其实就是猫的
分类
,☺
打开Google Colab后,首先第一步挂载Google Drive
!apt-get install -y -qq software-properties-common python-software-properties module-init-tools
!add-apt-repository -y ppa:alessandro-strada/ppa 2>&1 > /dev/null
!apt-get update -qq 2>&1 > /dev/null
!apt-get -y install -qq google-drive-ocamlfuse fuse
from google.colab import auth
auth.authenticate_user()
from oauth2client.client import GoogleCredentials
creds = GoogleCredentials.get_application_default()
import getpass
!google-drive-ocamlfuse -headless -id={creds.client_id} -secret={creds.client_secret} < /dev/null 2>&1 | grep URL
vcode = getpass.getpass()
!echo {vcode} | google-drive-ocamlfuse -headless -id={creds.client_id} -secret={creds.client_secret}
!mkdir -p drive
!google-drive-ocamlfuse drive
import os
path = "/content/drive/assignment2(C1W2)"
os.chdir(path)
os.listdir(path)
通过以上操作,完成了Google Drive的挂载,现在可以继续进行后面的工作了
在接下来的工作里,需要用的包有:
标记
了猫和非猫的m_train
图片标记
了猫和非猫的m_test
图片(209, 64, 64, 3)
# (样本量,像素值,像素值,RGB)
Number of training examples: m_train = 209
Number of testing examples: m_test = 50
Height/Width of each image: num_px = 64
Each image is of size: (64, 64, 3)
train_set_x shape: (209, 64, 64, 3)
train_set_y shape: (1, 209)
test_set_x shape: (50, 64, 64, 3)
test_set_y shape: (1, 50)
矩阵转换
将一个图像的复杂的矩阵(64,64,4)转换为一个numpy array(12288),
train_set_x_flatten = train_set_x_orig.reshape(train_set_x_orig.shape[0],64*64*3).T
test_set_x_flatten = test_set_x_orig.reshape(test_set_x_orig.shape[0],64*64*3).T
标准化数据集
mark
:why?
train_set_x = train_set_x_flatten/255.
test_set_x = test_set_x_flatten/255.
如何建立一个神经网络?
Sigmoid实现 :
# GRADED FUNCTION: sigmoid
def sigmoid(z):
"""
Compute the sigmoid of z
Arguments:
z -- A scalar or numpy array of any size.
Return:
s -- sigmoid(z)
"""
### START CODE HERE ### (≈ 1 line of code)
s = np.true_divide(1,1+np.true_divide(1,np.exp(z)))
### END CODE HERE ###
return s
初始化w和b
# GRADED FUNCTION: initialize_with_zeros
def initialize_with_zeros(dim):
"""
This function creates a vector of zeros of shape (dim, 1) for w and initializes b to 0.
Argument:
dim -- size of the w vector we want (or number of parameters in this case)
Returns:
w -- initialized vector of shape (dim, 1)
b -- initialized scalar (corresponds to the bias)
"""
### START CODE HERE ### (≈ 1 line of code)
w = np.zeros((dim,1))
b = 0
### END CODE HERE ###
assert(w.shape == (dim, 1))
assert(isinstance(b, float) or isinstance(b, int))
return w, b
Here are the two formulas you will be using:
(7) ∂ J ∂ w = 1 m X ( A − Y ) T \frac{\partial J}{\partial w} = \frac{1}{m}X(A-Y)^T\tag{7} ∂w∂J=m1X(A−Y)T(7)
(8) ∂ J ∂ b = 1 m ∑ i = 1 m ( a ( i ) − y ( i ) ) \frac{\partial J}{\partial b} = \frac{1}{m} \sum_{i=1}^m (a^{(i)}-y^{(i)})\tag{8} ∂b∂J=m1i=1∑m(a(i)−y(i))(8)
# GRADED FUNCTION: propagate
def propagate(w, b, X, Y):
"""
Implement the cost function and its gradient for the propagation explained above
Arguments:
w -- weights, a numpy array of size (num_px * num_px * 3, 1)
b -- bias, a scalar
X -- data of size (num_px * num_px * 3, number of examples)
Y -- true "label" vector (containing 0 if non-cat, 1 if cat) of size (1, number of examples)
Return:
cost -- negative log-likelihood cost for logistic regression
dw -- gradient of the loss with respect to w, thus same shape as w
db -- gradient of the loss with respect to b, thus same shape as b
Tips:
- Write your code step by step for the propagation. np.log(), np.dot()
"""
m = X.shape[1]
# FORWARD PROPAGATION (FROM X TO COST)
### START CODE HERE ### (≈ 2 lines of code)
A = sigmoid(np.dot(w.T,X)+b)
cost = -(np.dot(Y,np.log(A).T)+np.dot(1-Y,np.log(1-A).T))/m
### END CODE HERE ###
# BACKWARD PROPAGATION (TO FIND GRAD)
### START CODE HERE ### (≈ 2 lines of code)
dw = np.dot(X,(A-Y).T)/m
db = np.sum(A-Y)/m
### END CODE HERE ###
assert(dw.shape == w.shape)
assert(db.dtype == float)
cost = np.squeeze(cost)
assert(cost.shape == ())
grads = {"dw": dw,
"db": db}
return grads, cost
更新参数
# GRADED FUNCTION: optimize
def optimize(w, b, X, Y, num_iterations, learning_rate, print_cost = False):
"""
This function optimizes w and b by running a gradient descent algorithm
Arguments:
w -- weights, a numpy array of size (num_px * num_px * 3, 1)
b -- bias, a scalar
X -- data of shape (num_px * num_px * 3, number of examples)
Y -- true "label" vector (containing 0 if non-cat, 1 if cat), of shape (1, number of examples)
num_iterations -- number of iterations of the optimization loop
learning_rate -- learning rate of the gradient descent update rule
print_cost -- True to print the loss every 100 steps
Returns:
params -- dictionary containing the weights w and bias b
grads -- dictionary containing the gradients of the weights and bias with respect to the cost function
costs -- list of all the costs computed during the optimization, this will be used to plot the learning curve.
Tips:
You basically need to write down two steps and iterate through them:
1) Calculate the cost and the gradient for the current parameters. Use propagate().
2) Update the parameters using gradient descent rule for w and b.
"""
costs = []
for i in range(num_iterations):
# Cost and gradient calculation (≈ 1-4 lines of code)
### START CODE HERE ###
grads, cost = propagate(w,b,X,Y)
### END CODE HERE ###
# Retrieve derivatives from grads
dw = grads["dw"]
db = grads["db"]
# update rule (≈ 2 lines of code)
### START CODE HERE ###
w = w-learning_rate*dw
b = b-learning_rate*db
### END CODE HERE ###
# Record the costs
if i % 100 == 0:
costs.append(cost)
# Print the cost every 100 training examples
if print_cost and i % 100 == 0:
print ("Cost after iteration %i: %f" %(i, cost))
params = {"w": w,
"b": b}
grads = {"dw": dw,
"db": db}
return params, grads, costs
实现预测
# GRADED FUNCTION: predict
def predict(w, b, X):
'''
Predict whether the label is 0 or 1 using learned logistic regression parameters (w, b)
Arguments:
w -- weights, a numpy array of size (num_px * num_px * 3, 1)
b -- bias, a scalar
X -- data of size (num_px * num_px * 3, number of examples)
Returns:
Y_prediction -- a numpy array (vector) containing all predictions (0/1) for the examples in X
'''
m = X.shape[1]
Y_prediction = np.zeros((1,m))
w = w.reshape(X.shape[0], 1)
# Compute vector "A" predicting the probabilities of a cat being present in the picture
### START CODE HERE ### (≈ 1 line of code)
A = sigmoid(np.dot(w.T,X)+b)
### END CODE HERE ###
for i in range(A.shape[1]):
# Convert probabilities A[0,i] to actual predictions p[0,i]
### START CODE HERE ### (≈ 4 lines of code)
Y_prediction[0,i] = int(A[0,i]>0.5)
### END CODE HERE ###
assert(Y_prediction.shape == (1, m))
return Y_prediction
模型集成
# GRADED FUNCTION: model
def model(X_train, Y_train, X_test, Y_test, num_iterations = 2000, learning_rate = 0.5, print_cost = False):
"""
Builds the logistic regression model by calling the function you've implemented previously
Arguments:
X_train -- training set represented by a numpy array of shape (num_px * num_px * 3, m_train)
Y_train -- training labels represented by a numpy array (vector) of shape (1, m_train)
X_test -- test set represented by a numpy array of shape (num_px * num_px * 3, m_test)
Y_test -- test labels represented by a numpy array (vector) of shape (1, m_test)
num_iterations -- hyperparameter representing the number of iterations to optimize the parameters
learning_rate -- hyperparameter representing the learning rate used in the update rule of optimize()
print_cost -- Set to true to print the cost every 100 iterations
Returns:
d -- dictionary containing information about the model.
"""
### START CODE HERE ###
# initialize parameters with zeros (≈ 1 line of code)
w, b = np.zeros((X_train.shape[0],1)),0
# Gradient descent (≈ 1 line of code)
parameters, grads, costs = optimize(w,b,X_train,Y_train,num_iterations,learning_rate,print_cost)
# Retrieve parameters w and b from dictionary "parameters"
w = parameters["w"]
b = parameters["b"]
# Predict test/train set examples (≈ 2 lines of code)
Y_prediction_test = predict(w,b,X_test)
Y_prediction_train = predict(w,b,X_train)
### END CODE HERE ###
# Print train/test Errors
print("train accuracy: {} %".format(100 - np.mean(np.abs(Y_prediction_train - Y_train)) * 100))
print("test accuracy: {} %".format(100 - np.mean(np.abs(Y_prediction_test - Y_test)) * 100))
d = {"costs": costs,
"Y_prediction_test": Y_prediction_test,
"Y_prediction_train" : Y_prediction_train,
"w" : w,
"b" : b,
"learning_rate" : learning_rate,
"num_iterations": num_iterations}
return d
略微失望的是今天实现的图片分类器准确度真的是一般!!!
在实现过程当中主要碰到的问题便是维度的问题
,在模型集成过程时,把w用样本个数来进行初始化了。
numpy函数记得不牢,查阅了几个函数:
np.true_divide:注意区分
np.divide()和np.true_divide()的区别
广播机制:np_arr - 1
,和1 - np_arr
都是会进行广播的,刚开始愚蠢的给倒过来了,以为不可以
为什么要进行标准化
?