Numpy基础
Numpy基础
s =1/(1+math.exp(-x)) #不能用作向量
import numpy as np
# numpy可以用作向量
# example of np.exp
x = np.array([1, 2, 3])
print(np.exp(x)) # result is (exp(1), exp(2), exp(3))
实现一个sigmod
import numpy as np # this means you can access numpy functions by writing np.function() instead of numpy.function()
def sigmoid(x):
"""
Compute the sigmoid of x
Arguments:
x -- A scalar or numpy array of any size
Return:
s -- sigmoid(x)
"""
### START CODE HERE ### (≈ 1 line of code)
s = 1 / (1 + np.exp(-x))
### END CODE HERE ###
return s
x = np.array([1, 2, 3])
sigmoid(x)
out[9]:array([0.73105858, 0.88079708, 0.95257413])
Sigmoid gradient
𝑠𝑖𝑔𝑚𝑜𝑖𝑑_𝑑𝑒𝑟𝑖𝑣𝑎𝑡𝑖𝑣𝑒(𝑥)=𝜎′(𝑥)=𝜎(𝑥)(1−𝜎(𝑥))
# GRADED FUNCTION: sigmoid_derivative
def sigmoid_derivative(x):
"""
Compute the gradient (also called the slope or derivative) of the sigmoid function with respect to its input x.
You can store the output of the sigmoid function into variables and then use it to calculate the gradient.
Arguments:
x -- A scalar or numpy array
Return:
ds -- Your computed gradient.
"""
### START CODE HERE ### (≈ 2 lines of code)
s = sigmoid(x)
ds = s * (1 - s)
### END CODE HERE ###
return ds
x = np.array([1, 2, 3])
print ("sigmoid_derivative(x) = " + str(sigmoid_derivative(x)))
输出：sigmoid_derivative(x) = [0.19661193 0.10499359 0.04517666]
重塑数组
深度学习中两个常用的numpy函数是np.shape和np.reshape()。
-X.shape用于获取矩阵/向量X的shape（维度）。
-X.reshape（...）用于将X重塑为其他尺寸。
例如，在计算机科学中，图像由shape为(length,height,depth=3)的3D数组表示。但是，当你读取图像作为算法的输入时，会将其转换为维度为(length∗height∗3,1)的向量。换句话说，将3D阵列“展开”或重塑为1D向量。
# GRADED FUNCTION: image2vector
def image2vector(image):
"""
Argument:
image -- a numpy array of shape (length, height, depth)
Returns:
v -- a vector of shape (length*height*depth, 1)
"""
### START CODE HERE ### (≈ 1 line of code)
v = image.reshape(image.shape[0] * image.shape[1] * image.shape[2], 1)
### END CODE HERE ###
return v
# This is a 3 by 3 by 2 array, typically images will be (num_px_x, num_px_y,3) where 3 represents the RGB values
image = np.array([[[ 0.67826139, 0.29380381],
[ 0.90714982, 0.52835647],
[ 0.4215251 , 0.45017551]],
[[ 0.92814219, 0.96677647],
[ 0.85304703, 0.52351845],
[ 0.19981397, 0.27417313]],
[[ 0.60659855, 0.00533165],
[ 0.10820313, 0.49978937],
[ 0.34144279, 0.94630077]]])
print ("image2vector(image) = " + str(image2vector(image)))
输出
image2vector(image) = [[0.67826139]
[0.29380381]
[0.90714982]
[0.52835647]
[0.4215251 ]
[0.45017551]
[0.92814219]
[0.96677647]
[0.85304703]
[0.52351845]
[0.19981397]
[0.27417313]
[0.60659855]
[0.00533165]
[0.10820313]
[0.49978937]
[0.34144279]
[0.94630077]]
行标准化
我们在机器学习和深度学习中使用的另一种常见技术是对数据进行标准化。 由于归一化后梯度下降的收敛速度更快，通常会表现出更好的效果。 通过归一化，也就是将x更改为
\[\frac{x}{\| x\|}
\]
（将x的每个行向量除以其范数）。举例：
\[x =
\begin{bmatrix}
0 & 3 & 4 \\
2 & 6 & 4 \\
\end{bmatrix}\tag{3}
\]
\[\| x\| = np.linalg.norm(x, axis = 1, keepdims = True) = \begin{bmatrix}
5 \\
\sqrt{56} \\
\end{bmatrix}\tag{4}
\]
\[x\_normalized = \frac{x}{\| x\|} = \begin{bmatrix}
0 & \frac{3}{5} & \frac{4}{5} \\
\frac{2}{\sqrt{56}} & \frac{6}{\sqrt{56}} & \frac{4}{\sqrt{56}} \\
\end{bmatrix}\tag{5}
\]
练习：执行 normalizeRows（）来标准化矩阵的行。 将此函数应用于输入矩阵x之后，x的每一行应为单位长度（即长度为1）向量。
# GRADED FUNCTION: normalizeRows
def normalizeRows(x):
"""
Implement a function that normalizes each row of the matrix x (to have unit length).
Argument:
x -- A numpy matrix of shape (n, m)
Returns:
x -- The normalized (by row) numpy matrix. You are allowed to modify x.
"""
### START CODE HERE ### (≈ 2 lines of code)
# Compute x_norm as the norm 2 of x. Use np.linalg.norm(..., ord = 2, axis = ..., keepdims = True)
x_norm = np.linalg.norm(x, axis = 1, keepdims = True)
# Divide x by its norm.
x = x / x_norm
### END CODE HERE ###
return x
x = np.array([
[0, 3, 4],
[1, 6, 4]])
print("normalizeRows(x) = " + str(normalizeRows(x)))
normalizeRows(x) = [[0. 0.6 0.8 ]
[0.13736056 0.82416338 0.54944226]]
广播和softmax函数
练习: 使用numpy实现softmax函数。 你可以将softmax理解为算法需要对两个或多个类进行分类时使用的标准化函数.
\[\text{for } x \in \mathbb{R}^{1\times n} \text{, } softmax(x) = softmax(\begin{bmatrix}
x_1 &&
x_2 &&
... &&
x_n
\end{bmatrix}) = \begin{bmatrix}
\frac{e^{x_1}}{\sum_{j}e^{x_j}} &&
\frac{e^{x_2}}{\sum_{j}e^{x_j}} &&
... &&
\frac{e^{x_n}}{\sum_{j}e^{x_j}}
\end{bmatrix}
\]
\[softmax(x) = softmax\begin{bmatrix}
x_{11} & x_{12} & x_{13} & \dots & x_{1n} \\
x_{21} & x_{22} & x_{23} & \dots & x_{2n} \\
\vdots & \vdots & \vdots & \ddots & \vdots \\
x_{m1} & x_{m2} & x_{m3} & \dots & x_{mn}
\end{bmatrix} = \begin{bmatrix}
\frac{e^{x_{11}}}{\sum_{j}e^{x_{1j}}} & \frac{e^{x_{12}}}{\sum_{j}e^{x_{1j}}} & \frac{e^{x_{13}}}{\sum_{j}e^{x_{1j}}} & \dots & \frac{e^{x_{1n}}}{\sum_{j}e^{x_{1j}}} \\
\frac{e^{x_{21}}}{\sum_{j}e^{x_{2j}}} & \frac{e^{x_{22}}}{\sum_{j}e^{x_{2j}}} & \frac{e^{x_{23}}}{\sum_{j}e^{x_{2j}}} & \dots & \frac{e^{x_{2n}}}{\sum_{j}e^{x_{2j}}} \\
\vdots & \vdots & \vdots & \ddots & \vdots \\
\frac{e^{x_{m1}}}{\sum_{j}e^{x_{mj}}} & \frac{e^{x_{m2}}}{\sum_{j}e^{x_{mj}}} & \frac{e^{x_{m3}}}{\sum_{j}e^{x_{mj}}} & \dots & \frac{e^{x_{mn}}}{\sum_{j}e^{x_{mj}}}
\end{bmatrix} = \begin{pmatrix}
softmax\text{(first row of x)} \\
softmax\text{(second row of x)} \\
... \\
softmax\text{(last row of x)} \\
\end{pmatrix}
\]
def softmax(x):
"""Calculates the softmax for each row of the input x.
Your code should work for a row vector and also for matrices of shape (n, m).
Argument:
x -- A numpy matrix of shape (n,m)
Returns:
s -- A numpy matrix equal to the softmax of x, of shape (n,m)
"""
### START CODE HERE ### (≈ 3 lines of code)
# Apply exp() element-wise to x. Use np.exp(...).
x_exp = np.exp(x)
# Create a vector x_sum that sums each row of x_exp. Use np.sum(..., axis = 1, keepdims = True).
x_sum = np.sum(x_exp, axis = 1, keepdims = True)
# Compute softmax(x) by dividing x_exp by x_sum. It should automatically use numpy broadcasting.
s = x_exp / x_sum
### END CODE HERE ###
return s
In [17]:
x = np.array([
[9, 2, 5, 0, 0],
[7, 5, 0, 0 ,0]])
print("softmax(x) = " + str(softmax(x)))
输出：
softmax(x) = [[9.80897665e-01 8.94462891e-04 1.79657674e-02 1.21052389e-04
1.21052389e-04]
[8.78679856e-01 1.18916387e-01 8.01252314e-04 8.01252314e-04
8.01252314e-04]]
0.9808976649146187
注意：
-如果你在上方输出 x_exp，x_sum和s的维度并重新运行练习单元，则会看到x_sum的纬度为（2,1），而x_exp和s的维度为（2,5）。 x_exp/x_sum 可以使用python广播。
恭喜你！ 你现在已经对python numpy有了很好的理解，并实现了一些将在深度学习中用到的功能。
向量化
x1 = [9, 2, 5, 0, 0, 7, 5, 0, 0, 0, 9, 2, 5, 0, 0]
x2 = [9, 2, 2, 9, 0, 9, 2, 5, 0, 0, 9, 2, 5, 0, 0]
### VECTORIZED DOT PRODUCT OF VECTORS ###
tic = time.process_time()
dot = np.dot(x1,x2)
toc = time.process_time()
print ("dot = " + str(dot) + "\n ----- Computation time = " + str(1000*(toc - tic)) + "ms")
### VECTORIZED OUTER PRODUCT ###
tic = time.process_time()
outer = np.outer(x1,x2)
toc = time.process_time()
print ("outer = " + str(outer) + "\n ----- Computation time = " + str(1000*(toc - tic)) + "ms")
### VECTORIZED ELEMENTWISE MULTIPLICATION ###
tic = time.process_time()
mul = np.multiply(x1,x2)
toc = time.process_time()
print ("elementwise multiplication = " + str(mul) + "\n ----- Computation time = " + str(1000*(toc - tic)) + "ms")
### VECTORIZED GENERAL DOT PRODUCT ###
tic = time.process_time()
dot = np.dot(W,x1)
toc = time.process_time()
print ("gdot = " + str(dot) + "\n ----- Computation time = " + str(1000*(toc - tic)) + "ms")
输出：
dot = 278
----- Computation time = 0.0ms
outer = [[81 18 18 81 0 81 18 45 0 0 81 18 45 0 0]
[18 4 4 18 0 18 4 10 0 0 18 4 10 0 0]
[45 10 10 45 0 45 10 25 0 0 45 10 25 0 0]
[ 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0]
[ 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0]
[63 14 14 63 0 63 14 35 0 0 63 14 35 0 0]
[45 10 10 45 0 45 10 25 0 0 45 10 25 0 0]
[ 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0]
[ 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0]
[ 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0]
[81 18 18 81 0 81 18 45 0 0 81 18 45 0 0]
[18 4 4 18 0 18 4 10 0 0 18 4 10 0 0]
[45 10 10 45 0 45 10 25 0 0 45 10 25 0 0]
[ 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0]
[ 0 0 0 0 0 0 0 0 0 0 0 0 0 0 0]]
----- Computation time = 0.0ms
elementwise multiplication = [81 4 10 0 0 63 10 0 0 0 81 4 25 0 0]
----- Computation time = 0.0ms
gdot = [ 21.57937154 22.58814194 13.70092277]
----- Computation time = 0.0ms
实现L1和L2损失函数
练习：实现L1损失函数的Numpy向量化版本。 我们会发现函数abs（x）（x的绝对值）很有用。
提示：
-损失函数用于评估模型的性能。 损失越大，预测(y^) 与真实值(y)的差异也就越大。 在深度学习中，我们使用诸如Gradient Descent之类的优化算法来训练模型并最大程度地降低成本。
-L1损失函数定义为：
\[\begin{align*} & L_1(\hat{y}, y) = \sum_{i=0}^m|y^{(i)} - \hat{y}^{(i)}| \end{align*}\tag{6}
\]
# GRADED FUNCTION: L1
def L1(yhat, y):
"""
Arguments:
yhat -- vector of size m (predicted labels)
y -- vector of size m (true labels)
Returns:
loss -- the value of the L1 loss function defined above
"""
### START CODE HERE ### (≈ 1 line of code)
loss = np.sum(np.abs(yhat-y))
### END CODE HERE ###
return loss
yhat = np.array([.9, 0.2, 0.1, .4, .9])
y = np.array([1, 0, 0, 1, 1])
print("L1 = " + str(L1(yhat,y)))
输出：L1 = 1.1
练习：实现L2损失函数的Numpy向量化版本。 有好几种方法可以实现L2损失函数，但是还是np.dot（）函数更好用。 如果
\[x = [x_1, x_2, ..., x_n]
\]
则np.dot（x，x）=
\[\sum_{j=0}^n x_j^{2}
\]
-L2损失函数定义为：
\[\begin{align*} & L_2(\hat{y},y) = \sum_{i=0}^m(y^{(i)} - \hat{y}^{(i)})^2 \end{align*}\tag{7}
\]
# GRADED FUNCTION: L2
def L2(yhat, y):
"""
Arguments:
yhat -- vector of size m (predicted labels)
y -- vector of size m (true labels)
Returns:
loss -- the value of the L2 loss function defined above
"""
### START CODE HERE ### (≈ 1 line of code)
loss = np.dot(yhat-y,yhat-y)
### END CODE HERE ###
return loss
yhat = np.array([.9, 0.2, 0.1, .4, .9])
y = np.array([1, 0, 0, 1, 1])
print("L2 = " + str(L2(yhat,y)))
输出：L2 = 0.43
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