Petr,

Thank you for your assistance; however, your code does not produce the 
correct results in the following two cases;

ID,Q1,Q2,Q3
17,Option 1, Option 3, Option 2
24,Option 2, Option 3, Option 1

In both cases it chooses a preference of Option 1, when the correct 
answer is None

# 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
# Test data that shows all possible responses to the questions
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 10:21 AM, PIKAL Petr wrote:
>
> Hi
>
> you are still rather cryptic. If you said you want to extract 
> prevalent choices in each row it would save me (and you) a lot of time.
>
> Import data and make necessary chnages
>
> survey.results <- read.csv("SamplePairedComparisonData.csv")
>
> 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
>
> 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"))
>
> Then make the object matrix without first column
>
> sr<-survey.results[,-1]
>
> sr<-as.matrix(sr)
>
> Elaborate evaluation function
>
> fff <- function(x) names(which.max(table(x)))
>
> Apply function to matrix
>
> result <-  apply(sr,1, fff)
>
> Bind result with original data
>
> survey.results <- cbind(survey.results, result)
>
> And voila, you shall have reqiured values.
>
> Regards
>
> Petr
>
> *From:*Walter Anderson [mailto:wandrson01@gmail.com]
> *Sent:* Monday, December 09, 2013 4:37 PM
> *To:* PIKAL Petr
> *Subject:* Re: [R] Need help figuring out sapply (and similar 
> functions) with multiple parameter user defined function
>
> Peter.
>
> First, let me thank you for your assistance on this.  Your ideas below 
> for eliminating the three functions I had that 'adjusted' the data was 
> received and appreciated.  I included that approach in my revised 
> script.  I am only left with a for loop and my final preference 
> determining code.  I am including the current script and test data 
> (all possible choices included-27).
>
> >From my review of all of the responses, I don't see an alternative to 
> the if then block that tests the state of the three questions or the 
> for loop that executes it.  I made some attempts to use the ifelse 
> statement mentioned in the responses, but couldn't get anything to work.
>
> Again, thanks for your assistance and time on this!
>
> Walter
>


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