• Logistic回归介绍

Logistic回归适用于二值相应变量（0/1）。模型假设 Y 服从二项分布，线性模型的拟合形式：

其中 π = μy是 Y的条件均（即给定一系列 X 值时 Y=1的概率），（π/1-π）为 Y =1 时的优势比，log(π/1-π)为对数优势比，或logit。本例中，log(π/1-π)为连接函数，概率分布为二项分布，可用如下代码拟合Logistic回归模型  

glm(Y~X1+X2+X3,family = binomial(link ="logit"),data =mydata)

例

当通过一系列连续型和/或类别型预测变量来预测二值的结果变量时，Logistic回归是一个非常有用的工具

#使用AER包中的数据框Affairs为例，探究婚外情的回归过程> data(Affairs,package = "AER")#导入包中的数据，在函数中也有require(包名)> summary(Affairs)             #先看下描述性统计，知道整体的情况    affairs          gender         age         yearsmarried    children  religiousness     education       occupation    Min.   : 0.000   female:315   Min.   :17.50   Min.   : 0.125   no :171   Min.   :1.000   Min.   : 9.00   Min.   :1.000   1st Qu.: 0.000   male  :286   1st Qu.:27.00   1st Qu.: 4.000   yes:430   1st Qu.:2.000   1st Qu.:14.00   1st Qu.:3.000   Median : 0.000                Median :32.00   Median : 7.000             Median :3.000   Median :16.00   Median :5.000   Mean   : 1.456                Mean   :32.49   Mean   : 8.178             Mean   :3.116   Mean   :16.17   Mean   :4.195   3rd Qu.: 0.000                3rd Qu.:37.00   3rd Qu.:15.000             3rd Qu.:4.000   3rd Qu.:18.00   3rd Qu.:6.000   Max.   :12.000                Max.   :57.00   Max.   :15.000             Max.   :5.000   Max.   :20.00   Max.   :7.000       rating      Min.   :1.000   1st Qu.:3.000   Median :4.000   Mean   :3.932   3rd Qu.:5.000   Max.   :5.000  > table(Affairs$affairs) # 生成交叉表格，会自动统计每类的次数  0   1   2   3   7  12 451  34  17  19  42  38 #Logistic回归是对二值型结果的统计，所以先将数据转化为因子> Affairs$affairs[Affairs$affairs > 0] <- 1   #[Affairs$affairs > 0]为真时，赋值为1> Affairs$affairs[Affairs$affairs == 0] <- 0> Affairs$ynaffair <- factor(Affairs$affairs,levels = c(0,1),labels=c("No,Yes"))  #转化为因子> table(Affairs$ynaffair)#在使用table看下结果No,Yes1 No,Yes2       451     150 #拟合Logistic模型> fit.full <- glm(ynaffair ~ gender + age + yearsmarried + children + +                   religiousness + education + occupation +rating,+                 data=Affairs,family=binomial())> summary(fit.full)Call:glm(formula = ynaffair ~ gender + age + yearsmarried + children +     religiousness + education + occupation + rating, family = binomial(),     data = Affairs)Deviance Residuals:     Min       1Q   Median       3Q      Max  -1.5713  -0.7499  -0.5690  -0.2539   2.5191  Coefficients:              Estimate Std. Error z value Pr(>|z|)    (Intercept)    1.37726    0.88776   1.551 0.120807    gendermale     0.28029    0.23909   1.172 0.241083       #无“*”号表示不显著，即 p>0.05age           -0.04426    0.01825  -2.425 0.015301 *     #"*"越多表示越显著yearsmarried   0.09477    0.03221   2.942 0.003262 ** childrenyes    0.39767    0.29151   1.364 0.172508    religiousness -0.32472    0.08975  -3.618 0.000297 ***education      0.02105    0.05051   0.417 0.676851    occupation     0.03092    0.07178   0.431 0.666630    rating        -0.46845    0.09091  -5.153 2.56e-07 ***---Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1(Dispersion parameter for binomial family taken to be 1)    Null deviance: 675.38  on 600  degrees of freedomResidual deviance: 609.51  on 592  degrees of freedomAIC: 627.51Number of Fisher Scoring iterations: 4

从结果中可以看到，性别、孩子、学历职业等对方程都不显著，可以剔除这些再拟合简单的模型，然后两个模型进行比较，看下简单模型是否合理

#剔除显著的变量，再拟合> fit.reduced <- glm(ynaffair ~ age + yearsmarried + religiousness + +                      rating, data=Affairs, family=binomial())> summary(fit.reduced)Call:glm(formula = ynaffair ~ age + yearsmarried + religiousness +     rating, family = binomial(), data = Affairs)Deviance Residuals:     Min       1Q   Median       3Q      Max  -1.6278  -0.7550  -0.5701  -0.2624   2.3998  Coefficients:              Estimate Std. Error z value Pr(>|z|)    (Intercept)    1.93083    0.61032   3.164 0.001558 ** age           -0.03527    0.01736  -2.032 0.042127 *  yearsmarried   0.10062    0.02921   3.445 0.000571 ***religiousness -0.32902    0.08945  -3.678 0.000235 ***rating        -0.46136    0.08884  -5.193 2.06e-07 ***---Signif. codes:  0 ‘***’ 0.001 ‘**’ 0.01 ‘*’ 0.05 ‘.’ 0.1 ‘ ’ 1(Dispersion parameter for binomial family taken to be 1)    Null deviance: 675.38  on 600  degrees of freedomResidual deviance: 615.36  on 596  degrees of freedomAIC: 625.36                                              #发现 简单模型的AIC值比之前的模型的要小，说明是可行的，然后我们也可以用anova()对两次拟合模型进行比较Number of Fisher Scoring iterations: 4

由于两个模型嵌套(fit.reduced是fit.full的一个子集)可以使用anova()进行比较， 对于广义线性模型，可以卡方检验

##使用anova()对两个嵌套模型进行比较，广义线性回归使用Chisp（卡方检验）> anova(fit.full,fit.reduced,test="Chisq")Analysis of Deviance TableModel 1: ynaffair ~ gender + age + yearsmarried + children + religiousness +     education + occupation + ratingModel 2: ynaffair ~ age + yearsmarried + religiousness + rating  Resid. Df Resid. Dev Df Deviance Pr(>Chi)1       592     609.51                     2       596     615.36 -4  -5.8474   0.2108   #卡方值不显著（p=0.217）表明四个预测变量的新模型与九个完整预测变量的模型拟合程度一样好