R associative memory doesn't work as expected - r

Question

I am trying to use associative memory and ddply to add a column to a data frame. For example:
First, I have defined association and a function that uses association to calculate product of two elements of a row (property damage and multiplier) to get actual damage in dollars. Here,"B" means Billion, "m|M" means MIllions, etc.
validMultiplierLetter <- c("B", "h", "H", "k", "K", "m", "M")
Multiplier <- c(1000000000, 100, 100, 1000, 1000, 1000000, 1000000)
names(Multiplier) <- validMultiplierLetter
The function ploss (property loss) is:
ploss <- function(pd,pm) {
if (pm %in% validMultiplierLetter) pd*Multiplier[pm]
else 0
}
here is a sample data frame with columns pd (property damage) and pm (multiplier) and ddply code to create a pl (property loss) column, which is a product of property damage and the associated value of multiplier. Invalid multipliers are equivalent to 0 (e.g., "+").
tdf <- data.frame(pd = c(5, 10, 15, 20, 25), pm = c("B", "m", "K", "+", "h"))
tldf <- ddply(tdf, .(pd, pm), transform, pl = ploss(pd,pm))
I get the following output when I execute the code above - you can see that the right multiplier was not used for the rows.
> tldf
pd pm pl
1 5 B 500
2 10 m 10000
3 15 K 15000
4 20 + 0
5 25 h 2500
Strangely though, when you pass constant, the multiplier works correctly. But, when you pass a variable (whose value is same as the constant), for some reason you get an incorrect result.
> Multiplier["B"]
B
1e+09
> tdf$pm[1]
[1] B
Levels: + B h K m
> Multiplier[tdf$pm[1]]
h
100
Any explanation of why this happens and how to fix it is greatly appreciated. Thanks.

Answer 1

The problem is that tdf$pm is a factor. When presented a factor, [ will use the factor levels rather than the character values:
x <- 10:15
names(x) <- LETTERS[1:6]
x
## A B C D E F
## 10 11 12 13 14 15
x[c('A','F')] # Lookup by name
## A F
## 10 15
x[factor(c('A','F'))] # Lookup by integer
## A B
## 10 11
This is fixed by using as.character around the factor, so that a character vector is presented to [:
x[as.character(factor(c('A','F')))]
## A F
## 10 15
For your problem, you can coerce to character in the transform function:
ddply(tdf, .(pd, pm), transform, pl = ploss(pd,as.character(pm)))
## pd pm pl
## 1 5 B 5.0e+09
## 2 10 m 1.0e+07
## 3 15 K 1.5e+04
## 4 20 + 0.0e+00
## 5 25 h 2.5e+03
In addition, you could vectorize your ploss function in the obvious way and do the job directly with transform:
ploss <- function(pd,pm) {
ifelse(pm %in% validMultiplierLetter, pd*Multiplier[pm], 0)
}
transform(tdf, pl=ploss(pd, as.character(pm)))
## pd pm pl
## 1 5 B 5.0e+09
## 2 10 m 1.0e+07
## 3 15 K 1.5e+04
## 4 20 + 0.0e+00
## 5 25 h 2.5e+03
And of course, the as.character coercion could be within the function ploss, so it isn't required in the transform call:
ploss <- function(pd,pm) {
ifelse(pm %in% validMultiplierLetter, pd*Multiplier[as.character(pm)], 0)
}

Answer 2

The problem I see is that, if you're using the default R options, tdf$pm is a factor, not a character. You can check this with class(tdf$pm). What's happening here is that "B" is really a mask for 2 (following the order in the printout: Levels: + B h K m), so pd has the value of 2 as far as [ is concerned, and Multiplier[2] is 100 as you've assigned.
When you call data.frame (or read.table) you need to add the argument stringsAsFactors = FALSE, or change the corresponding global option with the options function.