使用Numpy的最大池反向传播

Question

我正在使用Numpy实现CNN，我找不到一种方法来高效地实现max池的反向传播，就像我为前向传播所做的那样。这就是我在前向传播中所做的：

def run(self, x, is_training=True):
    """
    Applying MaxPooling on `x`
    :param x: input - [n_batches, channels, height, width]
    :param is_training: a boolean indicating whether training or not
    :return: output of the MaxPooling on `x`
    """
    n_batch, ch_x, h_x, w_x = x.shape
    h_poolwindow, w_poolwindow = self.pool_size

    out_h = int((h_x - h_poolwindow) / self.stride) + 1
    out_w = int((w_x - w_poolwindow) / self.stride) + 1

    windows = as_strided(x,
                         shape=(n_batch, ch_x, out_h, out_w, *self.pool_size),
                         strides=(x.strides[0], x.strides[1],
                                  self.stride * x.strides[2],
                                  self.stride * x.strides[3],
                                  x.strides[2], x.strides[3])
                         )
    out = np.max(windows, axis=(4, 5))

    if is_training:
        self.cache['X'] = x
    return out

我当前反向传播的实现：

def backprop(self, dA_prev):
    """
    Backpropagation in a max-pooling layer
    :return: the derivative of the cost layer with respect to the current layer
    """
    x = self.cache['X']
    n_batch, ch_x, h_x, w_x = x.shape
    h_poolwindow, w_poolwindow = self.pool_size

    dA = np.zeros(shape=x.shape)  # dC/dA --> gradient of the input
    for n in range(n_batch):
        for ch in range(ch_x):
            curr_y = out_y = 0
            while curr_y + h_poolwindow <= h_x:
                curr_x = out_x = 0
                while curr_x + w_poolwindow <= w_x:
                    window_slice = x[n, ch, curr_y:curr_y + h_poolwindow, curr_x:curr_x + w_poolwindow]
                    i, j = np.unravel_index(np.argmax(window_slice), window_slice.shape)
                    dA[n, ch, curr_y + i, curr_x + j] = dA_prev[n, ch, out_y, out_x]

                    curr_x += self.stride
                    out_x += 1

                curr_y += self.stride
                out_y += 1
    return dA

我能把它矢量化吗？

Answer 1

我设法通过将前向传播更改为：

windows = as_strided(x,
                     shape=(n_batch, ch_x, out_h, out_w, *self.pool_size),
                     strides=(x.strides[0], x.strides[1],
                              self.stride * x.strides[2],
                              self.stride * x.strides[3],
                              x.strides[2], x.strides[3])
                     )
out = np.max(windows, axis=(4, 5))

maxs = out.repeat(2, axis=2).repeat(2, axis=3)
x_window = x[:, :, :out_h * self.stride, :out_w * self.stride]
mask = np.equal(x_window, maxs).astype(int)

if is_training:
    self.cache['X'] = x
    self.cache['mask'] = mask
return out

并将反向传播更改为：

mask = self.cache['mask']
dA = dA_prev.repeat(h_poolwindow, axis=2).repeat(w_poolwindow, axis=3)
dA = np.multiply(dA, mask)
pad = np.zeros(x.shape)
pad[:, :, :dA.shape[2], :dA.shape[3]] = dA
return pad