从R中的glm中提取系数

Question

I have performed a logistic regression with the following result:

ssi.logit.single.age["coefficients"]
# $coefficients
#  (Intercept)          age 
# -3.425062382  0.009916508 

I need to pick up the coefficient for age , and currently I use the following code:

ssi.logit.single.age["coefficients"][[1]][2]

It works, but I don't like the cryptic code here, can I use the name of the coefficient (ie (Intercept) or age )

Answer 1

有一个名为coef的提取函数可以从模型中获取系数：

coef(ssi.logit.single.age)["age"]

Answer 2

I've found it, from here

Look at the data structure produced by summary()

> names(summary(lm.D9))
  [1] "call"          "terms"         "residuals"     "coefficients"
  [5] "aliased"       "sigma"         "df"            "r.squared"
  [9] "adj.r.squared" "fstatistic"    "cov.unscaled"

Now look at the data structure for the coefficients in the summary:

> summary(lm.D9)$coefficients
             Estimate Std. Error   t value     Pr(>|t|)
(Intercept)    5.032  0.2202177 22.850117 9.547128e-15
groupTrt      -0.371  0.3114349 -1.191260 2.490232e-01

> class(summary(lm.D9)$coefficients)
[1] "matrix"

> summary(lm.D9)$coefficients[,3]
(Intercept)    groupTrt
   22.850117   -1.191260