Again, I think the best solution will involve subsetting. If you are feeling ambitious, you can try to work this solution out on your own, but you will learn just as quickly by mentally working through the following proposed solution.
We know that our prize should be $0 if we have no cherries, $2 if we have one cherry, and $5 if we have two cherries. You can create a vector that contains this information. This will be a very simple lookup table:
c(0, 2, 5)Now, like in Case 1, you can subset the vector to retrieve the correct prize. In this case, the prize’s aren’t identified by a symbol name, but by the number of cherries present. Do we have that information? Yes, it is stored in cherries. We can use basic integer subsetting to get the correct prize from the prior lookup table, for example, c(0, 2, 5)[1].
Now, like in Case 1, you can subset the vector to retrieve the correct prize. In this case, the prizes aren’t identified by a symbol name, but by the number of cherries present. Do we have that information? Yes, it is stored in cherries. We can use basic integer subsetting to get the correct prize from the prior lookup table, for example, c(0, 2, 5)[1].
cherries isn’t exactly suited for integer subsetting because it could contain a zero, but that’s easy to fix. We can subset with cherries + 1. Now when cherries equals zero, we have:
cherries + 1
## 1