2020-5-31 吴恩达-NN&DL-w2 NN基础(课后作业)

参考https://zhuanlan.zhihu.com/p/31268885

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

1、神经元计算什么？

• 神经元先计算激活函数，再计算线性函数(z = Wx + b)
• 神经元先计算线性函数（z = Wx + b），再计算激活函数。（正确）
• 神经元计算函数g，函数g计算(Wx + b)。
• 在将输出应用于激活函数之前，神经元计算所有特征的平均值

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2、Which of these is the “Logistic Loss”?以下哪个是逻辑回归损失函数？

• $L^{(i)}({\hat{y}}^{(i)},y^{(i)})=|y^{(i)}-{\hat{y}}^{(i)}|$
• $L^{(i)}({\hat{y}}^{(i)},y^{(i)})=max(0,y^{(i)}-{\hat{y}}^{(i)})$
• $L^{(i)}({\hat{y}}^{(i)},y^{(i)})=-(y^{(i)}log({\hat{y}}^{(i)})+(1+y^{(i)})log(1-{\hat{y}}^{(i)}))$。（正确）
• $L^{(i)}({\hat{y}}^{(i)},y^{(i)})=|y^{(i)}-{\hat{y}}^{(i)}{|}^2$

参见2.3 logistic 回归损失函数 。采用平方误差可能导致优化非凸（局部最优，不是全局最优），而上面定义的损失函数可以得到全局最优结果。

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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?

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。 如何将其转换为列向量？

• x = img.reshape((1，32 * 32 * 3))
• x = img.reshape((32 * 32 , 3))
• x = img.reshape((32 * 32 * 3, 1))。正确
• x = img.reshape((3，32 * 32 ))

参见2.16 关于 python / numpy 向量的说明

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4、Consider the two following random arrays “a” and “b”: 有2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

What will be the shape of “c”? 请问数组c的维度是怎么样？

• c.shape = (3, 2)
• c.shape = (2, 1)
• c.shape = (2, 3)。（正确）
• The computation cannot happen because the sizes don’t match.It’s going to be “Error”!

参见2.15 Python 中的广播。
B（列向量）复制3次，然后和A的每一列相加。

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5、Consider the two following random arrays “a” and “b”:有2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

What will be the shape of “c”? 请问数组c的维度是怎么样？

• The computation cannot happen because the sizes don’t match.It’s going to be “Error”!（正确）
• c.shape = (4, 3)
• c.shape = (3, 3)
• c.shape = (4, 2)

数组按照元素相乘需要两个矩阵之间的维数相同，但是a和b维度不同，所以这将报错，无法计算。

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6、Suppose you have n_x input features per example. Recall that $X=[x^{(1)},x^{(2)}\dots x^{(m)}]$. What is the dimension of X?
假设你的每一个实例有$n_x$个输入特征，想一下在$X=[x^{(1)},x^{(2)}\dots x^{(m)}]$中，X的维度是多少？

• (m,1)
• (m,$n_x$)
• ($n_x$,m)。（正确）
• (1,m)

m个x向量横向堆叠。x是包含$n_x$个元素的列向量。

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7、Recall that np.dot(a,b) performs a matrix multiplication on a and b, whereas a*b performs an element-wise multiplication.
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。

Consider the two following random arrays “a” and “b”:
有2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)

What is the shape of c? 请问数组c的维度是怎么样？

• c.shape = (150, 150)
• The computation cannot happen because the sizes don’t match.It’s going to be “Error”!
• c.shape = (12288, 150)
• c.shape = (12288, 45)。（正确）

矩阵乘法，没什么好说的。

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8、Consider the following code snippet: 观察下面代码

# 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? 如何向量化？

• c = a + b.T。（正确）
• c = a.T + b
• c = a + b
• c = a.T + b.T

c的维度（3，4），a维度无需转置，b需要转置并广播。

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9、Consider the following code: 观察下面代码

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 * is an element-wise product so c.shape = (3, 3).（正确）
• This will invoke broadcasting, so b is copied three times to become (3,3), and * invokes a matrix multiplication operation of 3x3 matrices so c will be (3, 3).
• This will multiply a 3x3 matrix a with a 3x1 vector, thus resulting in a 3x1 vector. That is, c.shape= (3, 1).
• 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).

数组按照元素相乘。使用广播机制，b会被复制三次。

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10、Consider the following computation graph. 观察下面计算图

What is the output J? J是什么？

• J= (c - 1) * (b + a)
• J= (a - 1) * (b + c)。（正确）
• J= a * b + b * c + a * c
• J= (b - 1) * (c + a)

推导过程如下

J = u + v - w
  = a * b + a * c - (b + c)
  = a * (b + c) - (b + c)
  = (a - 1) * (b + c)