吴恩达深度学习笔记 第二章测验

参考自:https://blog.csdn.net/u013733326/article/details/79865858

第2周测验 - 神经网络基础

1. 神经元节点计算什么？

   • 【 】神经元节点先计算激活函数，再计算线性函数(z = Wx + b)

   • 【★】神经元节点先计算线性函数（z = Wx + b），再计算激活。

   • 【 】神经元节点计算函数g，函数g计算(Wx + b)。

   • 【 】在 将输出应用于激活函数之前，神经元节点计算所有特征的平均值

     请注意：神经元的输出是a = g（Wx + b），其中g是激活函数（sigmoid，tanh，ReLU，…）。

2. 下面哪一个是Logistic损失？

   • 点击这里.

   请注意：我们使用交叉熵损失函数。

3. 假设img是一个（32,32,3）数组，具有3个颜色通道：红色、绿色和蓝色的32x32像素的图像。 如何将其重新转换为列向量？

   x = img.reshape((32 * 32 * 3, 1))

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4. 看一下下面的这两个随机数组“a”和“b”：

   a = np.random.randn(2, 3) # a.shape = (2, 3)
   b = np.random.randn(2, 1) # b.shape = (2, 1)
   c = a + b

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   请问数组c的维度是多少？

   答： B（列向量）复制3次，以便它可以和A的每一列相加，所以：c.shape = (2, 3)

5. 看一下下面的这两个随机数组“a”和“b”：

   a = np.random.randn(4, 3) # a.shape = (4, 3)
   b = np.random.randn(3, 2) # b.shape = (3, 2)
   c = a * b

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   请问数组“c”的维度是多少？

   答：运算符 “*” 说明了按元素乘法来相乘，但是元素乘法需要两个矩阵之间的维数相同，所以这将报错，无法计算。

6. 假设你的每一个实例有n_x个输入特征，想一下在X=[x^(1), x^(2)…x^(m)]中，X的维度是多少？

   答： (n_x, m)

   请注意：一个比较笨的方法是当l=1的时候，那么计算一下Z (l) =W (l) A (l)  Z(l)=W(l)A(l) ，所以我们就有：

   • A (1)  A(1) = X
   • X.shape = (n_x, m)
   • Z (1)  Z(1) .shape = (n (1)  n(1) , m)
   • W (1)  W(1) .shape = (n (1)  n(1) , n_x)

7. 回想一下，np.dot（a，b）在a和b上执行矩阵乘法，而`a * b’执行元素方式的乘法。

   看一下下面的这两个随机数组“a”和“b”：

   a = np.random.randn(12288, 150) # a.shape = (12288, 150)
   b = np.random.randn(150, 45) # b.shape = (150, 45)
   c = np.dot(a, b)

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   请问c的维度是多少？

   答： c.shape = (12288, 45), 这是一个简单的矩阵乘法例子。

8. 看一下下面的这个代码片段：

   
   # a.shape = (3,4)
   
   
   # b.shape = (4,1)
   
   for i in range(3):
     for j in range(4):
       c[i][j] = a[i][j] + b[j]

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   请问要怎么把它们向量化？

   答：c = a + b.T

9. 看一下下面的代码：

   a = np.random.randn(3, 3)
   b = np.random.randn(3, 1)
   c = a * b

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   请问c的维度会是多少？
   答：这将会使用广播机制，b会被复制三次，就会变成(3,3)，再使用元素乘法。所以： c.shape = (3, 3).

10. 看一下下面的计算图：

    J = u + v - w
      = a * b + a * c - (b + c)
      = a * (b + c) - (b + c)
      = (a - 1) * (b + c)

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    答: (a - 1) * (b + c)
    博主注：由于弄不到图，所以很抱歉。

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Week 2 Quiz - Neural Network Basics

1. What does a neuron compute?

   • [ ] A neuron computes an activation function followed by a linear function (z = Wx + b)

   • [x] A neuron computes a linear function (z = Wx + b) followed by an activation function

   • [ ] A neuron computes a function g that scales the input x linearly (Wx + b)

   • [ ] A neuron computes the mean of all features before applying the output to an activation function

   Note: The output of a neuron is a = g(Wx + b) where g is the activation function (sigmoid, tanh, ReLU, …).

2. Which of these is the “Logistic Loss”?

   • Check here.

   Note: We are using a cross-entropy loss function.

3. Suppose img is a (32,32,3) array, representing a 32x32 image with 3 color channels red, green and blue. How do you reshape this into a column vector?

   • x = img.reshape((32 * 32 * 3, 1))

4. Consider the two following random arrays “a” and “b”:

   a = np.random.randn(2, 3) # a.shape = (2, 3)
   b = np.random.randn(2, 1) # b.shape = (2, 1)
   c = a + b

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   What will be the shape of “c”?

   b (column vector) is copied 3 times so that it can be summed to each column of a. Therefore, c.shape = (2, 3).

5. Consider the two following random arrays “a” and “b”:

   a = np.random.randn(4, 3) # a.shape = (4, 3)
   b = np.random.randn(3, 2) # b.shape = (3, 2)
   c = a * b

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   What will be the shape of “c”?

   “*” operator indicates element-wise multiplication. Element-wise multiplication requires same dimension between two matrices. It’s going to be an error.

6. Suppose you have n_x input features per example. Recall that X=[x^(1), x^(2)…x^(m)]. What is the dimension of X?

   (n_x, m)

   Note: A stupid way to validate this is use the formula Z^(l) = W^(l)A^(l) when l = 1, then we have

   • A^(1) = X
   • X.shape = (n_x, m)
   • Z^(1).shape = (n^(1), m)
   • W^(1).shape = (n^(1), n_x)

7. Recall that np.dot(a,b) performs a matrix multiplication on a and b, whereas a*b performs an element-wise multiplication.

   Consider the two following random arrays “a” and “b”:

   a = np.random.randn(12288, 150) # a.shape = (12288, 150)
   b = np.random.randn(150, 45) # b.shape = (150, 45)
   c = np.dot(a, b)

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   What is the shape of c?

   c.shape = (12288, 45), this is a simple matrix multiplication example.

8. Consider the following code snippet:

   
   # a.shape = (3,4)
   
   
   # b.shape = (4,1)
   
   for i in range(3):
     for j in range(4):
       c[i][j] = a[i][j] + b[j]

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   How do you vectorize this?

   c = a + b.T

9. Consider the following code:

   a = np.random.randn(3, 3)
   b = np.random.randn(3, 1)
   c = a * b

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   What will be c?

   This will invoke broadcasting, so b is copied three times to become (3,3), and 鈭� is an element-wise product so c.shape = (3, 3).

10. Consider the following computation graph.

    J = u + v - w
      = a * b + a * c - (b + c)
      = a * (b + c) - (b + c)
      = (a - 1) * (b + c)

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    Answer: (a - 1) * (b + c)