问题描述
我有兴趣在 AWS::Serverless::Api/DeFinitionBody
中参数矩阵的每一行中应用 l2
范数约束。到目前为止我尝试过的是
scipy.optimize.minimize
但计算参数 def l2_const(x):
x = x.reshape(r,c)
b = np.sqrt((x**2).sum(axis=1)) - 1
return np.broadcast_to(b[:,None],(r,c)).flatten()
x0 = np.random.random((r,c))
const = ({'type': 'eq','fun': l2_const},)
f_min = minimize(fun=cost,x0=x0,method='SLSQP',jac=gradient,constraints=const)
都是零。有谁知道如何正确实施这种类型的约束?
编辑 1:可以在我之前对 post 的回答中找到应用此类约束的示例。
编辑 2:您可以在下面找到一个完整的工作示例。当使用约束时,结果非常低。欢迎提出任何建议。
班级:
f_min.x
主要内容:
import numpy as np
from scipy.optimize import minimize
from sklearn import preprocessing
class myLR():
def __init__(self,reltol=1e-8,maxit=1000,opt_method=None,verbose=True,seed=0):
self.maxit = maxit
self.reltol = reltol
self.seed = seed
self.verbose = verbose
self.opt_method = opt_method
self.lbin = preprocessing.LabelBinarizer()
def w_2d(self,w,n_classes):
return np.reshape(w,(n_classes,-1))
def softmax(self,W,X):
a = np.exp(X @ W.T)
o = a / np.sum(a,axis=1,keepdims=True)
return o
def squared_norm(self,x):
x = np.ravel(x,order='K')
return np.dot(x,x)
def cost(self,X,T,n_samples,n_classes):
W = self.w_2d(W,n_classes)
log_O = np.log(self.softmax(W,X))
c = -(T * log_O).sum()
return c / n_samples
def gradient(self,n_classes)
O = self.softmax(W,X)
grad = -(T - O).T.dot(X)
return grad.ravel() / n_samples
def l1_constraint(self,x,n_classes,n_features):
x = x.reshape(n_classes,-1)
b = x.sum(axis=1) - 1
return np.broadcast_to(b[:,n_features)).flatten()
def fit(self,y=None):
n_classes = len(np.unique(y))
n_samples,n_features = X.shape
if n_classes == 2:
T = np.zeros((n_samples,n_classes),dtype=np.float64)
for i,cls in enumerate(np.unique(y)):
T[y == cls,i] = 1
else:
T = self.lbin.fit_transform(y)
np.random.seed(self.seed)
W_0 = np.random.random((n_classes,n_features))
const = ({'type': 'eq','fun': self.l1_constraint,'args': (n_classes,n_features,)},)
options = {'disp': self.verbose,'maxiter': self.maxit}
f_min = minimize(fun=self.cost,x0=W_0,args=(X,method=self.opt_method,constraints=const,jac=self.gradient,options=options)
self.coef_ = self.w_2d(f_min.x,n_classes)
self.W_ = self.coef_
return self
def predict_proba(self,X):
O = self.softmax(self.W_,X)
return O
def predict(self,X):
sigma = self.predict_proba(X)
y_pred = np.argmax(sigma,axis=1)
return y_pred
编辑 3:我用
替换了约束import numpy as np
from sklearn import datasets
import matplotlib.pyplot as plt
from sklearn.metrics import accuracy_score
from myLR import myLR
iris = datasets.load_iris()
X = iris.data[:,0:2]
y = iris.target
par_dict2 = {'reltol': 1e-6,'maxit': 20000,'verbose': 20,'seed': 0}
# Create different classifiers.
classifiers = {
'myLR': myLR(**par_dict2),}
n_classifiers = len(classifiers)
plt.figure(figsize=(3 * 2,n_classifiers * 2))
plt.subplots_adjust(bottom=.2,top=.95)
xx = np.linspace(3,9,100)
yy = np.linspace(1,5,100).T
xx,yy = np.meshgrid(xx,yy)
Xfull = np.c_[xx.ravel(),yy.ravel()]
accuracy_score
for index,(name,classifier) in enumerate(classifiers.items()):
classifier.fit(X,y)
coef_ = classifier.coef_
print(np.linalg.norm(coef_,axis=1))
y_pred = classifier.predict(X)
accuracy = accuracy_score(y,y_pred)
print("Accuracy (train) for %s: %0.1f%% " % (name,accuracy * 100))
# View probabilities:
probas = classifier.predict_proba(Xfull)
n_classes = np.unique(y_pred).size
for k in range(n_classes):
plt.subplot(n_classifiers,index * n_classes + k + 1)
plt.title("Class %d" % k)
if k == 0:
plt.ylabel(name)
imshow_handle = plt.imshow(probas[:,k].reshape((100,100)),extent=(3,1,5),origin='lower')
plt.xticks(())
plt.yticks(())
idx = (y_pred == k)
if idx.any():
plt.scatter(X[idx,0],X[idx,1],marker='o',c='w',edgecolor='k')
ax = plt.axes([0.15,0.04,0.7,0.05])
plt.title("Probability")
plt.colorbar(imshow_handle,cax=ax,orientation='horizontal')
plt.show()
它产生更好的结果。但是,计算出的分量 def l1_constraint(self,n_features):
x = x.reshape(n_classes,-1)
b = x.sum(axis=1) - 1
return b
和 x1
的总和不为 1?可以吗?
解决方法
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