Python：如何防止Scipy'；s通过改变初始猜测x0的形状来优化函数？
我正在尝试从Scipy实现优化算法。当我在没有输入雅可比梯度函数的情况下实现它时，它工作得很好。我相信当我输入梯度时，我遇到的问题是因为最小化函数本身正在改变初始值x0的形状。您可以从下面代码的输出中看到这一点 输入：Python：如何防止Scipy'；s通过改变初始猜测x0的形状来优化函数？,python,numpy,optimization,scipy,minimize,Python,Numpy,Optimization,Scipy,Minimize,我正在尝试从Scipy实现优化算法。当我在没有输入雅可比梯度函数的情况下实现它时，它工作得很好。我相信当我输入梯度时，我遇到的问题是因为最小化函数本身正在改变初始值x0的形状。您可以从下面代码的输出中看到这一点 输入： import numpy as np from costFunction import * import scipy.optimize as op def sigmoid(z): epsilon = np.finfo(z.dtype).eps g = 1/(
import numpy as np
from costFunction import *
import scipy.optimize as op
def sigmoid(z):
epsilon = np.finfo(z.dtype).eps
g = 1/(1+np.exp(-z))
g = np.clip(g,epsilon,1-epsilon)
return g
def costFunction(theta,X,y):
m = y.size
h = sigmoid(X@theta)
J = 1/(m)*(-y.T@np.log(h)-(1-y).T@np.log(1-h))
grad = 1/m*X.T@(h-y)
print ('Shape of theta is',np.shape(theta),'\n')
print ('Shape of gradient is',np.shape(grad),'\n')
return J, grad
X = np.array([[1, 3],[5,7]])
y = np.array([[1],[0]])
m,n = np.shape(X)
one_vec = np.ones((m,1))
X = np.hstack((one_vec,X))
initial_theta = np.zeros((n+1,1))
print ('Running costFunction before executing minimize function...\n')
cost, grad = costFunction(initial_theta,X,y) #To test the shape of gradient before calling minimize
print ('Executing minimize function...\n')
Result = op.minimize(costFunction,initial_theta,args=(X,y),method='TNC',jac=True,options={'maxiter':400})
Running costFunction before executing minimize function...
Shape of theta is (3, 1)
Traceback (most recent call last):
Shape of gradient is (3, 1)
Executing minimize function...
Shape of theta is (3,)
File "C:/Users/#####/minimizeshapechange.py", line 34, in <module>
Shape of gradient is (3, 2)
Result = op.minimize(costFunction,initial_theta,args=(X,y),method='TNC',jac=True,options={'maxiter':400})
File "C:\Users\#####\anaconda3\lib\site-packages\scipy\optimize\_minimize.py", line 453, in minimize
**options)
File "C:\Users\#####\anaconda3\lib\site-packages\scipy\optimize\tnc.py", line 409, in _minimize_tnc
xtol, pgtol, rescale, callback)
ValueError: tnc: invalid gradient vector from minimized function.
Process finished with exit code 1
在执行最小化函数之前运行costFunction。。。
θ的形状是（3，1）
回溯（最近一次呼叫最后一次）：
梯度形状为（3，1）
正在执行最小化函数。。。
θ的形状是（3，）
文件“C:/Users/#####/minimizeshapechange.py”，第34行，在
梯度形状为（3,2）
Result=op.minimize（costFunction，initial_theta，args=（X，y），method='TNC'，jac=True，options={maxiter'：400}）
文件“C:\Users\\\\\\\\\\\\\\\\\\\\\\\\\anaconda3\lib\site packages\scipy\optimize\\ u minimize.py”，第453行，最小化
**选项）
文件“C:\Users\######\anaconda3\lib\site packages\scipy\optimize\tnc.py”，第409行，在tnc中
xtol、pgtol、重新缩放、回调）
ValueError:tnc:最小化函数的梯度向量无效。
进程已完成，退出代码为1
- （1） 你的梯度被打破了！
- scipy期望一个与您的
相同的数组x0 - 您的渐变形状为
，而（3,2）
是预期的（n+1,1） - 与使用
（der=导数）的教程中给出的示例进行比较scipy.optimize.rosen\u der
- scipy期望一个与您的
- （2） 您的scipy版本似乎有点旧，因为我的（0.19.0）告诉我：
ValueError:tnc:最小化函数的梯度向量无效。
备注：上述代码是5年前更改/触摸/引入的。如果在使用列出的代码时（删除了导入costFunction），您确实没有遇到此错误，那么您似乎正在使用scipy
gradgrad.ravelinitial\u theta=np.zeros（（n+1））[：，np.newaxis]如果有人能澄清这些观点，那就太好了！谢谢你的成本函数代码错了，也许你应该看看
def costFunction(theta,X,y):
h_theta = sigmoid(X@theta)
J = (-y) * np.log(h_theta) - (1 - y) * np.log(1 - h_theta)
return np.mean(J)
请在单独单元格中以jpuiter in1等格式复制并通过
In 1
import pandas as pd
import numpy as np
import matplotlib.pyplot as plt
import seaborn as sns
%matplotlib inline
filepath =('C:/Pythontry/MachineLearning/dataset/couresra/ex2data1.txt')
data =pd.read_csv(filepath,sep=',',header=None)
#print(data)
X = data.values[:,:2] #(100,2)
y = data.values[:,2:3] #(100,1)
#print(np.shape(y))
#In 2
#%% ==================== Part 1: Plotting ====================
postive_value = data.loc[data[2] == 1]
#print(postive_value.values[:,2:3])
negative_value = data.loc[data[2] == 0]
#print(len(postive_value))
#print(len(negative_value))
ax1 = postive_value.plot(kind='scatter',x=0,y=1,s=50,color='b',marker="+",label="Admitted") # S is line width #https://matplotlib.org/api/_as_gen/matplotlib.axes.Axes.scatter.html#matplotlib.axes.Axes.scatter
ax2 = negative_value.plot(kind='scatter',x=0,y=1,s=50,color='y',ax=ax1,label="Not Admitted")
ax1.set_xlabel("Exam 1 score")
ax2.set_ylabel("Exam 2 score")
plt.show()
#print(ax1 == ax2)
#print(np.shape(X))
# In 3
#============ Part 2: Compute Cost and Gradient ===========
[m,n] = np.shape(X) #(100,2)
print(m,n)
additional_coulmn = np.ones((m,1))
X = np.append(additional_coulmn,X,axis=1)
initial_theta = np.zeros((n+1), dtype=int)
print(initial_theta)
# In4
#Sigmoid and cost function
def sigmoid(z):
g = np.zeros(np.shape(z));
g = 1/(1+np.exp(-z));
return g
def costFunction(theta, X, y):
J = 0;
#print(theta)
receive_theta = np.array(theta)[np.newaxis] ##This command is used to create the 1D array
#print(receive_theta)
theta = np.transpose(receive_theta)
#print(np.shape(theta))
#grad = np.zeros(np.shape(theta))
z = np.dot(X,theta) # where z = theta*X
#print(z)
h = sigmoid(z) #formula h(x) = g(z) whether g = 1/1+e(-z) #(100,1)
#print(np.shape(h))
#J = np.sum(((-y)*np.log(h)-(1-y)*np.log(1-h))/m);
J = np.sum(np.dot((-y.T),np.log(h))-np.dot((1-y).T,np.log(1-h)))/m
#J = (-y * np.log(h) - (1 - y) * np.log(1 - h)).mean()
#error = h-y
#print(np.shape(error))
#print(np.shape(X))
grad =np.dot(X.T,(h-y))/m
#print(grad)
return J,grad
#In5
[cost, grad] = costFunction(initial_theta, X, y)
print('Cost at initial theta (zeros):', cost)
print('Expected cost (approx): 0.693\n')
print('Gradient at initial theta (zeros): \n',grad)
print('Expected gradients (approx):\n -0.1000\n -12.0092\n -11.2628\n')
In6 # Compute and display cost and gradient with non-zero theta
test_theta = [-24, 0.2, 0.2]
#test_theta_value = np.array([-24, 0.2, 0.2])[np.newaxis] #This command is used to create the 1D row array
#test_theta = np.transpose(test_theta_value) # Transpose
#test_theta = test_theta_value.transpose()
[cost, grad] = costFunction(test_theta, X, y)
print('\nCost at test theta: \n', cost)
print('Expected cost (approx): 0.218\n')
print('Gradient at test theta: \n',grad);
print('Expected gradients (approx):\n 0.043\n 2.566\n 2.647\n')
#IN6
# ============= Part 3: Optimizing using range =============
import scipy.optimize as opt
#initial_theta_initialize = np.array([0, 0, 0])[np.newaxis]
#initial_theta = np.transpose(initial_theta_initialize)
print ('Executing minimize function...\n')
# Working models
#result = opt.minimize(costFunction,initial_theta,args=(X,y),method='TNC',jac=True,options={'maxiter':400})
result = opt.fmin_tnc(func=costFunction, x0=initial_theta, args=(X, y))
# Not working model
#costFunction(initial_theta,X,y)
#model = opt.minimize(fun = costFunction, x0 = initial_theta, args = (X, y), method = 'TNC',jac = costFunction)
print('Thetas found by fmin_tnc function: ', result);
print('Cost at theta found : \n', cost);
print('Expected cost (approx): 0.203\n');
print('theta: \n',result[0]);
print('Expected theta (approx):\n');
print(' -25.161\n 0.206\n 0.201\n');
对我有效的方法是将y重塑为向量（1-D）而不是矩阵（2-D数组）。我只是使用了下面的代码，然后重新运行SciPy的minimize函数，它就成功了
y=np.重塑（y，100）#例如，如果你的y变量有100个数据点。有点晚了，但我也开始使用Python实现anderw赋值，并投入了大量精力来解决上述问题。最后是我的作品 这对我很有帮助，但在函数调用中有一个更改，请参阅以下内容：-
result=op.fmin\tnc（func=costFunction，x0=initial\u theta，fprime=None，approx\u grad=True，args=（X，y））（n，）（n，1）执行优化函数前θ的形状：（3，）…costFunction内θ的形状是（3，）…梯度：[[0.0.][1.1.][1.1.]initial_theta=np.zeros（n+1）initial_theta = np.zeros((n+1))
initial_theta.shape
Gradient(initial_theta,X,y).shape
import scipy.optimize as opt
model = opt.minimize(fun = CostFunc, x0 = initial_theta, args = (X, y), method = 'TNC', jac = Gradient)
def costFunction(theta,X,y):
h_theta = sigmoid(X@theta)
J = (-y) * np.log(h_theta) - (1 - y) * np.log(1 - h_theta)
return np.mean(J)
In 1
import pandas as pd
import numpy as np
import matplotlib.pyplot as plt
import seaborn as sns
%matplotlib inline
filepath =('C:/Pythontry/MachineLearning/dataset/couresra/ex2data1.txt')
data =pd.read_csv(filepath,sep=',',header=None)
#print(data)
X = data.values[:,:2] #(100,2)
y = data.values[:,2:3] #(100,1)
#print(np.shape(y))
#In 2
#%% ==================== Part 1: Plotting ====================
postive_value = data.loc[data[2] == 1]
#print(postive_value.values[:,2:3])
negative_value = data.loc[data[2] == 0]
#print(len(postive_value))
#print(len(negative_value))
ax1 = postive_value.plot(kind='scatter',x=0,y=1,s=50,color='b',marker="+",label="Admitted") # S is line width #https://matplotlib.org/api/_as_gen/matplotlib.axes.Axes.scatter.html#matplotlib.axes.Axes.scatter
ax2 = negative_value.plot(kind='scatter',x=0,y=1,s=50,color='y',ax=ax1,label="Not Admitted")
ax1.set_xlabel("Exam 1 score")
ax2.set_ylabel("Exam 2 score")
plt.show()
#print(ax1 == ax2)
#print(np.shape(X))
# In 3
#============ Part 2: Compute Cost and Gradient ===========
[m,n] = np.shape(X) #(100,2)
print(m,n)
additional_coulmn = np.ones((m,1))
X = np.append(additional_coulmn,X,axis=1)
initial_theta = np.zeros((n+1), dtype=int)
print(initial_theta)
# In4
#Sigmoid and cost function
def sigmoid(z):
g = np.zeros(np.shape(z));
g = 1/(1+np.exp(-z));
return g
def costFunction(theta, X, y):
J = 0;
#print(theta)
receive_theta = np.array(theta)[np.newaxis] ##This command is used to create the 1D array
#print(receive_theta)
theta = np.transpose(receive_theta)
#print(np.shape(theta))
#grad = np.zeros(np.shape(theta))
z = np.dot(X,theta) # where z = theta*X
#print(z)
h = sigmoid(z) #formula h(x) = g(z) whether g = 1/1+e(-z) #(100,1)
#print(np.shape(h))
#J = np.sum(((-y)*np.log(h)-(1-y)*np.log(1-h))/m);
J = np.sum(np.dot((-y.T),np.log(h))-np.dot((1-y).T,np.log(1-h)))/m
#J = (-y * np.log(h) - (1 - y) * np.log(1 - h)).mean()
#error = h-y
#print(np.shape(error))
#print(np.shape(X))
grad =np.dot(X.T,(h-y))/m
#print(grad)
return J,grad
#In5
[cost, grad] = costFunction(initial_theta, X, y)
print('Cost at initial theta (zeros):', cost)
print('Expected cost (approx): 0.693\n')
print('Gradient at initial theta (zeros): \n',grad)
print('Expected gradients (approx):\n -0.1000\n -12.0092\n -11.2628\n')
In6 # Compute and display cost and gradient with non-zero theta
test_theta = [-24, 0.2, 0.2]
#test_theta_value = np.array([-24, 0.2, 0.2])[np.newaxis] #This command is used to create the 1D row array
#test_theta = np.transpose(test_theta_value) # Transpose
#test_theta = test_theta_value.transpose()
[cost, grad] = costFunction(test_theta, X, y)
print('\nCost at test theta: \n', cost)
print('Expected cost (approx): 0.218\n')
print('Gradient at test theta: \n',grad);
print('Expected gradients (approx):\n 0.043\n 2.566\n 2.647\n')
#IN6
# ============= Part 3: Optimizing using range =============
import scipy.optimize as opt
#initial_theta_initialize = np.array([0, 0, 0])[np.newaxis]
#initial_theta = np.transpose(initial_theta_initialize)
print ('Executing minimize function...\n')
# Working models
#result = opt.minimize(costFunction,initial_theta,args=(X,y),method='TNC',jac=True,options={'maxiter':400})
result = opt.fmin_tnc(func=costFunction, x0=initial_theta, args=(X, y))
# Not working model
#costFunction(initial_theta,X,y)
#model = opt.minimize(fun = costFunction, x0 = initial_theta, args = (X, y), method = 'TNC',jac = costFunction)
print('Thetas found by fmin_tnc function: ', result);
print('Cost at theta found : \n', cost);
print('Expected cost (approx): 0.203\n');
print('theta: \n',result[0]);
print('Expected theta (approx):\n');
print(' -25.161\n 0.206\n 0.201\n');
opt.fmin_tnc(func = costFunction, x0 = theta.flatten(),fprime = gradient, args = (X, y.flatten()))