Karen,
Thank you for your reply. That approach makes the code for determining
preference much clearer. Unfortunately, it appears I am still needing
to use a for loop to process each row.
For any who find these posts and may have similar questions, I am
including my current code and sample data to provide a working solution:
survey.results <- read.csv("SamplePairedComparisonData.csv")
# The following function evaluates which, if any, preference the
respondent has shown
get.pref <- function(x)
{
# names(which.max(table(x)))
n.1 <- sum(x==1) # Calculate the number of times the respondent chose
option 1
n.2 <- sum(x==2) # Calculate the number of times the respondent chose
option 2
n.3 <- sum(x==3) # Calculate the number of times the respondent chose
option 3
retVal <- "None"
if (n.1 >= 2) {
retVal <- "Option 1"
} else if (n.2 >= 2) {
retVal <- "Option 2"
} else if (n.3 >= 2) {
retVal <- "Option 3"
}
return(retVal)
}
# The data is expected to be organized as so:
# ID, Q1, Q2, Q3
# and that the questions are in the following format
# Q.1) Which do you prefer?
# 1) Option 1
# 2) Option 2
# Q.2) Which do you prefer?
# 1) Option 1
# 2) Option 3
# Q.3) Which do you prefer?
# 1) Option 2
# 2) Option 3
# And the next three lines reprocess the data table to organize the
questions so that the values for Q.2) are 1 or 3 and the values for Q.3)
are 2 or 3
survey.results[survey.results[,3]==2,3]<-3 # Convert column 3 (Q2) to a
value of 3 if the existing value is 2
# The order of the following two commands is important, don't change
survey.results[survey.results[,4]==2,4]<-3 # Convert column 4 (Q3) to a
value of 3 if the existing value is 1
survey.results[survey.results[,4]==1,4]<-2 # Convert column 4 (Q3) to a
value of 2 if the existing value is 1
for (indx in 1:length(survey.results$ID))
{
survey.results$Preference[indx] <- get.pref(survey.results[indx, 2:4])
}
# The following lines convert the numerical values in the fields to text
(factors)
# If the questions don't have empty (zero) values than the "None" should
be removed from those questions where zero isn't possible
survey.results$Q1 <- factor(survey.results$Q1,
labels=c("None","Option 1","Option 2"))
survey.results$Q2 <- factor(survey.results$Q2,
labels=c("None","Option 1","Option 3"))
survey.results$Q3 <- factor(survey.results$Q3,
labels=c("None","Option 2","Option 3"))
# survey.results$Preference <-
apply(as.matrix(survey.results[,-1]),1,get.pref)
and here is the sample data.
#The following sample data represents every possible response to the
three questions of the type described above
ID,Q1,Q2,Q3
1,0,0,0
2,0,0,1
3,0,0,2
4,0,1,0
5,0,1,1
6,0,1,2
7,0,2,0
8,0,2,1
9,0,2,2
10,1,0,0
11,1,0,1
12,1,0,2
13,1,1,0
14,1,1,1
15,1,1,2
16,1,2,0
17,1,2,1
18,1,2,2
19,2,0,0
20,2,0,1
21,2,0,2
22,2,1,0
23,2,1,1
24,2,1,2
25,2,2,0
26,2,2,1
27,2,2,2
On 12/09/2013 09:42 AM, Karen Keating wrote:
> Hi Walter,
> From your list of preferences, it seems that Option 2 is preferred if
> there are at least two 2's for the ID, and Option 1 is preferred if
> there are at least two 1's for the ID. If neither of these conditions
> are met, then there is no preference. If this interpretation is
> correct, I would suggest replacing your list of 'if' statements with
> the following:
>
> get.pref<-function(x) {
> n.1<-sum(x==1)
> n.2<-sum(x==2)
> pref='None'
> if(n.1>=2) {pref='Option 1'} else
> if(n.2>=2) {pref='Option 2'}
> pref
> }
> R-help@r-project.org <mailto:R-help@r-project.org> mailing list
> https://stat.ethz.ch/mailman/listinfo/r-help
> PLEASE do read the posting guide
> http://www.R-project.org/posting-guide.html
> and provide commented, minimal, self-contained, reproducible code.
>
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