Petr,
Thank you for you assistance. Particularly this last bit. It really
helped me understand exactly how your solution worked and why I was
confused by rapply processing rows. It was apply that could process
rows (and/or columns) of a matrix,
Anyway for anyone who wants the version of my script that incorporates
all of the suggested changes, here it is.
survey.results <- read.csv("SamplePairedComparisonData.csv")
# The following function evaluate which, if any, preference the
respondent has shown
get.pref <- function(x)
{
tb <- table(x)
if ((length(tb) == 3) & max((tb) == min(tb)))
retVal <- "None" # This opton checks for a couple of special
cases where the simple which.max function fails to
# detect that each of the three items has a single response.
else
# For all other conditions the following detects which response
received the most responses (2+) and records a NONE
# in any case that doesn't achieve that level of response.
retVal <- names(which.max(table(x)))
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 (Q4)
to a value of 2 if the existing value is 1
# 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
# The labels 'Option 1' and 'Option 2' can be replaced with more
appropriate text as needed.
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"))
# Take the survey data table and get rid of the first column (ID)
and convert the remaining three columns
# to a matrix. Then apply the get.pref function to each of the rows
in the matrix
survey.results$Preference <-
apply(as.matrix(survey.results[,-1]),1,get.pref)
On 12/10/2013 05:02 AM, PIKAL Petr wrote:
>
> In that case you need correct those cases inside the function.
> Everything else stays same.
>
> Something like
>
> fff <- function(x) {
>
> tb<-table(x)
>
> if(max(tb)==min(tb)) res <- "None" else res <- names(which.max(table(x)))
>
> }
>
> Regards
>
> Petr
>
>
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