I've spent several days compiling the following code (I apologize in advance
- this code is very inelegant, and I'm sure could be written much more
efficiently, but I've stuck with whatever method I could get to work -
sometimes the more efficient code I just couldn't get to work without an
error, because of my R inexperience).
My main motivation for writing the code is that I'd like to be able to
assess interaction effects, and so I want to be able to rotate through the
reference values so that I can assess different interaction effects than
simply the default ones reported by the single set of reference values given
in a single analysis. The code below generates a lot more than that, but
since I am such an R newbie (and therefore struggling to learn so many
different things at once) I thought that if I could just generate one bit
matrix with all the pairwise comparisons, I could then pick out the stuff I
actually want later (probably just manually at first, but later hopefully by
automating my code to return a matrix with just the comparisons I want).
Here is the code:
DV<-factors[,1]
IV<-factors[,2]
int<-factors[,3]
IV<-IV[!is.na(IV)]
int<-int[!is.na(int)]
#Limit our modification of reference values to categorical variables only by
redefining IV vector as IVcat
IVcat<-IV
for(p in 1:length(IV)){
if((class(dataset[[toString(IV[p])]])=="character")|(class(dataset[[toString(IV[p])]])=="factor")){
IVcat[p]<-IV[p]
}
else{IVcat[p]<-""}
}
IVcat<-IVcat[!is.na(IVcat)]
#Create vectors (IVvalslist[n] for nth factor) for each IVcat containing
each possible value for that IV
IVvalslist<-vector('list', length(IVcat))
for(i in 1:length(IVcat)){
assign(paste("IVvalues",i,sep=""),unique(dataset[[toString(IVcat[i])]])[!is.na(unique(dataset[[toString(IVcat[i])]]))])
IVvalslist[i]<-list(get(paste("IVvalues",i,sep="")))
}
#Create a data frame (refM) with every combination of values for each IVcat
refM<-expand.grid(IVvalslist)
#Loop through all possible reference values, and then run the model,
#and then compile the model summary output into a single matrix
#Go through each row of the matrix of possible reference value combinations
for(j in 1:nrow(refM)){
#Go through each reference value for each factor in that row, and assign
#that reference value for that factor
for(k in 1:length(IVcat)){
dataset[[ toString(IVcat[k]) ]]<-relevel(dataset[[ toString(IVcat[k])
]],ref=toString(refM[j,k]))
}
#Run model with new reference values from row j
model<-paste(paste(DV[1],"~1",sep=""),paste(IV,collapse="+"),paste(int,collapse="+"),sep="+")
modeloutput<-glm(model,family=binomial(logit),data=dataset)
#assigning all output from every possible combo of ref values to a single
matrix named coeffM
if(j==1){coeffM<-coef(summary(modeloutput))}
if(j>1){coeffM<-rbind(coeffM,coef(summary(modeloutput)))}
}
*I tested each step of the code individually, and each individual step runs
fine. However, when I run the whole set of code at once, I get the
following error message after the last line of R input above:
Warning in model.matrix.default(mt, mf, contrasts) :
variable 'retention' converted to a factor
Error in weights * y : non-numeric argument to binary operator
And if I ask for R to return coeffM or modeloutput, it tells me that no such
objects exist. *
The object factors is as follows:> factors
var1 var2 var3
1 retention method method*level
2 <NA> level <NA>
3 <NA> gpa <NA>
And I can't give the actual dataset (named dataset above) here for human
subject reasons, but here is a made-up sample of what it looks like:
id instructor method success retention level career
1 1001 NAME1 online 1 1 LL
career
2 1002 NAME2 face-to-face 1 1 UL
lib. arts
STEM required ethnicity
gender age finaid gpa
1 non-STEM elective Asian or Pacific Islander M
28 1.97
2 non-STEM maj. req. Asian or Pacific Islander F
21 none 3.01
experience credits yrsenrolled course
1 no online exp. 22 1 NAME1_MAT100
2 no online exp. 33 2 NAME2_ENG100
If anyone can help me figure out what is going wrong here, I'd be incredibly
grateful! I've tried searching repeatedly for this error, and the other
instances that I've found haven't seemed to apply to this situation?
But as
I am just learning R, I may be missing something really obvious.
Thanks again for taking the time to read my post!
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