【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
【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
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)
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★】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.
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。
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.
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 + bWhat 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 * bWhat 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, whereasa*bperforms 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.TConsider the following code:
a = np.random.randn(3, 3) b = np.random.randn(3, 1) c = a * bWhat 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).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)