R higher order function map on a dataframe - r

Question

In standard functional programming, Map takes a list l and a function F and returns a new list with F applied to every element. As an example consider:
F(x) = x^2 and the list l = [1, 2, 3, 4, 5]
Map(f, l) would produce the list: [1, 4, 9, 16, 25]
I would like to use this notion of Map on an R dataframe. I would like my function F(x) to compute x / rowSum(row that x belongs to in the dataframe).
Consider the data frame given by:
df <- data.frame()
for(i in 1:5)
{
df <- rbind(df, c(i, i+1, i+2, i+3, i+4))
}
colnames(df) <- c("a", "b", "c", "d", "e")
Which gives:
a b c d e
1 1 2 3 4 5
2 2 3 4 5 6
3 3 4 5 6 7
4 4 5 6 7 8
5 5 6 7 8 9
I would like Map(F, df) to produce:
[,1] [,2] [,3] [,4] [,5]
v1 0.06666667 0.1333333 0.2 0.2666667 0.3333333
v2 0.10000000 0.1500000 0.2 0.2500000 0.3000000
v3 0.12000000 0.1600000 0.2 0.2400000 0.2800000
v4 0.13333333 0.1666667 0.2 0.2333333 0.2666667
v5 0.14285714 0.1714286 0.2 0.2285714 0.2571429
which is a dataframe where F is applied to every entry x in df.
The only hard part is figuring out how to write F:
F <- function(x) x / rowSum( row in which x belongs to in dataframe)
Map(F, df)
How do I write F
EDIT Here is an iterative solution:
pStat <- data.frame()
for(i in 1: 5)
{
v <- df[i,] / rowSums(df[i,])
pStates <- rbind(pStates, v)
}

Answer 1

R's recycling rules work out of the box
df / rowSums(df)
A data.frame is a (column-oriented) list of equal-length vectors (try df[[2]], for instance, or str(df)), so Map(F, df) is acting as in other functional languages by applying F to each column. The use of rowSums implies that the data are all numeric; it is often appropriate and efficient to then use a matrix, where recycling still works out of the box.
m <- as.matrix(df)
m / rowSums(m)
One could use a closure (e.g., a function that returns a function) to provide constant arguments (rowSums(df)) to a (inefficient) Map solution that acts explicitly on each column
Ffactory <- function(df) { r = rowSums(df); function(x) x / r }
mapped <- Map(Ffactory(df), df)
remembering to coerce the list to a data frame
as.data.frame(mapped)