问题遇到的现象和发生背景
本人在进行成分数据的分析，跟着本网站的一个代码教程写的。前提是每日的时间总和是一定的，但是现在到最后两步遇见了问题。即回归方程已经建立，在用均数代入方程进行预测时出现问题。
遇到的现象和发生背景，请写出第一个错误信息
Error in model.frame.default(Terms, newdata, na.action = na.action, xlev = object$xlevels) :
变数的长度不一样('b')
用代码块功能插入代码，请勿粘贴截图。 不用代码块回答率下降 50%
attach(mydata)
library(compositions)
comp<-cbind(sleep,x1,SED,LPA,MPA,VPA)
comp<-acomp(comp)
comp.mean<-mean(comp)
lm <- lm(BMI ~ ilr(comp) + Age3)
mean.pred <- predict(lm, newdata=list(comp=comp.mean, Age3=mean(Age3)))
运行结果及详细报错内容
> mean.pred <- predict(lm, newdata=list(comp=comp.mean, Age3=mean(Age3)))
Error in model.frame.default(Terms, newdata, na.action = na.action, xlev = object$xlevels) :
变数的长度不一样('Age3')
> table(comp.mean)
comp.mean
0.0063532728884529 0.0243705383508746 0.0296751798496391 0.0528902119744366
1 1 1 1
0.361325194554032 0.525385602382565
1 1
> table(Age3.mean)
Age3.mean
29.48738170347
1
我的解答思路和尝试过的方法，不写自己思路的，回答率下降 60%
这是之前借鉴的代码。
# load the required R package, Compositions
library(Compositions)
# make the composition by binding the components together
comp <- cbind(sleep,SED,LPA,MVPA) #variables are in min/day
# tell R that comp is a compositional variable
comp <- acomp(comp)
# make the ilr multiple linear regression model. BF represents %body fat. # ilr() is the default isometric log ratio transformation included in the
# Compositions package.
lm <- lm(BF ~ ilr(comp) + ses) #ses, an example covariate
# determine the mean composition
comp.mean <- mean(comp)
# because comp has been designated as a compositional
# variable, R calls on mean.acomp() to calculate the
# compositional mean (i.e., geometric mean of each component,
# then adjusted so all components sum to 1). The mean is
# therefore expressed in proportions.
# predict %BF for the mean composition from above, keeping ses
# constant at its mean.
mean.pred <- predict(lm, newdata=list(comp=comp.mean, ses=mean(ses)))
# next, construct a new composition, where 30 min of sedentary
#time have been reallocated to sleep. However, here 30 min must
#be expressed as a proportion (remember, from above, that
#mean.acomp() adjusts the mean to proportions).
# The reallocated time is therefore 30/1440.
new.comp<- acomp(comp.mean+c(30/1440,-30/1440,0,0))
# Now, we predict %BF for the new composition (new.comp),
#keeping ses constant at its mean.
pred <- predict(lm, newdata=list(comp=new.comp, ses=mean(ses)))
# Finally, the estimated difference in %BF for the above time
#reallocation is (Equation 4):
pred- mean.pred
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想问一下这个问题怎么可以解决？
