Chapter 13 Ranks

Many questions take forms such as these:

• “Find the largest …”
• “Find the three largest …”
• “Find the smallest within each group …”

The functions min() and max() are reasonable candidates for carrying out such tasks, but they are limited. For instance, to find the name whose yearly count was the highest:

BabyNames %>% 
  filter(count == max(count))

The filter() function needs a criterion. The criterion count == max(count) [Click to see note.]Note the double equals sign ‘==’ used in evaluating the criterion passes through the case where the value of count matches the largest value of count. That will be the biggest case.

It’s also possible to ask for the cases that are almost as popular as the biggest, e.g. at least 90% as popular.

BabyNames %>% 
  filter(count > 0.90 * max(count))

Frequently the question will be framed in terms of the \(n\) biggest or smallest values, not as a fraction of the largest. To perform such tasks, a new transformation verb is helpful: rank(). The rank() function does something simple but powerful: it replaces each number in a set with where that number stands with respect to the others. For instance, look at the tiny data frame Set shown in Table 13.1. What’s the rank of the number 5 in the numbers variable.

Table 13.1: A data frame with one variable. We will refer to this as Set

The rank of the number 5 in the set is four. Why? 5 is the fourth smallest number in the set. (Three of the numbers in the set 2, 2, 4 are smaller than 5.) The smallest number in a set of \(n\) numbers will have rank 1; the largest will have rank \(n\).11 Unless there are ties. Table 13.2 shows the rank of each number in the set:

Set %>% mutate(the_rank = rank(numbers))

Table 13.2: The ranks of the values in numbers.

Note that the two 2’s have the same rank, as do the two 9’s. They’re tied.

Suppose you want to find the 3rd most popular name of all time. Use rank().

BabyNames %>% 
  group_by(name) %>%
  summarise(total = sum(count)) %>%
  filter(rank(desc(total)) == 3) 

Table 13.3: The third most popular name of all time

Or, to find the top three most popular names, replace == in the above by <=.

BabyNames %>% 
  group_by(name) %>%
  summarise(total = sum(count)) %>%
  filter( rank(desc(total)) <= 3) 

Table 13.4: The all-time top three most popular names in BabyNames.

When applied to grouped data, rank() will be calculated separately within each group. That is, the rank of a value will be with respect to the other cases in that group. For instance, here’s the third most popular name each year.

BabyNames %>% 
  group_by(year) %>% 
  filter(rank(desc(count)) == 3) 

Table 13.5: The third most popular name in each year

Sometimes, two or more numbers are tied in rank. The rank() function deals with these by assigning all the tied values the same rank, which is the mean of the ranks those values would have had if they were even slightly different. There are other rank-like transformation verbs that handle ties differently. For instance, row_number() breaks ties in favor of the first case encountered. (See Table 13.6.)

Set %>% 
  mutate(the_rank = rank(numbers), 
         ties_broken = row_number(numbers))

Table 13.6: row_number() breaks ties in rank.