Python 用scikit学习线性模型约束系数和
我正在做一个有1000个COEF的拉索夫。Statsmodels似乎无法处理如此多的COEF。所以我使用scikit学习。允许.fit_约束的Statsmodel(“coef1+coef2…=1”)。这将coefs之和限制为=1。我需要在Scikit中执行此操作。我还将截距保持在零Python 用scikit学习线性模型约束系数和,python,machine-learning,scikit-learn,regression,sklearn-pandas,Python,Machine Learning,Scikit Learn,Regression,Sklearn Pandas,我正在做一个有1000个COEF的拉索夫。Statsmodels似乎无法处理如此多的COEF。所以我使用scikit学习。允许.fit_约束的Statsmodel(“coef1+coef2…=1”)。这将coefs之和限制为=1。我需要在Scikit中执行此操作。我还将截距保持在零 from sklearn.linear_model import LassoCV LassoCVmodel = LassoCV(fit_intercept=False) LassoCVmodel.fit(x,y)
from sklearn.linear_model import LassoCV
LassoCVmodel = LassoCV(fit_intercept=False)
LassoCVmodel.fit(x,y)
任何帮助都将不胜感激 如评论中所述:文档和资料来源未表明sklearn支持此功能 我只是尝试了使用现成的凸优化解算器。这只是一个简单的原型方法,可能不适合您(未完全定义)的任务(样本量?) 一些评论:
- 实施/模型制定很容易
- 这个问题比我想象的更难解决
- 解算器ECOS有一般性问题
- 解算器SCS达到良好精度(比sklearn差)
- 但是:调整迭代以提高精度会破坏解算器
- 这个问题对于SCS来说是不可行的李>
- 基于SCS+bigM的公式(约束被张贴为目标内的惩罚条款)看起来可用;但可能需要调整
- 只有开源解算器经过测试,商业解算器可能更好
- 为了解决巨大的问题(性能比鲁棒性和准确性更重要),一种(加速的)投影随机梯度方法看起来很有希望
我很惊讶之前没有人在评论中提到这一点,但我认为你的问题陈述中存在概念上的误解 让我们从套索估计量的定义开始,例如,如Hastine、Tibshirani和Wainwright在《统计学习与稀疏套索和推广》中给出的: 给定N个预测器响应对{(xi,yi)}的集合
不支持的套索查找。您可能需要从头开始编写一些东西,修改sklearn或使用其他库。我可以使用sklearn.model_selection.GridSearchCV和fit_params吗?不。这没有任何意义。好的,谢谢@Saschaha如果这项任务对您很重要,您可以在cvxpy中轻松地表述此问题,并且解决应该足够快(虽然你没有说明你得到了多少样品)。哇,这非常有用。谢谢你花时间。
""" 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
""" 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