加入多个重叠立方体的矢量化方法

Question

我用滑动窗口对一个大的矩形图像做深入的学习。图像具有形状(高度、宽度)。

预测输出为形状(高度、宽度、prediction_probability)。我的预测是在重叠窗口中输出的，我需要将这些窗口加在一起，以便对整个输入图像逐像素地进行预测。窗口重叠超过2个像素(高度，宽度)。

在C++中，我以前做过这样的工作，创建一个大型的结果索引，然后将所有的ROI添加到一起。

#include <opencv2/core/core.hpp>
using namespace std;  

template <int numberOfChannels>
static void AddBlobToBoard(Mat& board, vector<float> blobData,
                                            int blobWidth, int blobHeight,
                                            Rect roi) {
 for (int y = roi.y; y < roi.y + roi.height; y++) {
    auto vecPtr = board.ptr< Vec <float, numberOfChannels> >(y);
    for (int x = roi.x; x < roi.x + roi.width; x++) {
      for (int channel = 0; channel < numberOfChannels; channel++) {
          vecPtr[x][channel] +=
            blobData[(band * blobHeight + y - roi.y) * blobWidth + x - roi.x];}}}

在Python中有一种矢量化的方法吗？

Answer 1

编辑：

@Kevin，如果你正在训练一个网络，你应该用一个完全连接的层来完成这个步骤。那就是说..。

我有一个非矢量化的解决方案，如果你想要的话。任何解决方案都是内存密集型的。在我的笔记本电脑上，它对CIFAR大小的灰色图像(32x32)起着快速的作用。也许关键的一步可以由聪明的人来引导。

首先，使用arr将测试数组win拆分为windows win。这是测试数据。

>>> import numpy as np
>>> from skimage.util.shape import view_as_windows as viewW
>>> arr = np.arange(20).reshape(5,4)
>>> win = viewW(arr, (3,3))
>>> arr # test data 
array([[ 0,  1,  2,  3],
       [ 4,  5,  6,  7],
       [ 8,  9, 10, 11],
       [12, 13, 14, 15],
       [16, 17, 18, 19]])
>>> win[0,0]==arr[:3,:3] # it works. 
array([[ True,  True,  True],
       [ True,  True,  True],
       [ True,  True,  True]])

现在，要重新组合，生成一个带有(5,4,6)形状的输出数组(5,4,6)。6是win中的窗口数，(5,4)是arr.shape。我们将沿着-1轴在每个切片中通过一个窗口填充这个数组。

# the array to be filled
out = np.zeros((5,4,6)) # shape of original arr stacked to the number of windows 

# now make the set of indices of the window corners in arr 
inds = np.indices((3,2)).T.reshape(3*2,2)

# and generate a list of slices. each selects the position of one window in out 
slices = [np.s_[i[0]:i[0]+3:1,i[1]:i[1]+3:1,j] for i,j in zip(inds,range(6))] 

# this will be the slow part. You have to loop through the slices. 
# does anyone know a vectorized way to do this? 
for (ii,jj),slc in zip(inds,slices):
    out[slices] = win[ii,jj,:,:]

现在，out数组在其正确的位置包含所有窗口，但在-1轴上分隔为窗格。要提取原始数组，可以将不包含零的所有元素平均在此轴上。

>>> out = np.true_divide(out.sum(-1),(out!=0).sum(-1))
>>> # this can't handle scenario where all elements in an out[i,i,:] are 0
>>> # so set nan to zero 

>>> out = np.nan_to_num(out)
>>> out 
array([[ 0.,  1.,  2.,  3.],
       [ 4.,  5.,  6.,  7.],
       [ 8.,  9., 10., 11.],
       [12., 13., 14., 15.],
       [16., 17., 18., 19.]])

你能想出一种用矢量化的方式对一组切片进行操作的方法吗？

合在一起：

def from_windows(win):
    """takes in an arrays of windows win and returns the original array from which they come"""
    a0,b0,w,w = win.shape # shape of window
    a,b = a0+w-1,b0+w-1 # a,b are shape of original image 
    n = a*b # number of windows 
    out = np.zeros((a,b,n)) # empty output to be summed over last axis 
    inds = np.indices((a0,b0)).T.reshape(a0*b0,2) # indices of window corners into out 
    slices = [np.s_[i[0]:i[0]+3:1,i[1]:i[1]+3:1,j] for i,j in zip(inds,range(n))] # make em slices 
    for (ii,jj),slc in zip(inds,slices): # do the replacement into out 
        out[slc] = win[ii,jj,:,:]
    out = np.true_divide(out.sum(-1),(out!=0).sum(-1)) # average over all nonzeros 
    out = np.nan_to_num(out) # replace any nans remnant from np.alltrue(out[i,i,:]==0) scenario
    return out # hope you've got ram 

而测试：

>>> arr = np.arange(32**2).reshape(32,32)
>>> win = viewW(arr, (3,3))
>>> np.alltrue(arr==from_windows(win))
True
>>> %timeit from_windows(win)
6.3 ms ± 117 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)

实际上，这还不够快，不能让你继续训练。