我正在使用1000个系数进行LassoCV。Statsmodels似乎无法处理这么多的系数。因此，我正在使用scikit learn。Statsmodel允许使用.fit_constrained("coef1 + coef2...=1")。这将约束系数之和等于1。我需要在Scikit中实现这个功能，并将截距保持为零。
from sklearn.linear_model import LassoCV
LassoCVmodel = LassoCV(fit_intercept=False)
LassoCVmodel.fit(x,y)
非常感谢您的帮助。
""" data """
from time import perf_counter as pc
import numpy as np
from sklearn import datasets
diabetes = datasets.load_diabetes()
A = diabetes.data
y = diabetes.target
alpha=0.1
print('Problem-size: ', A.shape)
def obj(x): # following sklearn's definition from user-guide!
return (1. / (2*A.shape[0])) * np.square(np.linalg.norm(A.dot(x) - y, 2)) + alpha * np.linalg.norm(x, 1)
""" sklearn """
print('\nsklearn classic l1')
from sklearn import linear_model
clf = linear_model.Lasso(alpha=alpha, fit_intercept=False)
t0 = pc()
clf.fit(A, y)
print('used (secs): ', pc() - t0)
print(obj(clf.coef_))
print('sum x: ', np.sum(clf.coef_))
""" cvxpy """
print('\ncvxpy + scs classic l1')
from cvxpy import *
x = Variable(A.shape[1])
objective = Minimize((1. / (2*A.shape[0])) * sum_squares(A*x - y) + alpha * norm(x, 1))
problem = Problem(objective, [])
t0 = pc()
problem.solve(solver=SCS, use_indirect=False, max_iters=10000, verbose=False)
print('used (secs): ', pc() - t0)
print(obj(x.value.flat))
print('sum x: ', np.sum(x.value.flat))
""" cvxpy -> sum x == 1 """
print('\ncvxpy + scs sum == 1 / 1st approach')
objective = Minimize((1. / (2*A.shape[0])) * sum_squares(A*x - y))
constraints = [sum(x) == 1]
problem = Problem(objective, constraints)
t0 = pc()
problem.solve(solver=SCS, use_indirect=False, max_iters=10000, verbose=False)
print('used (secs): ', pc() - t0)
print(obj(x.value.flat))
print('sum x: ', np.sum(x.value.flat))
""" cvxpy approach 2 -> sum x == 1 """
print('\ncvxpy + scs sum == 1 / 2nd approach')
M = 1e6
objective = Minimize((1. / (2*A.shape[0])) * sum_squares(A*x - y) + M*(sum(x) - 1))
constraints = [sum(x) == 1]
problem = Problem(objective, constraints)
t0 = pc()
problem.solve(solver=SCS, use_indirect=False, max_iters=10000, verbose=False)
print('used (secs): ', pc() - t0)
print(obj(x.value.flat))
print('sum x: ', np.sum(x.value.flat))
Problem-size: (442, 10)
sklearn classic l1
used (secs): 0.001451024380348898
13201.3508496
sum x: 891.78869298
cvxpy + scs classic l1
used (secs): 0.011165673357417458
13203.6549995
sum x: 872.520510561
cvxpy + scs sum == 1 / 1st approach
used (secs): 0.15350853891775978
13400.1272148
sum x: -8.43795102327
cvxpy + scs sum == 1 / 2nd approach
used (secs): 0.012579569383536493
13397.2932976
sum x: 1.01207061047
编辑
仅仅为了好玩，我使用了加速投影梯度方法实现了一个非常慢而且没有优化的原型求解器（代码中有备注！）。
尽管这是一种一阶方法，并且在这里表现缓慢（因为没有优化），但对于巨大问题来说，它应该能够更好地扩展。它应该有很多潜力！
警告：可能会被一些人视为高级数值优化 :-)。
编辑 2：我忘记在投影上添加非负约束条件（如果x可以为非负数，则sum(x) == 1没有太多意义！）。这使得求解变得更难（存在数值问题），显然应该使用其中的一种快速专用投影算法（我现在太懒了；我认为n * log n算法可用）。再次强调：这个APG求解器是一个原型，不适用于实际任务。
""" accelerated pg -> sum x == 1 """
def solve_pg(A, b, momentum=0.9, maxiter=1000):
""" remarks:
algorithm: accelerated projected gradient
projection: proj on probability-simplex
-> naive and slow using cvxpy + ecos
line-search: armijo-rule along projection-arc (Bertsekas book)
-> suffers from slow projection
stopping-criterion: naive
gradient-calculation: precomputes AtA
-> not needed and not recommended for huge sparse data!
"""
M, N = A.shape
x = np.zeros(N)
AtA = A.T.dot(A)
Atb = A.T.dot(b)
stop_count = 0
# projection helper
x_ = Variable(N)
v_ = Parameter(N)
objective_ = Minimize(0.5 * square(norm(x_ - v_, 2)))
constraints_ = [sum(x_) == 1]
problem_ = Problem(objective_, constraints_)
def gradient(x):
return AtA.dot(x) - Atb
def obj(x):
return 0.5 * np.linalg.norm(A.dot(x) - b)**2
it = 0
while True:
grad = gradient(x)
# line search
alpha = 1
beta = 0.5
sigma=1e-2
old_obj = obj(x)
while True:
new_x = x - alpha * grad
new_obj = obj(new_x)
if old_obj - new_obj >= sigma * grad.dot(x - new_x):
break
else:
alpha *= beta
x_old = x[:]
x = x - alpha*grad
# projection
v_.value = x
problem_.solve()
x = np.array(x_.value.flat)
y = x + momentum * (x - x_old)
if np.abs(old_obj - obj(x)) < 1e-2:
stop_count += 1
else:
stop_count = 0
if stop_count == 3:
print('early-stopping @ it: ', it)
return x
it += 1
if it == maxiter:
return x
print('\n acc pg')
t0 = pc()
x = solve_pg(A, y)
print('used (secs): ', pc() - t0)
print(obj(x))
print('sum x: ', np.sum(x))
acc pg
early-stopping @ it: 367
used (secs): 0.7714511330487027
13396.8642379
sum x: 1.00000000002
现在，这个 t 和 scikit-learn 中使用的 alpha 参数有什么关系呢？嗯，事实证明通过拉格朗日对偶性，每个 t 的值与一个 alpha 值之间存在一一对应的关系。
这意味着当您使用 LassoCV 时，由于您使用了一系列的 alpha 值，因此您根据定义使用了一系列允许的所有系数之和的值！
总之，所有系数之和等于一的条件等价于针对特定的 alpha 值使用 Lasso。