list 从 R 中的 glm 中提取系数
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extract coefficients from glm in R
提问by lokheart
I have performed a logistic regression with the following result:
我进行了逻辑回归，结果如下：
ssi.logit.single.age["coefficients"]
# $coefficients
# (Intercept) age
# -3.425062382 0.009916508
I need to pick up the coefficient for age, and currently I use the following code:
我需要获取 的系数age，目前我使用以下代码：
ssi.logit.single.age["coefficients"][[1]][2]
It works, but I don't like the cryptic code here, can I use the name of the coefficient (i.e. (Intercept)or age)
它有效，但我不喜欢这里的神秘代码，我可以使用系数的名称吗（即(Intercept)或age）
回答by James
There is an extraction function called coefto get coefficients from models:
有一个提取函数被调用coef来从模型中获取系数：
coef(ssi.logit.single.age)["age"]
回答by lokheart
I've found it, from here
我找到了，从这里
Look at the data structure produced by summary()
看summary()产生的数据结构
> names(summary(lm.D9))
[1] "call" "terms" "residuals" "coefficients"
[5] "aliased" "sigma" "df" "r.squared"
[9] "adj.r.squared" "fstatistic" "cov.unscaled"
Now look at the data structure for the coefficients in the summary:
现在查看摘要中系数的数据结构：
> summary(lm.D9)$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
