• Convolutional Neural Network-week1编程题(一步步搭建CNN模型)


    Convolutional Neural Networks: Step by Step

    implement convolutional (CONV) and pooling (POOL) layers in numpy, including both forward propagation and (optionally) backward propagation.

    Notation:

    • Superscript ([l]) denotes an object of the (l^{th}) layer.

      • Example: (a^{[4]}) is the (4^{th}) layer activation. (W^{[5]}) and (b^{[5]}) are the (5^{th}) layer parameters.
    • Superscript ((i)) denotes an object from the (i^{th}) example.

      • Example: (x^{(i)}) is the (i^{th}) training example input.
    • Lowerscript (i) denotes the (i^{th}) entry of a vector.

      • Example: (a^{[l]}_i) denotes the (i^{th}) entry of the activations in layer (l), assuming this is a fully connected (FC) layer.
    • (n_H), (n_W) and (n_C) denote respectively the height, width and number of channels of a given layer. If you want to reference a specific layer (l), you can also write (n_H^{[l]}), (n_W^{[l]}), (n_C^{[l]}).

    • (n_{H_{prev}}), (n_{W_{prev}}) and (n_{C_{prev}}) denote respectively the height, width and number of channels of the previous layer. If referencing a specific layer (l), this could also be denoted (n_H^{[l-1]}), (n_W^{[l-1]}), (n_C^{[l-1]}).

    1. Packages

    import numpy as np
    import h5py
    import matplotlib.pyplot as plt
    
    %matplotlib inline
    plt.rcParams['figure.figsize'] = (5.0, 4.0) # set default size of plots
    plt.rcParams['image.interpolation'] = 'nearest'
    plt.rcParams['image.cmap'] = 'gray'
    
    %load_ext autoreload
    %autoreload 2
    
    np.random.seed(1)
    

    2. Outline of Assignment

    • Convolution functions, including:

      • Zero Padding

      • Convolve window

      • Convolution forward

      • Convolution backward (optional)

    • Pooling functions, including:

      • Pooling forward

      • Create mask

      • Distribute value

      • Pooling backward (optional)

    Note: 每一步前向传播,都有对应的 反向传播,因此,你需要把每一步前向传播的parameters,存储到 cache中,用于反向传播.

    3. Convolutional Neural Networks

    一个卷积层(convolutional layer)将一个输入量转换成不同大小的输出量,如图:

    3.1 Zero-Padding

    Zero-padding adds zeros around the border of an image:

    Figure 1 : Zero-Padding:Image (3 channels, RGB) with a padding of 2.

    Zero-Padding的两个好处:

    • 允许你使用 CONV layer 而不必要减小 the height and width of the volumes.(尤其是搭建深层网络时)(Same convolutions)

    • 帮助我们保持图片边缘重要的信息. 没有Padding,很少有值,在下一层能够作为图片的边缘被像素值影响

    Exercise:实现函数,用0填充一批示例X的所有图像. Note if you want to pad the array "a" of shape ((5,5,5,5,5)) with pad = 1 for the 2nd dimension, pad = 3 for the 4th dimension and pad = 0 for the rest, you would do:

    a = np.pad(a, ((0,0), (1,1), (0,0), (3,3), (0,0)), 'constant', constant_values = (..,..))
    

    实现:

    # GRADED FUNCTION: zero_pad
    
    def zero_pad(X, pad):
        """
        Pad with zeros all images of the dataset X. The padding is applied to the height and width of an image, 
        as illustrated in Figure 1.
        
        Argument:
        X -- python numpy array of shape (m, n_H, n_W, n_C) representing a batch of m images
        pad -- integer, amount of padding around each image on vertical and horizontal dimensions
        
        Returns:
        X_pad -- padded image of shape (m, n_H + 2*pad, n_W + 2*pad, n_C)
        """
        
        ### START CODE HERE ### (≈ 1 line)
        # X_pad: (m, n_H + 2*pad, n_W + 2*pad, n_C)
        X_pad = np.pad(X, ((0, 0), (pad, pad), (pad, pad), (0, 0)), 'constant', constant_values=0)
        ### END CODE HERE ###
        
        return X_pad
    

    测试:

    np.random.seed(1)
    x = np.random.randn(4, 3, 3, 2)
    x_pad = zero_pad(x, 2)
    print ("x.shape =", x.shape)
    print ("x_pad.shape =", x_pad.shape)
    print ("x[1,1] =", x[1,1])
    print ("x_pad[1,1] =", x_pad[1,1])
    
    fig, axarr = plt.subplots(1, 2)
    axarr[0].set_title('x')
    axarr[0].imshow(x[0,:,:,0])
    axarr[1].set_title('x_pad')
    axarr[1].imshow(x_pad[0,:,:,0])
    

    输出:
    x.shape = (4, 3, 3, 2)
    x_pad.shape = (4, 7, 7, 2)
    x[1,1] = [[ 0.90085595 -0.68372786]
    [-0.12289023 -0.93576943]
    [-0.26788808 0.53035547]]
    x_pad[1,1] = [[0. 0.]
    [0. 0.]
    [0. 0.]
    [0. 0.]
    [0. 0.]
    [0. 0.]
    [0. 0.]]

    3.2 Single step of convolution

    在这一部分中,实现一个卷积的步骤,在该步骤中,将过滤器应用到输入的单个位置中。这将构建卷积单元:

    • 需要一个输入volume

    • 将滤波器应用到输入的每个位置

    • 输出一个不同大小的volume

    Figure 2 : Convolution operation 2x2的滤波器(filter) 和 步长(stride)为1 (stride = amount you move the window each time you slide)

    计算机图像应用中,左边矩阵中的每个值对应于单个像素值,我们通过3x3滤波器与图像卷积,将其值元素与原始矩阵相乘,然后将它们求和并添加偏差。将实现一个卷积步骤,对应于将滤波器应用于其中一个位置以获得单个实值输出

    稍后将将此函数应用于输入的多个位置,以实现完全卷积操作。

    Exercise:实现 conv_single_step().

    # GRADED FUNCTION: conv_single_step
    
    def conv_single_step(a_slice_prev, W, b):
        """
        Apply one filter defined by parameters W on a single slice (a_slice_prev) of the output activation 
        of the previous layer.
        
        Arguments:
        a_slice_prev -- slice of input data of shape (f, f, n_C_prev)
        W -- Weight parameters contained in a window - matrix of shape (f, f, n_C_prev)
        b -- Bias parameters contained in a window - matrix of shape (1, 1, 1)
        
        Returns:
        Z -- a scalar value, result of convolving the sliding window (W, b) on a slice x of the input data
        """
    
        ### START CODE HERE ### (≈ 2 lines of code)
        # Element-wise product between a_slice and W. Do not add the bias yet.
        s = np.multiply(a_slice_prev, W)
        # Sum over all entries of the volume s.
        Z = np.sum(s)
        # Add bias b to Z. Cast b to a float() so that Z results in a scalar value.
        Z = Z + float(b)
        ### END CODE HERE ###
    
        return Z
    

    测试:

    np.random.seed(1)
    a_slice_prev = np.random.randn(4, 4, 3)
    W = np.random.randn(4, 4, 3)
    b = np.random.randn(1, 1, 1)
    
    Z = conv_single_step(a_slice_prev, W, b)
    print("Z =", Z)
    

    输出:
    Z = -6.999089450680221

    3.3 Convolutional Neural Networks - Forward pass

    在前向传播中,你需要很多filters,并在输入上卷积,每次卷积,给你一个2D的矩阵输出,你将stack这些输出,组成一个3D volume:

    Exercise: 函数实现 在 input activation A_prev 上卷积filter W.

    • A_prev作为输(上一层m inputs激活的输出). 由W表示F filters/weights,b表示bias vector

    • 其中,每个filter都有自己的bias. 你可以访问包含 stride 和 padding的超参数字典

    Hint:

    1. 在matrix "a_prev"(shape(5,5,3))的左上角,选择一个2x2的slice,你需要: a_slice_prev = a_prev[0:2,0:2,:]

      • 你将使用 start/end indexes 定义 a_slice_prev
    2. 要定义 a_slice,你需要首先定义他的corners: vert_start, vert_end, horiz_start and horiz_end,下图展示每个Corner如何用 h,w,f,s 定义:

    Figure 3 : Definition of a slice using vertical and horizontal start/end (with a 2x2 filter) (This figure shows only a single channel)

    Reminder:
    卷积后的shape与input shape 有关的公式:

    [n_H = lfloor frac{n_{H_{prev}} - f + 2  imes pad}{stride} floor +1 ]

    [n_W = lfloor frac{n_{W_{prev}} - f + 2  imes pad}{stride} floor +1 ]

    [n_C = ext{number of filters used in the convolution} ]

    使用for-loop实现:

    # GRADED FUNCTION: conv_forward
    
    def conv_forward(A_prev, W, b, hparameters):
        """
        Implements the forward propagation for a convolution function
        
        Arguments:
        A_prev -- output activations of the previous layer, numpy array of shape (m, n_H_prev, n_W_prev, n_C_prev)
        W -- Weights, numpy array of shape (f, f, n_C_prev, n_C)
        b -- Biases, numpy array of shape (1, 1, 1, n_C)
        hparameters -- python dictionary containing "stride" and "pad"
            
        Returns:
        Z -- conv output, numpy array of shape (m, n_H, n_W, n_C)
        cache -- cache of values needed for the conv_backward() function
        """
        
        ### START CODE HERE ###
        # Retrieve dimensions from A_prev's shape (≈1 line)  
        (m, n_H_prev, n_W_prev, n_C_prev) = A_prev.shape
        
        # Retrieve dimensions from W's shape (≈1 line)
        (f, f, n_C_prev, n_C) = W.shape    # n_C: n_C个filter
        
        # Retrieve information from "hparameters" (≈2 lines)
        stride = hparameters['stride']
        pad = hparameters['pad']
        
        # Compute the dimensions of the CONV output volume using the formula given above. Hint: use int() to floor. (≈2 lines)
        n_H = int((n_H_prev - f + 2 * pad) / stride) + 1
        n_W = int((n_W_prev - f + 2 * pad) / stride) + 1
        
        # Initialize the output volume Z with zeros. (≈1 line)
        Z = np.zeros((m, n_H, n_W, n_C))   #  n_C: n_C个filter
        
        # Create A_prev_pad by padding A_prev
        A_prev_pad = zero_pad(A_prev, pad)
        
        for i in range(m):                               # loop over the batch of training examples
            a_prev_pad = A_prev_pad[i]                   # Select ith training example's padded activation
            for h in range(n_H):                           # loop over vertical axis of the output volume
                for w in range(n_W):                       # loop over horizontal axis of the output volume
                    for c in range(n_C):                   # loop over channels (= #filters) of the output volume
                        
                        # Find the corners of the current "slice" (≈4 lines)              
                        vert_start = h * stride
                        vert_end = vert_start + f
                        horiz_start = w * stride
                        horiz_end = horiz_start + f
                        
                        # Use the corners to define the (3D) slice of a_prev_pad (See Hint above the cell). (≈1 line)
                        a_slice_prev = a_prev_pad[vert_start:vert_end, horiz_start:horiz_end, :]
                        
                        # Convolve the (3D) slice with the correct filter W and bias b, to get back one output neuron. (≈1 line)
                        Z[i, h, w, c] = conv_single_step(a_slice_prev, W[...,c], b[...,c])  # 第c个filter的全部W,b
                                            
        ### END CODE HERE ###
        
        # Making sure your output shape is correct
        assert(Z.shape == (m, n_H, n_W, n_C))
        
        # Save information in "cache" for the backprop
        cache = (A_prev, W, b, hparameters)
        
        return Z, cache
    

    输出:

    np.random.seed(1)
    A_prev = np.random.randn(10,4,4,3)
    W = np.random.randn(2,2,3,8)
    b = np.random.randn(1,1,1,8)
    hparameters = {"pad" : 2,
                   "stride": 2}
    
    Z, cache_conv = conv_forward(A_prev, W, b, hparameters)
    print("Z's mean =", np.mean(Z))
    print("Z[3,2,1] =", Z[3,2,1])
    print("cache_conv[0][1][2][3] =", cache_conv[0][1][2][3])
    

    Z's mean = 0.048995203528855794
    Z[3,2,1] = [-0.61490741 -6.7439236 -2.55153897 1.75698377 3.56208902 0.53036437
    5.18531798 8.75898442]
    cache_conv[0][1][2][3] = [-0.20075807 0.18656139 0.41005165]

    Finally, CONV layer should also contain an activation, in which case we would add the following line of code:

    # Convolve the window to get back one output neuron
    Z[i, h, w, c] = ...
    # Apply activation
    A[i, h, w, c] = activation(Z[i, h, w, c])
    

    You don't need to do it here.

    4. Pooling layer

    池化层(Pooling layer)减小了输入的height和width,有助于减少计算,并且有助于特征检测在输入位置的不变,两种Pooling Layers:

    • Max-pooling layer: slides an ((f, f)) window over the input and stores the max value of the window in the output.

    • Average-pooling layer: slides an ((f, f)) window over the input and stores the average value of the window in the output.

    池化层(Pooling layers)没有反向传播训练参数,他们有 超参数:window size (f). 它指定计算fxf窗口max or average的 height和width.

    4.1 - Forward Pooling

    implement MAX-POOL and AVG-POOL, in the same function.

    Exercise: Implement the forward pass of the pooling layer. Follow the hints in the comments below.

    Reminder:
    As there's no padding, the formulas binding the output shape of the pooling to the input shape is:

    [n_H = lfloor frac{n_{H_{prev}} - f}{stride} floor +1 ]

    [n_W = lfloor frac{n_{W_{prev}} - f}{stride} floor +1 ]

    [n_C = n_{C_{prev}} ]

    # GRADED FUNCTION: pool_forward
    
    def pool_forward(A_prev, hparameters, mode = "max"):
        """
        Implements the forward pass of the pooling layer
        
        Arguments:
        A_prev -- Input data, numpy array of shape (m, n_H_prev, n_W_prev, n_C_prev)
        hparameters -- python dictionary containing "f" and "stride"
        mode -- the pooling mode you would like to use, defined as a string ("max" or "average")
        
        Returns:
        A -- output of the pool layer, a numpy array of shape (m, n_H, n_W, n_C)
        cache -- cache used in the backward pass of the pooling layer, contains the input and hparameters 
        """
        
        # Retrieve dimensions from the input shape
        (m, n_H_prev, n_W_prev, n_C_prev) = A_prev.shape
        
        # Retrieve hyperparameters from "hparameters"
        f = hparameters['f']
        stride = hparameters['stride']
        
        # Define the dimensions of the output    
        n_H = int((n_H_prev - f) / stride + 1)
        n_W = int((n_W_prev - f) / stride + 1)
        n_C = n_C_prev
        
        # Initialize output matrix A
        A = np.zeros((m, n_H, n_W, n_C))
        
        ### START CODE HERE ###
        for i in range(m):                         # loop over the training examples
            for h in range(n_H):                     # loop on the vertical axis of the output volume
                for w in range(n_W):                 # loop on the horizontal axis of the output volume
                    for c in range (n_C):            # loop over the channels of the output volume
                        
                        # Find the corners of the current "slice" (≈4 lines)
                        vert_start = h * stride
                        vert_end = vert_start + f
                        horiz_start = w * stride
                        horiz_end = horiz_start + f
                        
                        # Use the corners to define the current slice on the ith training example of A_prev, channel c. (≈1 line)
                        a_prev_slice = A_prev[i, vert_start:vert_end, horiz_start:horiz_end, c]
                        
                        # Compute the pooling operation on the slice. Use an if statment to differentiate the modes. Use np.max/np.mean.
                        if mode == "max":
                            A[i, h, w, c] = np.max(a_prev_slice)
                        elif mode == "average":
                            A[i, h, w, c] = np.mean(a_prev_slice)
        
        ### END CODE HERE ###
        
        # Store the input and hparameters in "cache" for pool_backward()
        cache = (A_prev, hparameters)
        
        # Making sure your output shape is correct
        assert(A.shape == (m, n_H, n_W, n_C))
        
        return A, cache
    

    测试:

    np.random.seed(1)
    A_prev = np.random.randn(2, 4, 4, 3)
    hparameters = {"stride" : 2, "f": 3}
    
    A, cache = pool_forward(A_prev, hparameters)
    print("mode = max")
    print("A =", A)
    print()
    A, cache = pool_forward(A_prev, hparameters, mode = "average")
    print("mode = average")
    print("A =", A)
    

    输出:
    mode = max
    A = [[[[1.74481176 0.86540763 1.13376944]]]
    [[[1.13162939 1.51981682 2.18557541]]]]

    mode = average
    A = [[[[ 0.02105773 -0.20328806 -0.40389855]]]
    [[[-0.22154621 0.51716526 0.48155844]]]]

    5. Backpropagation in convolutional neural networks

    5.1 Convolutional layer backward pass

    5.11 Computing dA

    [dA += sum _{h=0} ^{n_H} sum_{w=0} ^{n_W} W_c imes dZ_{hw} ag{1} ]

    • (W_c)是过滤器,(dZ_{hw})是卷积层第h行第w列的使用点乘计算后的输出Z的梯度。
    da_prev_pad[vert_start:vert_end, horiz_start:horiz_end, :] += W[:,:,:,c] * dZ[i, h, w, c]
    

    5.1.2 Computing dW

    This is the formula for computing (dW_c) ((dW_c) is the derivative of one filter) with respect to the loss:

    [dW_c += sum _{h=0} ^{n_H} sum_{w=0} ^ {n_W} a_{slice} imes dZ_{hw} ag{2} ]

    (dW_c)是一个过滤器的梯度,aslice是(Z_{ij})的激活值

    dW[:,:,:,c] += a_slice * dZ[i, h, w, c]
    

    5.1.3 - Computing db:

    This is the formula for computing (db) with respect to the cost for a certain filter (W_c):

    [db = sum_h sum_w dZ_{hw} ag{3} ]

    db[:,:,:,c] += dZ[i, h, w, c]
    

    Exercise: Implement the conv_backward function below. You should sum over all the training examples, filters, heights, and widths. You should then compute the derivatives using formulas 1, 2 and 3 above.

    def conv_backward(dZ, cache):
        """
        Implement the backward propagation for a convolution function
        
        Arguments:
        dZ -- gradient of the cost with respect to the output of the conv layer (Z), numpy array of shape (m, n_H, n_W, n_C)
        cache -- cache of values needed for the conv_backward(), output of conv_forward()
        
        Returns:
        dA_prev -- gradient of the cost with respect to the input of the conv layer (A_prev),
                   numpy array of shape (m, n_H_prev, n_W_prev, n_C_prev)
        dW -- gradient of the cost with respect to the weights of the conv layer (W)
              numpy array of shape (f, f, n_C_prev, n_C)
        db -- gradient of the cost with respect to the biases of the conv layer (b)
              numpy array of shape (1, 1, 1, n_C)
        """
        
        ### START CODE HERE ###
        # Retrieve information from "cache"
        (A_prev, W, b, hparameters) = cache
        
        # Retrieve dimensions from A_prev's shape
        (m, n_H_prev, n_W_prev, n_C_prev) = A_prev.shape
        
        # Retrieve dimensions from W's shape
        (f, f, n_C_prev, n_C) = W.shape
        
        # Retrieve information from "hparameters"
        stride = hparameters["stride"]
        pad = hparameters["pad"]
        
        # Retrieve dimensions from dZ's shape
        (m, n_H, n_W, n_C) = dZ.shape
        
        # Initialize dA_prev, dW, db with the correct shapes 
        dA_prev = np.zeros((m, n_H_prev, n_W_prev, n_C_prev))
        dW = np.zeros((f, f, n_C_prev, n_C))
        db = np.zeros((1, 1, 1, n_C))
    
        # Pad A_prev and dA_prev
        A_prev_pad = zero_pad(A_prev, pad)
        dA_prev_pad = zero_pad(dA_prev, pad)
        
        for i in range(m):                       # loop over the training examples
            
            # select ith training example from A_prev_pad and dA_prev_pad
            a_prev_pad = A_prev_pad[i]
            da_prev_pad = dA_prev_pad[i]
            
            for h in range(n_H):                   # loop over vertical axis of the output volume
                for w in range(n_W):               # loop over horizontal axis of the output volume
                    for c in range(n_C):           # loop over the channels of the output volume
                        
                        # Find the corners of the current "slice"
                        vert_start = h * stride
                        vert_end = vert_start + f
                        horiz_start = w * stride
                        horiz_end = horiz_start + f
                        
                        # Use the corners to define the slice from a_prev_pad
                        a_slice = a_prev_pad[vert_start:vert_end, horiz_start:horiz_end, :]
    #                     a_slice = A_prev_pad[i, vert_start:vert_end, horiz_start:horiz_end, :]
    
                        # Update gradients for the window and the filter's parameters using the code formulas given above
                        da_prev_pad[vert_start:vert_end, horiz_start:horiz_end, :] += W[:,:,:,c] * dZ[i, h, w, c]
    #                     dA_prev_pad[i, vert_start:vert_end, horiz_start:horiz_end, :] += W[:,:,:,c] * dZ[i, h, w, c]
                        dW[:,:,:,c] += a_slice * dZ[i, h, w, c]
                        db[:,:,:,c] += dZ[i, h, w, c]
                        
            # Set the ith training example's dA_prev to the unpaded da_prev_pad (Hint: use X[pad:-pad, pad:-pad, :])
            dA_prev[i, :, :, :] = da_prev_pad[pad:-pad, pad:-pad, :]
    #         dA_prev[i, :, :, :] = dA_prev_pad[i, pad:-pad, pad:-pad, :]
        ### END CODE HERE ###
        
        # Making sure your output shape is correct
        assert(dA_prev.shape == (m, n_H_prev, n_W_prev, n_C_prev))
        
        return dA_prev, dW, db
    

    测试:

    np.random.seed(1)
    dA, dW, db = conv_backward(Z, cache_conv)
    print("dA_mean =", np.mean(dA))
    print("dW_mean =", np.mean(dW))
    print("db_mean =", np.mean(db))
    

    dA_mean = 1.4524377775388075
    dW_mean = 1.7269914583139097
    db_mean = 7.839232564616838

    5.2 Pooling layer - backward pass

    5.2.1 Max pooling - backward pass

    创建掩码矩阵(保存最大值位置)

    [ X = egin{bmatrix} 1 && 3 \ 4 && 2 end{bmatrix} quad ightarrow quad M =egin{bmatrix} 0 && 0 \ 1 && 0 end{bmatrix} ag{4}]

    ps: 4是最大值,则mask中相应位置为1; 其他值不是最大值,则为0

    Exercise: Implement create_mask_from_window(). This function will be helpful for pooling backward.
    Hints:

    • np.max() may be helpful. It computes the maximum of an array.
    • If you have a matrix X and a scalar x: A = (X == x) will return a matrix A of the same size as X such that:
    A[i,j] = True if X[i,j] = x
    A[i,j] = False if X[i,j] != x
    
    • Here, you don't need to consider cases where there are several maxima in a matrix.
    def create_mask_from_window(x):
        """
        Creates a mask from an input matrix x, to identify the max entry of x.
        
        Arguments:
        x -- Array of shape (f, f)
        
        Returns:
        mask -- Array of the same shape as window, contains a True at the position corresponding to the max entry of x.
        """
        
        ### START CODE HERE ### (≈1 line)
        mask = (x == np.max(x))
        ### END CODE HERE ###
        
        return mask
    

    测试:

    np.random.seed(1)
    x = np.random.randn(2,3)
    mask = create_mask_from_window(x)
    print('x = ', x)
    print("mask = ", mask)
    

    x = [[ 1.62434536 -0.61175641 -0.52817175]
    [-1.07296862 0.86540763 -2.3015387 ]]
    mask = [[ True False False]
    [False False False]]

    Why do we keep track of the position of the max?

    • It's because this is the input value that ultimately influenced the output, and therefore the cost.

    • Backprop is computing gradients with respect to the cost, so anything that influences the ultimate cost should have a non-zero gradient.

    • So, backprop will "propagate" the gradient back to this particular input value that had influenced the cost.

    5.2.2 Average pooling - backward pass

    均值池化层的反向传播:

    [ dZ = 1 quad ightarrow quad dZ =egin{bmatrix} 1/4 && 1/4 \ 1/4 && 1/4 end{bmatrix} ag{5}]

    This implies that each position in the (dZ) matrix contributes equally to output because in the forward pass, we took an average.

    Exercise: Implement the function below to equally distribute a value dz through a matrix of dimension shape. Hint

    def distribute_value(dz, shape):
        """
        Distributes the input value in the matrix of dimension shape
        
        Arguments:
        dz -- input scalar
        shape -- the shape (n_H, n_W) of the output matrix for which we want to distribute the value of dz
        
        Returns:
        a -- Array of size (n_H, n_W) for which we distributed the value of dz
        """
        
        ### START CODE HERE ###
        # Retrieve dimensions from shape (≈1 line)
        (n_H, n_W) = shape
        
        # Compute the value to distribute on the matrix (≈1 line)
        average = dz / (n_H * n_W)
        
        # Create a matrix where every entry is the "average" value (≈1 line)
        a = np.ones(shape) * average
        ### END CODE HERE ###
        
        return a
    

    测试:

    a = distribute_value(2, (2,2))
    print('distributed value =', a)
    

    distributed value = [[0.5 0.5]
    [0.5 0.5]]

    5.2.3 Putting it together: Pooling backward

    compute backward propagation on a pooling layer.

    Exercise: Implement the pool_backward function in both modes ("max" and "average").

    • You will once again use 4 for-loops (iterating over training examples, height, width, and channels).

    • You should use an if/elif statement to see if the mode is equal to 'max' or 'average'.

    • If it is equal to 'average' you should use the distribute_value() function you implemented above to create a matrix of the same shape as a_slice.

    • Otherwise, the mode is equal to 'max', and you will create a mask with create_mask_from_window() and multiply it by the corresponding value of dZ.

    def pool_backward(dA, cache, mode = "max"):
        """
        Implements the backward pass of the pooling layer
        
        Arguments:
        dA -- gradient of cost with respect to the output of the pooling layer, same shape as A
        cache -- cache output from the forward pass of the pooling layer, contains the layer's input and hparameters 
        mode -- the pooling mode you would like to use, defined as a string ("max" or "average")
        
        Returns:
        dA_prev -- gradient of cost with respect to the input of the pooling layer, same shape as A_prev
        """
        
        ### START CODE HERE ###
        
        # Retrieve information from cache (≈1 line)
        (A_prev, hparameters) = cache
        
        # Retrieve hyperparameters from "hparameters" (≈2 lines)
        stride = hparameters['stride']
        f = hparameters['f']
        
        # Retrieve dimensions from A_prev's shape and dA's shape (≈2 lines)
        m, n_H_prev, n_W_prev, n_C_prev = A_prev.shape
        m, n_H, n_W, n_C = dA.shape
        
        # Initialize dA_prev with zeros (≈1 line)
        dA_prev = np.zeros(A_prev.shape)
        
        for i in range(m):                       # loop over the training examples
            # select training example from A_prev (≈1 line)
            a_prev = A_prev[i]
            for h in range(n_H):                   # loop on the vertical axis
                for w in range(n_W):               # loop on the horizontal axis
                    for c in range(n_C):           # loop over the channels (depth)
                        # Find the corners of the current "slice" (≈4 lines)
                        vert_start = h * stride
                        vert_end = vert_start + f
                        horiz_start = w * stride
                        horiz_end = horiz_start + f
                        
                        # Compute the backward propagation in both modes.
                        if mode == "max":
                            # Use the corners and "c" to define the current slice from a_prev (≈1 line)
                            a_prev_slice = a_prev[vert_start:vert_end, horiz_start:horiz_end, c]
    #                         a_prev_slice = A_prev[i, vert_start:vert_end, horiz_start:horiz_end, c]
                            # Create the mask from a_prev_slice (≈1 line)
                            mask = create_mask_from_window(a_prev_slice)
                            # Set dA_prev to be dA_prev + (the mask multiplied by the correct entry of dA) (≈1 line)
                            dA_prev[i, vert_start:vert_end, horiz_start:horiz_end, c] += np.multiply(mask, dA[i, h, w, c])
                            
                        elif mode == "average":
                            # Get the value a from dA (≈1 line)
                            da = dA[i, h, w, c]
                            # Define the shape of the filter as fxf (≈1 line)
                            shape = (f, f)
                            # Distribute it to get the correct slice of dA_prev. i.e. Add the distributed value of da. (≈1 line)
                            dA_prev[i, vert_start:vert_end, horiz_start:horiz_end, c] += distribute_value(da, shape)
                            
        ### END CODE ###
        
        # Making sure your output shape is correct
        assert(dA_prev.shape == A_prev.shape)
        
        return dA_prev
    

    测试:

    np.random.seed(1)
    A_prev = np.random.randn(5, 5, 3, 2)
    hparameters = {"stride" : 1, "f": 2}
    A, cache = pool_forward(A_prev, hparameters)
    dA = np.random.randn(5, 4, 2, 2)
    
    dA_prev = pool_backward(dA, cache, mode = "max")
    print("mode = max")
    print('mean of dA = ', np.mean(dA))
    print('dA_prev[1,1] = ', dA_prev[1,1])  
    print()
    dA_prev = pool_backward(dA, cache, mode = "average")
    print("mode = average")
    print('mean of dA = ', np.mean(dA))
    print('dA_prev[1,1] = ', dA_prev[1,1]) 
    

    mode = max
    mean of dA = 0.14571390272918056
    dA_prev[1,1] = [[ 0. 0. ]
    [ 5.05844394 -1.68282702]
    [ 0. 0. ]]

    mode = average
    mean of dA = 0.14571390272918056
    dA_prev[1,1] = [[ 0.08485462 0.2787552 ]
    [ 1.26461098 -0.25749373]
    [ 1.17975636 -0.53624893]]

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  • 原文地址:https://www.cnblogs.com/douzujun/p/13137696.html
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