Assign value to new column if another column has exact string - r

Question

I have a df that looks like this.
Date Winner
4/12 Tom
4/13 Abe
4/14 George
4/15 Tom
I would like to add new columns that assign a 1 if if the name appears in the winner column and 0 if the name did not appear and vice versa. Ideally the df would look like this as a result
Date Winner Tom_Win Tom_Lose Abe_Win Abe_Lose George_Win George Lose
4/12 Tom 1 0 0 1 0 1
4/13 Abe 0 1 1 0 0 1
4/14 George 0 1 0 1 1 0
4/15 Tom 1 0 0 1 0 1
Is there an easy way to accomplish this?

Answer 1

This is extremely simple to do if you use the model.matrix functions, it will create N dummy columns with 0 when the name does not appear and one when it does (exactly as you requested), the code below:
(assuming your data is called db)
> winners <- model.matrix(~Winner - 1, data=db)
> winners
WinnerAbe WinnerGeorge WinnerTom
1 0 0 1
2 1 0 0
3 0 1 0
4 0 0 1
This bit is to compute the columns with the losing values
winners <- as.data.frame(winners)
winners$loserAbe <- as.numeric(!winners$WinnerAbe) #naturally you have to
#do this for every column you need
WinnerAbe WinnerGeorge WinnerTom loserAbe
1 0 0 1 1
2 1 0 0 0
3 0 1 0 1
4 0 0 1 1
winners$Date <- db$Date #this last bit so you don't lose the date.

Answer 2

Using mtabulate from qdapTools package we can do the following three steps,
library(qdapTools)
d1 <- mtabulate(d3$Winner)
d2 <- setNames(data.frame(sapply(d1, function(i) ifelse(i == 1, 0, 1))),
paste0(names(d1), '_Lose'))
cbind(d3$Date, d1, d2)
# d3$Date Abe George Tom Abe_Lose George_Lose Tom_Lose
#1 4/12 0 0 1 1 1 0
#2 4/13 1 0 0 0 1 1
#3 4/14 0 1 0 1 0 1
#4 4/15 0 0 1 1 1 0
DATA
str(d3)
'data.frame': 4 obs. of 2 variables:
$ Date : Factor w/ 4 levels "4/12","4/13",..: 1 2 3 4
$ Winner: Factor w/ 3 levels "Abe","George",..: 3 1 2 3

Answer 3

I'm sure there is a better way than this but this works in base R and it's fairly simple:
If your data looks like this:
df <- data.frame(Date = c("4/12","4/13","4/14","4/15"),Winner = c("Tom","Abe","George","Tom"))
Append the extra columns like so:
xcols <- c(paste0(unique(df$Winner), '_Win'), paste0(unique(df$Winner), '_Lose'))
df[ , xcols] <- 0
Now make a character vector with instructions to give the points for every player.
evl <- unlist(lapply(unique(df$Winner), function(x){paste0('df[', which(df$Winner == x), ',', which(names(df) == paste0(x, '_Win')), '] <- 1')}))
And execute the code:
eval(parse(text = evl))

Answer 4

df <- data.frame(
Date = c("4/12", "4/13","4/14", "4/15"),
Winner = c("Tom", "Abe", "George", "Tom")
)
df2 <- do.call(cbind,
lapply(seq_along(levels(df$Winner)), function(x) {
win <- ifelse(df$Winner == levels(df$Winner)[x], 1, 0)
lose <- ifelse(df$Winner == levels(df$Winner)[x], 0, 1)
dat <- cbind(win, lose)
colnames(dat) <- c(paste(levels(df$Winner)[x], "win", sep = "_"), paste(levels(df$Winner)[x], "lose", sep = "_"))
dat
})
)
cbind(df, df2)
> cbind(df, df2)
Date Winner Abe_win Abe_lose George_win George_lose Tom_win Tom_lose
1 4/12 Tom 0 1 0 1 1 0
2 4/13 Abe 1 0 0 1 0 1
3 4/14 George 0 1 1 0 0 1
4 4/15 Tom 0 1 0 1 1 0