【问题标题】:Why does the choice of solver result in different weight in sklearn (LogisticRegression)为什么求解器的选择会导致 sklearn 中的权重不同(LogisticRegression)
【发布时间】:2018-04-30 12:35:52
【问题描述】:

为什么不同的求解器会在 sklearn 中对简单问题产生不同的学习权重? 这闻起来不像是数值不稳定性——看起来不同的求解器旨在收敛于不同的权重集。 怎么回事?

这似乎与正则化有关,因为随着 C 的增加效果消失

%matplotlib inline
import time
import pandas as pd
import numpy as np
from sklearn.linear_model.logistic import LogisticRegression

def linear(X,b): return X.dot(b.T)
def sigmoid(X,b): return 1/(1+np.exp(-linear(X,b)))

b = np.array([1.,1.])

def mk(X,y,model):
    model.fit(X,y)
    t = time.clock()
    c = np.concatenate((model.coef_[0,:], model.intercept_[[0]]))
    return c, time.clock()-t

max_iter = 1000
tol=0.00000001
for C in [0.0001, 0.001, 0.1, 1., 10., 1000.]:
    for n in [200, 20000]:
        x = np.arange(-5., 5, 10./n)
        Xb = np.vstack([x, np.ones(len(x))]).T
        p = sigmoid(Xb,b)
        y = np.random.binomial(1, p)
        X = x.reshape((len(x), 1))
        print C, n
        for solver in ["liblinear", "newton-cg", "lbfgs", "sag"]:
            print "  {:>14}".format(solver), mk(X,y, LogisticRegression(C=C, solver=solver, penalty='l2', max_iter=max_iter, tol=tol))

收敛系数如下。使用 B=[1., 1.]

生成训练数据
C      n
0.0001 200
       liblinear (array([ 0.01858615,  0.00219137]), 1.399999999995849e-05)
       newton-cg (array([ 0.01867489,  0.44809872]), 1.5000000000098268e-05)
           lbfgs (array([ 0.01867489,  0.44809872]), 1.5000000000098268e-05)
             sag (array([ 0.01867489,  0.448096  ]), 1.3000000000040757e-05)
0.0001 20000
       liblinear (array([ 0.49062599,  0.14814571]), 2.6999999999999247e-05)
       newton-cg (array([ 0.51257455,  0.59982733]), 2.6000000000081513e-05)
           lbfgs (array([ 0.51257471,  0.59982645]), 2.8999999999834714e-05)
             sag (array([ 0.51257455,  0.59982753]), 1.3000000000040757e-05)
0.001 200
       liblinear (array([ 0.15311436,  0.0154459 ]), 1.2999999999596668e-05)
       newton-cg (array([ 0.15456163,  0.34240309]), 1.3000000000040757e-05)
           lbfgs (array([ 0.15456166,  0.34240284]), 1.3000000000040757e-05)
             sag (array([ 0.15456163,  0.34240308]), 1.100000000020529e-05)
0.001 20000
       liblinear (array([ 0.81980482,  0.58214727]), 3.799999999998249e-05)
       newton-cg (array([ 0.86826695,  0.87335432]), 3.0999999999892225e-05)
           lbfgs (array([ 0.86826513,  0.87335497]), 3.2000000000032e-05)
             sag (array([ 0.86826695,  0.87335433]), 2.9999999999752447e-05)
0.1 200
       liblinear (array([ 0.80161404,  0.39606419]), 2.3999999999801958e-05)
       newton-cg (array([ 0.82243805,  0.57384594]), 2.9000000000056758e-05)
           lbfgs (array([ 0.82243822,  0.57384605]), 3.0000000000196536e-05)
             sag (array([ 0.82243805,  0.57384595]), 2.4999999999941735e-05)
0.1 20000
       liblinear (array([ 0.98022283,  0.97426356]), 3.799999999998249e-05)
       newton-cg (array([ 0.98180586,  0.98060186]), 2.3999999999801958e-05)
           lbfgs (array([ 0.98181404,  0.98060174]), 3.1000000000336314e-05)
             sag (array([ 0.98180586,  0.98060185]), 3.0999999999892225e-05)
1.0 200
       liblinear (array([ 0.77546388,  0.68083769]), 2.3999999999801958e-05)
       newton-cg (array([ 0.78084612,  0.71273566]), 1.1999999999900979e-05)
           lbfgs (array([ 0.7808462 ,  0.71273586]), 1.3000000000040757e-05)
             sag (array([ 0.78084612,  0.71273566]), 1.0000000000065512e-05)
1.0 20000
       liblinear (array([ 1.01286416,  1.01699803]), 3.1000000000336314e-05)
       newton-cg (array([ 1.013046  ,  1.01769302]), 1.6000000000016e-05)
           lbfgs (array([ 1.0130471 ,  1.01769538]), 3.0999999999892225e-05)
             sag (array([ 1.013046  ,  1.01769301]), 2.0999999999826713e-05)
10.0 200
       liblinear (array([ 1.19187232,  1.52139602]), 9.999999999621423e-06)
       newton-cg (array([ 1.19737431,  1.53737059]), 4.300000000023729e-05)
           lbfgs (array([ 1.19737519,  1.53736531]), 1.8999999999991246e-05)
             sag (array([ 1.19737429,  1.53737061]), 1.700000000059987e-05)
10.0 20000
       liblinear (array([ 0.9993963,  0.9748259]), 9.19999999995369e-05)
       newton-cg (array([ 0.9994126 ,  0.97489034]), 3.0999999999892225e-05)
           lbfgs (array([ 0.99941332,  0.97489202]), 3.10000000007804e-05)
             sag (array([ 0.99941261,  0.97489035]), 1.4000000000180535e-05)
1000.0 200
       liblinear (array([ 1.31759982,  1.48775989]), 9.000000000369823e-06)
       newton-cg (array([ 1.3176617 ,  1.48792683]), 1.4000000000180535e-05)
           lbfgs (array([ 1.31766183,  1.48792601]), 4.499999999918458e-05)
             sag (array([ 1.31766164,  1.48792671]), 1.100000000064938e-05)
1000.0 20000
       liblinear (array([ 0.99608584,  0.96883584]), 4.099999999951365e-05)
       newton-cg (array([ 0.99608601,  0.96883648]), 3.300000000017178e-05)
           lbfgs (array([ 0.99608665,  0.96883797]), 2.9999999999752447e-05)
             sag (array([ 0.99608603,  0.96883652]), 1.5000000000320313e-05)

而且,...所有版本似乎都根据人口规模收敛到不同的系数。几乎就好像成本没有被人口规模标准化,而有效的正则化是 C/n。是这样吗?

【问题讨论】:

  • 在我看来,除了 liblinear 之外的所有东西都在收敛到相同的权重,达到一定的数值​​精度(一阶与二阶方法;惊讶地看到 sag 工作得这么好)。基于 liblinear 的方法可能有点hacky,因为内部设计决策与正则化有关(我主要是在猜测)。再说一句:max_iter 效果在 newton-cg/lbfgs 和 liblinear/sag 之间有很大的不同,我会说(但不确定它是否对你的简单案例很重要)
  • 是的——那么 liblinear 是怎么回事?它是 sk.LogisticRegression 的默认值。也很奇怪:他们似乎在优化cost(X,y,w) + (1/C)*reg(w),而不是cost(X,y,w)/samplesize + (1/C)*reg(w)。这意味着随着样本量的增加,用于正则化的特定 C 的强度会降低。这与大多数正则化的描述不同。不明显这是有益的有害的,但肯定很难将 sklearn 与其他库进行苹果对苹果的比较。

标签: scikit-learn logistic-regression


【解决方案1】:

我发现不同的求解器支持不同的成本和梯度函数。

这是我发现不同变体优化的内容:

sklearn.LR(solver=liblinear): L + lam*Rb
sklearn.LR(solver=others):    L + lam*R
statsmodels.GLM(bionomial):   L/n + lam*Rb

地点:

lam = 1/C
L = logloss
n = training sample size
R = square of L2 norm of feature weights
Rb =square of L2 norm of feature weights and intercept

... sklearn 没有按样本大小对损失进行归一化有点奇怪。

【讨论】:

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