Reputation: 1045
Given x <- cbind(c(10,15,20,20,25,30,30,30,35,40,40,40,40,45),rep(c('M','F'),7))
, I want to calculate the rank sums of of categories M and F automatically, without doing it by hand. The thing I couldn't figure out is how to adjust the rank numbers when there is a tie. In this case, #3 and #4 are both 20 and thus share the rank value of 3.5 (instead of 3 and 4). Likewise #6 ~ #8 have the rank value of 7, and #10 ~ #13 have 11.5. Without this adjustment, the sums would be wrong.
#Wrong
sum(which(x[,2]=='F')) # =56
sum(which(x[,2]=='M')) # =49
#Right
sum(1,3.5,5,7,9,11.5,11.5) # =56.5
sum(2,3.5,7,7,11.5,11.5,14) # =48.5
I've tried table()
and duplicated()
, but couldn't figure out how to piece things together. Any ideas?
EDIT: My thanks to konvas for suggesting rank()
, which works in addition to bgoldst's solution.
Upvotes: 3
Views: 1898
Reputation: 31161
Base R
, you can use ave
:
setNames(unique(ave(rank(x[,1]), x[,2], FUN=sum)), unique(x[,2]))
# M F
# 48.5 56.5
Upvotes: 1
Reputation: 5951
With dplyr
library(dplyr)
x <- cbind(c(10,15,20,20,25,30,30,30,35,40,40,40,40,45),rep(c('M','F'),7))
data.frame(x) %>% mutate(rank=rank(X1)) %>% group_by(X2) %>% summarise(sum(rank))
Upvotes: 1
Reputation: 35314
You can sum()
the rank()
with aggregate()
:
x <- data.frame(age=c(10,15,20,20,25,30,30,30,35,40,40,40,40,45),sex=rep(c('M','F'),7));
aggregate(rank(age)~sex, x, sum );
## sex rank(age)
## 1 F 56.5
## 2 M 48.5
Upvotes: 4