R-“不符合论证”的错误

时间:2017-12-07 22:16:49

标签: r

我希望从有序logit模型中获得预测概率。

代码:

if [ "$(egrep -n 'end balance: [0-9.]' ./simulations/*.html)" -gt 1000 ]; then
echo ' end balance bigger than 1000 '
else
echo ' end balance less than 1000 '
fi

运行此操作后,会显示警告:

library(plyr)
library(forcats)
library(MASS)
library(tidyverse)
library(dplyr)

zz <- "DV IV1 IV2  year
1          51         0      1987
2          49         1     1988  
3          47         0      1990
5          50         0      1991
5          54         1      1992
1          23         0      1993
3          15         0      1994"

Data <- read.table(text=zz, header = TRUE)
Data <- mutate(Data, DV=as.ordered(DV))

order <- polr(DV ~ IV1 + IV2 + IV1*IV2 + factor(year), data=Data,
           method = "logistic", Hess = TRUE)

然后我跑:

Warning message:
In polr(DV ~ IV1 + IV2 + IV1 * IV2 + factor(year), data = Data,  :
 design appears to be rank-deficient, so dropping some coefs

R回应:

colMeans(predict(order, type="probs", 
             newdata=mutate(Data, IV2=1, IV1=50)), na.rm=TRUE)

我尝试了没有“因子(年)”的线性模型,然后错误消失了。但是“因素(年)”应该被包含在模型中以解决时间序列问题。如何在没有“不一致的参数”错误的情况下包含它?

0 个答案:

没有答案