I'm trying to order this dataframe by population and date, so I'm using the order() and rank() functions:
> df <- data.frame(idgeoville = c(5, 8, 4, 3, 4, 5, 8, 8),
date = c(rep(1950, 4), rep(2000, 4)),
population = c(500, 450, 350, 350, 650, 500, 500, 450))
> df
idgeoville date population
1 5 1950 500
2 8 1950 450
3 4 1950 350
4 3 1950 350
5 4 2000 650
6 5 2000 500
7 8 2000 500
8 8 2000 450
With ties.method = "first" I have no problem, finally I'm producing this dataframe:
idgeoville date population rank
1 5 1950 500 1
2 8 1950 450 2
3 4 1950 350 3
4 3 1950 350 4
5 4 2000 650 1
6 5 2000 500 2
7 8 2000 500 3
8 8 2000 450 4
But in fact, I want a dataframe with equal ranking for equal population rank, like this:
idgeoville date population rank
1 5 1950 500 1
2 8 1950 450 2
3 4 1950 350 3
4 3 1950 350 3
5 4 2000 650 1
6 5 2000 500 2
7 8 2000 500 2
8 8 2000 450 3
How can I resolve this problem with R? With a custom ties.method() or another R tricks?
I believe there is no option to do it with rank; here is a custom function that will do what you want, but it may be too slow if your data is huge:
Rank<-function(d) {
j<-unique(rev(sort(d)));
return(sapply(d,function(dd) which(dd==j)));
}
More simple way:
pop.rank <- as.numeric(factor(population))
This answers a slightly different question, namely how to sort a data.frame object based on multiple columns. To do this, you could use the function sort_df in package reshape:
> library(reshape)
> sort_df(df,vars=c('date','population'))
idgeoville date population
3 4 1950 350
4 3 1950 350
2 8 1950 450
1 5 1950 500
8 8 2000 450
6 5 2000 500
7 8 2000 500
5 4 2000 650
来源：https://stackoverflow.com/questions/3197999/how-i-can-create-a-new-ties-method-with-the-r-rank-function