Cbind / Rbind With Ifelse Condition

Here is the code I am working with:

x <- c("Yes","No","No","Yes","Maybe")
y <- t(1:10)
z <- t(11:20)

rbind.data.frame(ifelse(x == "Yes",y,z))

      

        
        
        
      

    

This creates

    X1L X12L X13L X4L X15L
1   1   12   13   4   15

      

        
        
        
      

    

Desired output:

x
1   Yes   1    2    3    4    5    6    7    8    9    10
2    No   11   12   13   14   15   16   17   18   19    20
3    No   11   12   13   14   15   16   17   18   19    20
4   Yes   1    2    3    4    5    6    7    8    9    10
5 Maybe   11   12   13   14   15   16   17   18   19    20

      

        
        
        
      

    

I thought I could use an ifelse statement with rbind.data.frame () or cbind.data.frame () function. Therefore, if x == "Yes"

, then it will be combined with the vector "y". As shown in the first line of the desired output. Conversely, if x!="Yes"

, then it will be combined with the vector "z". How would I do it. I also thought that indexing with the () function might be possible, but I couldn't think of how I would use it.

UPDATE ANOTHER QUESTION

Here is the code I am working with:

    a <- c(1,0,1,0,0,0,0)
    b <- 1:7 

    t(sapply(a, function(test) if(test==0) b else 0))

Which produces 

        [,1] [,2]      [,3] [,4]      [,5]      [,6]      [,7]     
    [1,] 0    Integer,7 0    Integer,7 Integer,7 Integer,7 Integer,7

      

        
        
        
      

    

Can any of you explain this? The code below works, but I was wondering why the sapply function was not working. Also how to save the matrix created below?

`row.names<-`(t(t(rbind(b,0))[,(a!='1')+1L]),x) 

      

        
        
        
      

    

The answer to my last question

For the sapply function to work, you need a vector to enter the else statement, so

        a <- c(1,0,1,0,0,0,0)
        b <- 1:7 
        c <- rep(0, times = length(a))
        t(sapply(a, function(test) if(test==0) b else c))

      

        
        
        
      

    

This now gives the correct conclusion.