【深度学习】吴恩达深度学习-Course1神经网络与深度学习-第二周神经网络基础作业
视频链接：【中英字幕】吴恩达深度学习课程第一课 — 神经网络与深度学习
本文题目来源：
- 【中英】【吴恩达课后测验】Course 1 - 神经网络和深度学习 -第二周测验
- 吴恩达深度学习第一部分第二周作业打卡
目录
- 英文习题
- 中文习题
- 答案
英文习题
1.What does a neuron compute?
A. A neuron computes the mean of all features before applying the output to an activation function
B. A neuron computes a linear function (z = Wx + b) followed by an activation function
C. A neuron computes an activation function followed by a linear function (z = Wx + b)
D. A neuron computes a function g that scales the input x linearly (Wx + b)
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”?
A. L(i)(y^ (i),y(i))=−(y(i)log(y^ (i))+(1−y(i))log(1−y^(i)))
B. L(i)(y^ (i),y(i))=max(0,y(i)−y^(i))
C. L(i)(y^ (i),y(i))=∣y(i)−y^(i)∣2
D. L(i)(y^ (i),y(i))=∣y(i)−y^(i)∣
Note: We are using a cross-entropy loss function.(Click Here)
(上面看不了可下载这个免费平替版)
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?
A. x = img.reshape((32 * 32,3))
B. x = img.reshape((3,32 * 32))
C. x = img.reshape((32 * 32 * 3,1))
D. x = img.reshape((1,32 * 32,*3))
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”?
A. The computation cannot happen because the sizes don’t match. It’s going to be “Error”!
B. c.shape = (3, 2)
C. c.shape = (2, 3)
D. c.shape = (2, 1)
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”?
A. c.shape = (3, 3)
B. c.shape = (4,2)
C.The computation cannot happen because the sizes don’t match. It’s going to be “Error”!
D.c.shape = (4, 3)
6. Suppose you have nx input features per example. Recall that X=[x(1)x(2)…x(m)]. What is the dimension of X?
A. (m,1)
B. (1,m)
C. (m,nx)
D. (nx,m)
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 is the shape of c?
A. The computation cannot happen because the sizes don’t match. It’s going to be “Error”!
B. c.shape = (12288, 45)
C. c.shape = (12288, 150)
D. c.shape = (150,150)
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?
A. c = a + b.T
B. c = a.T + b.T
C. c = a.T + b
D. c = a + b
9.Consider the following code:
a = np.random.randn(3, 3)
b = np.random.randn(3, 1)
c = a*b
What will be c? (If you’re not sure, feel free to run this in python to find out).
A. This will invoke broadcasting, so b is copied three times to become (3,3), and ∗ is an element-wise product so c.shape will be (3, 3)
B. This will invoke broadcasting, so b is copied three times to become (3, 3), and ∗ invokes a matrix multiplication operation of two 3x3 matrices so c.shape will be (3, 3)
C. This will multiply a 3x3 matrix a with a 3x1 vector, thus resulting in a 3x1 vector. That is, c.shape = (3,1).
D. It will lead to an error since you cannot use “*” to operate on these two matrices. You need to instead use np.dot(a,b)
10.Consider the following computation graph.
What is the output J?
A. J = (c - 1)(b + a)
B. J = (a - 1) (b + c)
C. J = ab + bc + a*c
D. J = (b - 1) * (c + a)
中文习题
1.神经元节点计算什么？
A.神经元节点先计算激活函数，再计算线性函数(z = Wx + b)
B.神经元节点先计算线性函数（z = Wx + b），再计算激活。
C.神经元节点计算函数g，函数g计算(Wx + b)。
D.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？
A. L(i)(y^ (i),y(i))=−(y(i)log(y^ (i))+(1−y(i))log(1−y^(i)))
B. L(i)(y^ (i),y(i))=max(0,y(i)−y^(i))
C. L(i)(y^ (i),y(i))=∣y(i)−y^(i)∣2
D. L(i)(y^ (i),y(i))=∣y(i)−y^(i)∣
请注意：我们使用交叉熵损失函数。(点击此处)
(上面看不了可下载这个免费平替版)
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？
A. x = img.reshape((32 * 32,3))
B. x = img.reshape((3,32 * 32))
C. x = img.reshape((32 * 32 * 3,1))
D. x = img.reshape((1,32 * 32, *3))
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
数组c的维度是多少？
A.计算无法发生，因为大小不匹配，会发生“错误”!
B. c.shape = (3, 2)
C. c.shape = (2, 3)
D. c.shape = (2, 1)
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？
A. c.shape = (3, 3)
B. c.shape = (4,2)
C.计算无法发生，因为大小不匹配，会发生“错误”!
D.c.shape = (4, 3)
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？
A. (m,1)
B. (1,m)
C. (m,nx)
D. (nx,m)
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)
数组c的维度是多少？
A. 计算无法发生，因为大小不匹配，会发生“错误”!
B. c.shape = (12288, 45)
C. c.shape = (12288, 150)
D. c.shape = (150,150)
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？
A. c = a + b.T
B. c = a.T + b.T
C. c = a.T + b
D. c = a + b
这题答案的翻译有点乱，可以自己去看看英文版本的
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？ (如果你不确定，就跑一下这段代码试试).
A. 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).
B. 这会引发广播机制，所以b会被复制三次编程(3,3)，并且∗ 调用两个3x3矩阵的矩阵乘法运算，使c.shape为3,3）
C. 这将使3x3矩阵a与3x1向量相乘，从而得到3x1向量。也就是说，c.shape=（3,1)
D. 由于不能使用“*”对这两个矩阵进行操作，因此将导致错误。您需要使用np.dot（a，b）
10.看一下下面的计算图
输出的J是什么？
A. J = (c - 1)(b + a)
B. J = (a - 1) (b + c)
C. J = ab + bc + ac
D. J = (b - 1) * (c + a)
答案
- B
- A
- C
- C（由python的broadcasting，b会自动扩充为(2,3)以能够和a相加）
- C
- D
- B
- A（为什么不选C？注意，给出的for循环中是a[i][j]+b[j]，而不是a[j][i]+b[j]！，如果代码给的是后者，则选C）
- A（当我们使用“ * ”进行计算的时候，会触发b广播称为shape为(3 , 3)的矩阵，此时a与b相乘即对应元素相乘，即a中的元素与b中对应位置的元素进行相乘。当使用np.dot进行计算时，就类似于线性代数里面的计算方式，可以写一段代码来测试一下，如下：）
import numpy as np
a = np.random.randn(3, 3)
b = np.random.randn(3, 1)
print("a:")
print(a)
print("b:")
print(b)
c = a*b
print("a*b:")
print(c)
c = np.dot(a, b)
print("np.dot(a, b):")
print(c)
结果如下：
a:
[[-0.03228798 1.28358425 0.66830909]
[ 0.84179787 0.4902092 -0.09530794]
[ 0.35068597 -0.60616291 -0.47161416]]
b:
[[-0.21431806]
[ 0.71400772]
[ 0.41949772]]
a*b:
[[ 0.0069199 -0.27509529 -0.14323071]
[ 0.60105018 0.35001315 -0.06805061]
[ 0.14711196 -0.25428396 -0.19784106]]
np.dot(a, b):
[[ 1.2037631]
[ 0.1296192]
[-0.7058044]]
- B