Question

当前正在通过斯坦福CS131的免费在线课程学习计算机视觉和机器学习。遇到了一些沉重的数学公式，并想知道是否有人可以只知道图像的高度，宽度和核的高度和宽度，向我解释如何为卷积算法实现朴素的4个嵌套的for循环。通过在线研究，我提出了这个解决方案。

image_padded = np.zeros((image.shape[0] + 2, image.shape[1] + 2))
image_padded[1:-1, 1:-1] = image
for x in range(image.shape[1]):  # Loop over every pixel of the image
    for y in range(image.shape[0]):
        # element-wise multiplication of the kernel and the image
        out[y, x] = (kernel * image_padded[y:y + 3, x:x + 3]).sum()

我能够根据使用这种算法的一些网站示例来理解这一点，但是，我似乎无法理解4个嵌套的for循环是如何做到的。而且，如果可以的话，可以将公式分解为更容易理解的公式，然后再从网上找到给定的数学公式。

编辑：只是为了澄清一下，我留下的代码段在某种程度上可以正常工作，我试图提出一种解决方案，该解决方案的优化程度较低，对初学者更友好，例如此代码在询问什么：

def conv_nested(image, kernel):
    """A naive implementation of convolution filter.

    This is a naive implementation of convolution using 4 nested for-loops.
    This function computes convolution of an image with a kernel and outputs
    the result that has the same shape as the input image.

    Args:
        image: numpy array of shape (Hi, Wi)
        kernel: numpy array of shape (Hk, Wk)

    Returns:
        out: numpy array of shape (Hi, Wi)
    """
    Hi, Wi = image.shape
    Hk, Wk = kernel.shape
    out = np.zeros((Hi, Wi))
    ### YOUR CODE HERE

    ### END YOUR CODE

    return out

Answer 1

scipy.signal.correlate2d是您的朋友。

演示

我将您的代码包装在名为naive_correlation的函数中：

import numpy as np

def naive_correlation(image, kernel):
    image_padded = np.zeros((image.shape[0] + 2, image.shape[1] + 2))
    image_padded[1:-1, 1:-1] = image
    out = np.zeros_like(image)
    for x in range(image.shape[1]):image
        for y in range(image.shape[0]):
            out[y, x] = (kernel * image_padded[y:y + 3, x:x + 3]).sum()
    return out

请注意，由于out尚未初始化，因此您的代码段将引发错误。

In [67]: from scipy.signal import correlate2d

In [68]: img = np.array([[3, 9, 5, 9],
    ...:                 [1, 7, 4, 3],
    ...:                 [2, 1, 6, 5]])
    ...: 

In [69]: kernel = np.array([[0, 1, 0],
    ...:                    [0, 0, 0],
    ...:                    [0, -1, 0]])
    ...: 

In [70]: res1 = correlate2d(img, kernel, mode='same')

In [71]: res1
Out[71]: 
array([[-1, -7, -4, -3],
       [ 1,  8, -1,  4],
       [ 1,  7,  4,  3]])

In [72]: res2 = naive_correlation(img, kernel)

In [73]: np.array_equal(res1, res2)
Out[73]: True

如果您希望执行卷积而不是相关运算，则可以使用convolve2d。

编辑

这是您要寻找的吗？

def explicit_correlation(image, kernel):
    hi, wi= image.shape
    hk, wk = kernel.shape
    image_padded = np.zeros(shape=(hi + hk - 1, wi + wk - 1))    
    image_padded[hk//2:-hk//2, wk//2:-wk//2] = image
    out = np.zeros(shape=image.shape)
    for row in range(hi):
        for col in range(wi):
            for i in range(hk):
                for j in range(wk):
                    out[row, col] += image_padded[row + i, col + j]*kernel[i, j]
    return out

卷积算法的幼稚实现

1 个答案:

演示

编辑