本文介绍了从R中的方差分析(glm)中提取残留偏差的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着小编来一起学习吧!
问题描述
我在R中安装了glm模型,并拿了方差分析表.我需要提取残差"列.但是它会产生一个错误.以下是代码:
I fitted a glm model in R and took the anova table. I need to extract the "Residual Deviance" column. But it generates an error. Here are the codes:
创建数据:
counts <- c(18,17,15,20,10,20,25,13,12)
outcome <- gl(3,1,9)
treatment <- gl(3,3)
符合GLM:
glm.D93 <- glm(counts ~ outcome + treatment, family = quasipoisson(link = "log"))
方差分析表:
av.1=anova(glm.D93)
av.1
Analysis of Deviance Table
Model: quasipoisson, link: log
Response: counts
Terms added sequentially (first to last)
Df Deviance Resid. Df Resid. Dev
NULL 8 10.5814
outcome 2 5.4523 6 5.1291
treatment 2 0.0000 4 5.1291
现在,我需要提取"Resid.Dev"列.所以我尝试了str
Now I need to extract "Resid. Dev" column. So I tried str
> str(av.1)
Classes ‘anova’ and 'data.frame': 3 obs. of 4 variables:
$ Df : int NA 2 2
$ Deviance : num NA 5.45 0
$ Resid. Df : int 8 6 4
$ Resid. Dev: num 10.58 5.13 5.13
- attr(*, "heading")= chr "Analysis of Deviance Table\n\nModel: quasipoisson, link: log\n\nResponse: counts\n\nTerms added sequentially (first to last)\n\"| __truncated__
最后,我提取了Resid. Dev
,但它给了我一个错误:
Finally I extracted Resid. Dev
but it gives me an error:
> av.1$Resid. Dev
Error: unexpected symbol in "av.1$Resid. Dev"
推荐答案
使用引号
> av.1$"Resid. Dev"
[1] 10.581446 5.129141 5.129141
等价
av.1[["Resid. Dev"]]
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