r dataframe using rank - r

Question

I would like to rank the row of a dataframe (with 30 columns) which has numerical values ranking from -inf to +inf.
This is what I have:
df <- structure(list(StockA = c("-5", "3", "6"),
StockB = c("2", "-1", "3"),
StockC = c("-3", "-4", "4")),
.Names = c( "StockA","StockB", "StockC"),
class = "data.frame", row.names = c(NA, -3L))
> df
StockA StockB StockC
1 -5 2 -3
2 3 -1 -4
3 6 3 4
This is what I would like to have:
> df_rank
StockA StockB StockC
1 3 1 2
2 1 2 3
3 1 3 2
I am using this command:
> rank(df[1,])
StockA StockB StockC
2 3 1
The resulting rank variables are not correct though as you can see.

Answer 1

rank() assigns the lowest rank to the smallest value.
So the short answer to your question is to use rank of the vector multiplied by -1:
rank (-c(-5, 2, -3) )
[1] 1 3 2
Here is the full code:
# data frame definition. The numbers should actually be integers as pointed out
# in comments, otherwise the rank command will sort them as strings
# So in the real word you should define them as integers,
# but to go with your data I will convert them to integers in the next step
df <- structure(list(StockA = c("-5", "3", "6"),
StockB = c("2", "-1", "3"),
StockC = c("-3", "-4", "4")),
.Names = c( "StockA","StockB", "StockC"),
class = "data.frame", row.names = c(NA, -3L))
# since you plan to rank them not as strings, but numbers, you need to convert
# them to integers:
df[] <- lapply(df,as.integer)
# apply will return a matrix or a list and you need to
# transpose the result and convert it back to a data.frame if needed
result <- as.data.frame(t( apply(df, 1, FUN=function(x){ return(rank(-x)) }) ))
result
# StockA StockB StockC
# 3 1 2
# 1 2 3
# 1 3 2