内容简介:http://stackoverflow.com/questions/33316686/r-plotting-predictions-of-mass-polr-ordinal-model
我使用MASS的polr函数(在这种情况下是给不同种类的奶酪的数据),在序数数据上使用比例赔率累积logit模型:
data=read.csv("https://www.dropbox.com/s/psj74dx8ohnrdlp/cheese.csv?dl=1")
data$response=factor(data$response, ordered=T) # make response into ordered factor
head(data)
cheese response count
1 A 1 0
2 A 2 0
3 A 3 1
4 A 4 7
5 A 5 8
6 A 6 8
library(MASS)
fit=polr(response ~ cheese, weights=count, data=data, Hess=TRUE, method="logistic")
为了绘制模型的预测,我使用了一个效果图
library(effects)
library(colorRamps)
plot(allEffects(fit),ylab="Response",type="probability",style="stacked",colors=colorRampPalette(c("white","red"))(9))
我想知道,如果从预期的方式报告的效果包,也可以绘制像每种奶酪的平均喜好与95%conf间隔这一点?
编辑:原来我也问过如何获得Tukey posthoc测试,但同时我发现那些可以使用
library(multcomp) summary(glht(fit, mcp(cheese = "Tukey")))
或使用包lsmeans as
summary(lsmeans(fit, pairwise ~ cheese, adjust="tukey", mode = "linear.predictor"),type="response")
Russ Lenth只是向我指出,只需使用选项模式=“平均值”(?模型)的lsmeans即可获得平均偏好和95%置信区间(同样适用于使用包序号的clm或clmm模型) :
df=summary(lsmeans(fit, pairwise ~ cheese, mode = "mean"),type="response")$lsmeans cheese mean.class SE df asymp.LCL asymp.UCL A 6.272828 0.1963144 NA 5.888058 6.657597 B 3.494899 0.2116926 NA 3.079989 3.909809 C 4.879440 0.2006915 NA 4.486091 5.272788 D 7.422159 0.1654718 NA 7.097840 7.746478
这给了我正在寻找的情节:
library(ggplot2)
library(ggthemes)
ggplot(df, aes(cheese, mean.class)) + geom_bar(stat="identity", position="dodge", fill="steelblue", width=0.6) +
geom_errorbar(aes(ymin=asymp.LCL, ymax=asymp.UCL), width=0.15, position=position_dodge(width=0.9)) +
theme_few(base_size=18) + xlab("Type of cheese") + ylab("Mean preference") +
coord_cartesian(ylim = c(1, 9)) + scale_y_continuous(breaks=1:9)
http://stackoverflow.com/questions/33316686/r-plotting-predictions-of-mass-polr-ordinal-model
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