【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 - 好文

【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

上一篇：【课程1 - 第一周测验】 <https://blog.csdn.net/u013733326/article/details/79862336>
※※※※※【回到目录】 <https://blog.csdn.net/u013733326/article/details/79827273>※※※※※下一篇：
【课程1 - 第二周编程作业】 <https://blog.csdn.net/u013733326/article/details/79639509>

第2周测验 - 神经网络基础

*
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，…）。

*
下面哪一个是Logistic损失？

* 点击这里
<https://en.wikipedia.org/wiki/Cross_entropy#Cross-entropy_error_function_and_logistic_regression>
.
请注意：我们使用交叉熵损失函数。

*
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))
*
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？

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)

*
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？

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。

*
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)
*
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？

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

*
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？

答：c = a + b.T

*
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？
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).

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

上一篇：【第1周测验-深度学习简介】 <https://blog.csdn.net/u013733326/article/details/79862336>
※※※※※【回到目录】 <https://blog.csdn.net/u013733326/article/details/79827273>※※※※※下一篇：
【具有神经网络思维的Logistic回归】
<https://blog.csdn.net/u013733326/article/details/79639509>

Week 2 Quiz - Neural Network Basics

*
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, …).

*
Which of these is the “Logistic Loss”?

* Check here
<https://en.wikipedia.org/wiki/Cross_entropy#Cross-entropy_error_function_and_logistic_regression>
.
Note: We are using a cross-entropy loss function.

*
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))
*
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”?

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

*
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”?

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

*
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)
*
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?

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

*
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?

c = a + b.T

*
Consider the following code:
a = np.random.randn(3, 3) b = np.random.randn(3, 1) c = a * b
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 soc.shape = (3, 3).

*
Consider the following computation graph.
J = u + v - w = a * b + a * c - (b + c) = a * (b + c) - (b + c) = (a - 1) *
(b + c)
Answer: (a - 1) * (b + c)

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