吴恩达-深度学习入门-第二周课后测验题
前情须知
1、本文参考CSDN博主何宽老师的文章，仅用于个人学习使用，将答案部分单独摘出放在最后，方便进行自我检测。
参考文章链接: https://blog.csdn.net/u013733326/article/details/79865858
2、第二周分为测验题和编程题两部分
目录
- 前情须知
- 一、中文题目
- 二、英文题目
- 三、答案
一、中文题目
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
2、 下面哪一个是Logistic损失？
未找到图，略.
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？
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
请问数组c的维度是多少？
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
请问数组c的维度是多少？
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？
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)
请问c的维度是多少？
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]
请问要怎么把它们向量化？
9、 看一下下面的代码：
a = np.random.randn(3, 3)
b = np.random.randn(3, 1)
c = a * b
请问c的维度会是多少？
10、 看一下下面的计算图，写出最终的式子：
注：无图，直接看答案.
J = u + v - w
= a * b + a * c - (b + c)
= a * (b + c) - (b + c)
= (a - 1) * (b + c)
二、英文题目
1、 What does a neuron compute?
[ ] A neuron computes an activation function followed by a linear function (z = Wx + b)
[ ] 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
2、 Which of these is the “Logistic Loss”?
此题未找到图，略。
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?
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
What will be the shape of “c”?
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
What will be the shape of “c”?
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?
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)
What will be the shape of “c”?
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]
How do you vectorize this?
9、 Consider the following code:
a = np.random.randn(3, 3)
b = np.random.randn(3, 1)
c = a * b
What will be c?
10、 Consider the following computation graph.
Note：The calculation chart is not found, the answer is as follows:
J = u + v - w
= a * b + a * c - (b + c)
= a * (b + c) - (b + c)
= (a - 1) * (b + c)
三、答案
1、
[x] A neuron computes a linear function (z = Wx + b) followed by an activation function
Note: The output of a neuron is a = g(Wx + b) where g is the activation function (sigmoid, tanh, ReLU, …).
【★】神经元节点先计算线性函数（z = Wx + b），再计算激活。
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、
Figure not found, omitted.
Note: We are using a cross-entropy loss function.
3、
x = img.reshape((32 * 32 * 3, 1))
4、
b (column vector) is copied 3 times so that it can be summed to each column of a. Therefore,c.shape = (2, 3).
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、
“*” operator indicates element-wise multiplication. Element-wise multiplication requires same dimension between two matrices. It’s going to be an error.
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、
答：(n_x, m)
Note: A stupid way to validate this is use the formula Z^(l) = W(l)A(l) whenl=1, then we have
请注意：一个比较笨的方法是当l=1的时候，那么计算一下，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、
c.shape = (12288, 45), this is a simple matrix multiplication example.
答： c.shape = (12288, 45), 这是一个简单的矩阵乘法例子。
8、
答：c = a + b.T
9、
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、
J = u + v - w
= a * b + a * c - (b + c)
= a * (b + c) - (b + c)
= (a - 1) * (b + c)
答: (a - 1) * (b + c).