【中文】【吴恩达课后编程作业】Course 1 - 神经网络和深度学习 - 第二周作业
【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
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神经元节点计算什么?
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【 】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)。
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【 】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,...)。
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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。
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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。
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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\(Z^{(l)} = W^{(l)}A^{(l)}\),所以我们就有:- \(A^{(1)}\) = X
- X.shape = (n_x, m)
- \(Z^{(1)}\).shape = (\(n^{(1)}\), m)
- \(W^{(1)}\).shape = (\(n^{(1)}\), n_x)
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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
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What does a neuron compute?
Note: The output of a neuron is a = g(Wx + b) where g is the activation function (sigmoid, tanh, ReLU, ...).
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Which of these is the "Logistic Loss"?
- Check here.
Note: We are using a cross-entropy loss function.
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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))
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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.
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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)
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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.T -
Consider 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)

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