How to rank within groups in R?
Solution 1
You can do this pretty cleanly with dplyr
library(dplyr)
df %>%
group_by(customer_name) %>%
mutate(my_ranks = order(order(order_values, order_dates, decreasing=TRUE)))
Source: local data frame [5 x 4]
Groups: customer_name
customer_name order_dates order_values my_ranks
1 John 2010-11-01 15 3
2 Bob 2008-03-25 12 1
3 Alex 2009-11-15 5 1
4 John 2012-08-06 15 2
5 John 2015-05-07 20 1
Solution 2
The top rated answer (by cdeterman) is actually incorrect. The order function provides the location of the 1st, 2nd, 3rd, etc ranked values not the ranks of the values in their current order.
Let’s take a simple example where we want to rank, starting with the largest, grouping by customer name. I have included a manual ranking so we can check the values
> df
customer_name order_values manual_rank
1 John 2 5
2 John 5 2
3 John 9 1
4 John 1 6
5 John 4 3
6 John 3 4
7 Lucy 4 4
8 Lucy 9 1
9 Lucy 6 3
10 Lucy 2 6
11 Lucy 8 2
12 Lucy 3 5
If I run the code suggested by cdeterman I get the following incorrect ranks:
> df %>%
+ group_by(customer_name) %>%
+ mutate(my_ranks = order(order_values, decreasing=TRUE))
Source: local data frame [12 x 4]
Groups: customer_name [2]
customer_name order_values manual_rank my_ranks
<fctr> <dbl> <dbl> <int>
1 John 2 5 3
2 John 5 2 2
3 John 9 1 5
4 John 1 6 6
5 John 4 3 1
6 John 3 4 4
7 Lucy 4 4 2
8 Lucy 9 1 5
9 Lucy 6 3 3
10 Lucy 2 6 1
11 Lucy 8 2 6
12 Lucy 3 5 4
Order is used to re-order dataframes into decreasing or increasing order. What we actually want is to run the order function twice, with the second order function giving us the actual ranks we want.
> df %>%
+ group_by(customer_name) %>%
+ mutate(good_ranks = order(order(order_values, decreasing=TRUE)))
Source: local data frame [12 x 4]
Groups: customer_name [2]
customer_name order_values manual_rank good_ranks
<fctr> <dbl> <dbl> <int>
1 John 2 5 5
2 John 5 2 2
3 John 9 1 1
4 John 1 6 6
5 John 4 3 3
6 John 3 4 4
7 Lucy 4 4 4
8 Lucy 9 1 1
9 Lucy 6 3 3
10 Lucy 2 6 6
11 Lucy 8 2 2
12 Lucy 3 5 5
Solution 3
This can be achieved with ave and rank. ave passes the proper groups to rank. The result from rank is reversed due to the requested order:
with(x, ave(as.numeric(order_dates), customer_name, FUN=function(x) rev(rank(x))))
## [1] 3 1 1 2 1
Solution 4
In base R you can do this with the slightly unwieldy
transform(df,rank=ave(1:nrow(df),customer_name,
FUN=function(x) order(order_values[x],order_dates[x],decreasing=TRUE)))
customer_name order_dates order_values rank 1 John 2010-11-01 15 3 2 Bob 2008-03-25 12 1 3 Alex 2009-11-15 5 1 4 John 2012-08-06 15 2 5 John 2015-05-07 20 1
where order is provided both the primary and tie-breaker values for each group.
Solution 5
df %>%
group_by(customer_name) %>%
arrange(customer_name,desc(order_values)) %>%
mutate(rank2=rank(order_values))
Saul Feliz
Updated on August 07, 2021Comments
-
Saul Feliz over 2 years
OK, check out this data frame...
customer_name order_dates order_values 1 John 2010-11-01 15 2 Bob 2008-03-25 12 3 Alex 2009-11-15 5 4 John 2012-08-06 15 5 John 2015-05-07 20Lets say I want to add an order variable that Ranks the highest order value, by name, by max order date, using the last order date at the tie breaker. So, ultimately the data should look like this:
customer_name order_dates order_values ranked_order_values_by_max_value_date 1 John 2010-11-01 15 3 2 Bob 2008-03-25 12 1 3 Alex 2009-11-15 5 1 4 John 2012-08-06 15 2 5 John 2015-05-07 20 1Where everyone's single order gets 1, and all subsequent orders are ranked based on the value, and the tie breaker is the last order date getting priority. In this example, John's 8/6/2012 order gets the #2 rank because it was placed after 11/1/2010. The 5/7/2015 order is 1 because it was the biggest. So, even if that order was placed 20 years ago, it should be the #1 Rank because it was John's highest order value.
Does anyone know how I can do this in R? Where I can Rank within a group of specified variables in a data frame?
Thanks for your help!