python – 添加多个重叠立方体的矢量化方式

2019年7月28日 245次阅读

我正在使用滑动窗口对大型矩形图像进行深度学习.图像具有形状(高度,宽度).

预测输出是形状的ndarray(高度,宽度,预测概率).我的预测是在重叠窗口中输出的,我需要将它们加在一起以获得整个输入图像的逐像素预测.窗口在(高度,宽度)中重叠超过2个像素.

在C之前,我已经完成了这样的事情,创建了一个大的结果索引,然后将所有的ROI加在一起.

#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];}}}

是否有一种在Python中执行此操作的矢量化方法？

最佳答案 编辑：

@Kevin IMO无论如何,如果您正在训练网络,您应该使用完全连接的层执行此步骤.那说..

如果你想要一些东西可以使用,我有一个非矢量化的解决方案.任何解决方案都将耗费内存.在我的笔记本电脑上,它可以快速分类为CIFAR大小的灰色图像(32×32).也许关键步骤可以由聪明的人进行矢量化.

首先使用skimage将测试数组arr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]])

现在重新组合,生成一个带有形状的输出阵列(5,4,6). 6是win中的窗口数,(5,4)是arr.shape.我们将沿-1轴在每个切片中的一个窗口填充此数组.

# 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数组包含所有窗口的正确位置,但是分隔成-1轴的窗格.要提取原始数组,您可以平均该轴下的所有元素,这些元素不包含零.

>>> 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.]])

你能想出一种以矢量化方式操作切片数组的方法吗？

全部一起：

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

和测试：

>>> 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,这对你来说并不够快

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