0.297343   0.178945
   0.141980   0.109375   0.144351   0.231000   0.157065

在abvove：

• ABS 是向量（长度），其计算公式与矢量乘法（ A（:)'* A（绝对值：）其实就是和（A [i] [j]的^ 2））
• 然后，我们用它来归一化矢量所以这将是长度为1。

• abs is the absolute value of the vector (its length), which is calculated with vectorized multiplications (A(:)' * A(:) is actually sum(A[i][j]^2))
• Then we use it to normalize the vector so it will be of length 1.

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