我正在执行多项式回归并测试系数的线性组合。但是当我试图测试系数的线性组合时,我遇到了一些问题。
LnModel_1 <- lm(formula = PROF ~ UI_1+UI_2+I(UI_1^2)+UI_1:UI_2+I(UI_2^2))
summary(LnModel_1)
它输出以下值:
Call:
lm(formula = PROF ~ UI_1 + UI_2 + I(UI_1^2) + UI_1:UI_2 + I(UI_2^2))
Residuals:
Min 1Q Median 3Q Max
-3.4492 -0.5405 0.1096 0.4226 1.7346
Coefficients:
Estimate Std. Error t value Pr(>|t|)
(Intercept) 4.66274 0.06444 72.354 < 2e-16 ***
UI_1 0.25665 0.07009 3.662 0.000278 ***
UI_2 0.25569 0.09221 2.773 0.005775 **
I(UI_1^2) -0.15168 0.04490 -3.378 0.000789 ***
I(UI_2^2) -0.08418 0.05162 -1.631 0.103643
UI_1:UI_2 -0.02849 0.05453 -0.522 0.601621
然后我使用名称(coef())来提取系数名称
names(coef(LnModel_1))
output:
[1] "(Intercept)" "UI_1" "UI_2" "I(UI_1^2)"
"I(UI_2^2)""UI_1:UI_2"
出于某些原因,当我使用glht()时,它会在UI_2 ^ 2上给出错误
slope <- glht(LnModel_1, linfct = c("UI_2+ UI_1:UI_2*2.5+ 2*2.5*I(UI_2^2) =0
") )
Output:
Error: multcomp:::expression2coef::walkCode::eval: within ‘UI_2^2’, the term
‘UI_2’ must not denote an effect. Apart from that, the term must evaluate to
a real valued constant
不知道为什么它会给我这个错误信息。如何将I(UI_2 ^ 2)系数输入到glht()
非常感谢你
问题似乎是I(UI^2)
可以被解释为R中的表达,就像你在这里做的那样LnModel_1 <- lm(formula = PROF ~ UI_1+UI_2+I(UI_1^2)+UI_1:UI_2+I(UI_2^2))
因此,您应该指出要在字符串中评估R
的string
:
slope <- glht(LnModel_1, linfct = c("UI_2+ UI_1:UI_2*2.5+ 2*2.5*\`I(UI_2^2)\` =0
") )
检查我的例子(因为我无法重现你的问题):
library(multcomp)
cars <- copy(mtcars)
setnames(cars, "disp", "UI_2")
model <- lm(mpg~I(UI_2^2),cars)
names(coef(model))
slope <- glht(model, linfct = c("2*2.5*\`I(UI_2^2)\` =0") )