Hi You probably missed this
Instead of those tricky ifs (uff uff) you can use either of these df[df[,2]==2,2]<-3 df$C<-as.numeric(as.character(factor(df$C, labels=c(0,2,3)))) df A B C 1 0 3 3 2 0 1 2 3 0 3 2 4 1 0 3 5 1 0 3 6 2 3 2 7 1 3 2 8 2 3 3 9 1 1 0 10 0 0 3 And here I am lost again. Please, can you ***clearly*** state the way how do you want to choose preferences based on values in those three columns? Regards Petr Petr Pikal From: Walter Anderson [mailto:wandrso...@gmail.com] Sent: Friday, December 06, 2013 5:32 PM To: PIKAL Petr Subject: RE: [R] Need help figuring out sapply (and similar functions) with multiple parameter user defined function I posted the first 6 rows of my data in the first post. On Dec 6, 2013 10:31 AM, "PIKAL Petr" <petr.pi...@precheza.cz<mailto:petr.pi...@precheza.cz>> wrote: Hi So first step is over. Anyway, is there any problem with using dput as I suggested? Instead of using your date I need to generate my own. A<-sample(0:2, 10, replace=T) B<-sample(0:2, 10, replace=T) C<-sample(0:2, 10, replace=T) df<-data.frame(A,B,C) df[df[,2]==2,2]<-3 df$C<-as.numeric(as.character(factor(df$C, labels=c(0,2,3)))) df A B C 1 0 3 3 2 0 1 2 3 0 3 2 4 1 0 3 5 1 0 3 6 2 3 2 7 1 3 2 8 2 3 3 9 1 1 0 10 0 0 3 > -----Original Message----- > From: Walter Anderson > [mailto:wandrso...@gmail.com<mailto:wandrso...@gmail.com>] > Sent: Friday, December 06, 2013 5:11 PM > To: PIKAL Petr; r-help@r-project.org<mailto:r-help@r-project.org> > Subject: Re: [R] Need help figuring out sapply (and similar functions) > with multiple parameter user defined function > > 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 Instead of those tricky ifs (uff uff) you can use either of these df[df[,2]==2,2]<-3 df$C<-as.numeric(as.character(factor(df$C, labels=c(0,2,3)))) df A B C 1 0 3 3 2 0 1 2 3 0 3 2 4 1 0 3 5 1 0 3 6 2 3 2 7 1 3 2 8 2 3 3 9 1 1 0 10 0 0 3 And here I am lost again. Please, can you clearly state the way how do you want to choose preferences based on values in those three columns. Regards Petr > > 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-boun...@r-project.org> > >> [mailto:r-help-bounces@r-<mailto:r-help-bounces@r-> > >> project.org<http://project.org>] On Behalf Of Walter Anderson > >> Sent: Friday, December 06, 2013 4:44 PM > >> To: r-help@r-project.org<mailto: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]] > >> > >> ______________________________________________ > >> 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. [[alternative HTML version deleted]] ______________________________________________ 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.
