Thank you for your response!
I am attempting to determine a preference from the answers to three
binomial questions;
q.1) 1 or 2 q.2) 1 or 3 q.3) 2 or 3
However, the questions are coded with either a 1 or 2 (though no answer
is also possible) and the first three functions (q#.ans) convert those
values to the 1,2,or 3 shown above
and generate one of the following result for each row of the table; 0 -
no preference, or 1,2,3 which indicates the preference indicated by the
question
The if's implement the following state conditions:
# ID A B C Preference
# 1 0 0 0 None
# 2 0 0 1 None
# 3 0 0 2 None
# 4 0 1 0 None
# 5 0 1 1 Option 1
# 6 0 1 2 None
# 7 0 2 0 None
# 8 0 2 1 None
# 9 0 2 2 Option 2
# 10 1 0 0 None
# 11 1 0 1 Option 1
# 12 1 0 2 None
# 13 1 1 0 Option 1
# 14 1 1 1 Option 1
# 15 1 1 2 Option 1
# 16 1 2 0 None
# 17 1 2 1 Option 1
# 18 1 2 2 Option 2
# 19 2 0 0 None
# 20 2 0 1 None
# 21 2 0 2 Option 2
# 22 2 1 0 None
# 23 2 1 1 Option 1
# 24 2 1 2 Option 2
# 25 2 2 0 Option 2
# 26 2 2 1 Option 2
# 27 2 2 2 Option 2
The if statement only implements those values from the state machine
that show a preference (ID's 5,9,11,13-15,17-18,21,23-27)
On 12/06/2013 09:59 AM, PIKAL Petr wrote:
Hi
The warning is due to fact that "if" takes only single scalar value not an
entire vector.
Maybe you shall explain more clearly what result do you expect.
I bet that there is vectorised solution to your problem but I am lost in your
ifs and cannot follow what shall be the output.
Please use
dput(head(df))
when showing input data and clearly describe intended result.
Regards
Petr
-----Original Message-----
From: r-help-boun...@r-project.org [mailto:r-help-bounces@r-
project.org] On Behalf Of Walter Anderson
Sent: Friday, December 06, 2013 4:44 PM
To: r-help@r-project.org
Subject: [R] Need help figuring out sapply (and similar functions) with
multiple parameter user defined function
I am having trouble understanding how to use sapply (or similar
functions) with a user defined function with multiple parameters.
I have the following functions defined
q1.ans <- function(x)
{
retVal = 0
if (x == 1) {
retVal = 1
} else if (x ==2) {
retVal = 2
}
return (retVal)
}
q2.ans <- function(x)
{
retVal = 0
if (x == 1) {
retVal = 1
} else if (x ==2) {
retVal = 3
}
return (retVal)
}
q3.ans <- function(x)
{
retVal = 0
if (x == 1) {
retVal = 2
} else if (x ==2) {
retVal = 3
}
return (retVal)
}
evaluate.questions <- function(q.1,q.2,q.3)
{
a <- q1.ans(q.1)
b <- q2.ans(q.2)
c <- q3.ans(q.3)
retVal = 0 # Set default value to be no preference
# The following code only implements those values from the state
machine that show a preference (ID's 5,9,11,13-15,17-18,21,23-27)
if (a == 0) {
if (b == 1) {
if (c == 1) {
retVal = 1 # State machine ID 5
}
} else if (b == 2) {
if (c == 2) {
retVal = 2 # State machine ID 9
}
}
} else if (a == 1) {
if (b == 0) {
if (c == 1) {
retVal = 1 # State machine ID 11
}
} else if (b == 1) {
retVal = 1 # State machine ID's 13-15, value of C doesn't
matter
} else if (b == 2) {
if (c == 1) {
retVal = 1 # State machine ID 17
} else if (c == 2) {
retVal = 2 # State machine ID 18
}
}
} else if (a == 2) {
if (b == 0) {
if (c == 2) {
retVal = 2 # State machine ID 21
}
} else if (b == 1) {
if (c == 1) {
retVal = 1 # State machine ID 23
} else if (c == 2) {
retVal = 2 # State machine ID 24
}
} else if (b == 2) {
retVal = 2 # State machine ID's 25-27, value of C doesn't
matter
}
}
return (retVal)
}
And a data set that looks like this:
ID,Q1,Q2,Q3
1,2,2,2
2,2,1,1
3,1,1,1
4,1,2,2
5,2,2,1
6,1,2,1
...
I have been researching and it appears that I should be using the
sapply function to apply the evaluate.question function above to each
row in the data frame like this
preferences <- sapply(df, evaluate.questions, function(x,y,z)
evaluate.questions(df['Q1'],df['Q2'],df['Q3']))
Unfortunately this doesn't work and the problem appears that the sapply
function is not feeding the parameters to the evaluate.questions
function as I expect. Can someone provide some guidance on what I am
doing wrong?
This is the error message I am getting:
Error in x --1 :
Comparison (1) is possible only for atomic and list types In
addition: warning messages:
In if (x == 1) { :
the condition has length > 1 and only the first element will be used
[[alternative HTML version deleted]]
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______________________________________________
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PLEASE do read the posting guide http://www.R-project.org/posting-guide.html
and provide commented, minimal, self-contained, reproducible code.
