做几个二元逻辑回归而不是做多项回归是否可行？从这个问题：多项逻辑回归与一对一二元逻辑回归我看到多项回归可能具有较低的标准误差。
但是，我想使用的包还没有推广到多项回归（ncvreg：http ://cran.r-project.org/web/packages/ncvreg/ncvreg.pdf ），所以我想知道我是否可以简单地做几个二元逻辑回归。
几种逻辑回归与多项回归
机器算法验证
r
物流
多项分布
2022-03-28 21:16:58
2个回答
使用多项 logit 模型，您可以施加所有预测概率加起来为 1 的约束。当您使用单独的二进制 logit 模型时，您不能再施加该约束，它们毕竟是在单独的模型中估计的。所以这将是这两个模型之间的主要区别。
正如您在下面的示例中看到的（在 Stata 中，因为这是我最了解的程序），模型往往相似但不相同。我会特别小心推断预测的概率。
// some data preparation
. sysuse nlsw88, clear
(NLSW, 1988 extract)
.
. gen byte occat = cond(occupation < 3 , 1, ///
> cond(inlist(occupation, 5, 6, 8, 13), 2, 3)) ///
> if !missing(occupation)
(9 missing values generated)
. label variable occat "occupation in categories"
. label define occat 1 "high" ///
> 2 "middle" ///
> 3 "low"
. label value occat occat
.
. gen byte middle = (occat == 2) if occat !=1 & !missing(occat)
(590 missing values generated)
. gen byte high = (occat == 1) if occat !=2 & !missing(occat)
(781 missing values generated)
// a multinomial logit model
. mlogit occat i.race i.collgrad , base(3) nolog
Multinomial logistic regression Number of obs = 2237
LR chi2(6) = 218.82
Prob > chi2 = 0.0000
Log likelihood = -2315.9312 Pseudo R2 = 0.0451
-------------------------------------------------------------------------------
occat | Coef. Std. Err. z P>|z| [95% Conf. Interval]
--------------+----------------------------------------------------------------
high |
race |
black | -.4005801 .1421777 -2.82 0.005 -.6792433 -.121917
other | .4588831 .4962591 0.92 0.355 -.5137668 1.431533
|
collgrad |
college grad | 1.495019 .1341625 11.14 0.000 1.232065 1.757972
_cons | -.7010308 .0705042 -9.94 0.000 -.8392165 -.5628451
--------------+----------------------------------------------------------------
middle |
race |
black | .6728568 .1106792 6.08 0.000 .4559296 .889784
other | .2678372 .509735 0.53 0.599 -.7312251 1.266899
|
collgrad |
college grad | .976244 .1334458 7.32 0.000 .714695 1.237793
_cons | -.517313 .0662238 -7.81 0.000 -.6471092 -.3875168
--------------+----------------------------------------------------------------
low | (base outcome)
-------------------------------------------------------------------------------
// separate logits:
. logit high i.race i.collgrad , nolog
Logistic regression Number of obs = 1465
LR chi2(3) = 154.21
Prob > chi2 = 0.0000
Log likelihood = -906.79453 Pseudo R2 = 0.0784
-------------------------------------------------------------------------------
high | Coef. Std. Err. z P>|z| [95% Conf. Interval]
--------------+----------------------------------------------------------------
race |
black | -.5309439 .1463507 -3.63 0.000 -.817786 -.2441017
other | .2670161 .5116686 0.52 0.602 -.735836 1.269868
|
collgrad |
college grad | 1.525834 .1347081 11.33 0.000 1.261811 1.789857
_cons | -.6808361 .0694323 -9.81 0.000 -.816921 -.5447512
-------------------------------------------------------------------------------
. logit middle i.race i.collgrad , nolog
Logistic regression Number of obs = 1656
LR chi2(3) = 90.13
Prob > chi2 = 0.0000
Log likelihood = -1098.9988 Pseudo R2 = 0.0394
-------------------------------------------------------------------------------
middle | Coef. Std. Err. z P>|z| [95% Conf. Interval]
--------------+----------------------------------------------------------------
race |
black | .6942945 .1114418 6.23 0.000 .4758725 .9127164
other | .3492788 .5125802 0.68 0.496 -.6553598 1.353918
|
collgrad |
college grad | .9979952 .1341664 7.44 0.000 .7350339 1.260957
_cons | -.5287625 .0669093 -7.90 0.000 -.6599023 -.3976226
-------------------------------------------------------------------------------
您可以尝试“one vs. all”方法，在这种方法中，您可以训练与您拥有的类一样多的二元分类器。对于每个分类器，正样本是属于该类的样本，其余的是负样本，因此每个逻辑分类器都会为您提供具体样本属于该类的条件概率。
现在，在分类时，您将每个新样本分配给相应分类器为您提供最高概率的类。
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