Dplyr：如何分组（全部）

Question

As I saw somewhere, when there are multiple layers of group_by(), one summarise() peels off a layer of grouping. In sql, there's "group by all". I wonder if there's a way to cancel all grouping in dplyr (so that, eg, we can get max of all, rather than of each group)

Example:

library(dplyr)
library(car)

mtcars %>%
  select(cyl, gear, carb) %>%
  group_by(cyl, gear) %>%
  summarise(count = n()) %>%
  arrange(desc(count))

Output:

Source: local data frame [8 x 3]
Groups: cyl

  cyl gear count
1   4    4     8
2   4    5     2
3   4    3     1
4   6    4     4
5   6    3     2
6   6    5     1
7   8    3    12
8   8    5     2

So the data was group_by(cyl, gear): two layers of grouping. The summarise() counted how many cars in each (cyl, gear) group, and then peels off the group_by(gear) layer. Now the data is group_by(cyl). As you can see the descent order only works for each cyl. (Descent in line 1-3 for cyl==4, descent in line 4-6 for cyl==6,...). How can we get a descent of all 8 lines? (Line 7 should be the first line.)

Another example of how summarise() peels of grouping:

mtcars %>%
  select(cyl, gear, carb) %>%
  group_by(cyl, gear) %>%
  summarise(count = n())

Output:
Source: local data frame [8 x 3]
Groups: cyl

  cyl gear count
1   4    3     1
2   4    4     8
3   4    5     2
4   6    3     2
5   6    4     4
6   6    5     1
7   8    3    12
8   8    5     2

---
mtcars %>%
  select(cyl, gear, carb) %>%
  group_by(cyl, gear) %>%
  summarise(count = n()) %>%
  summarise(count1 = max(count))


Output:
Source: local data frame [3 x 2]

  cyl count1
1   4      8
2   6      4
3   8     12

---

mtcars %>%
  select(cyl, gear, carb) %>%
  group_by(cyl, gear) %>%
  summarise(count = n()) %>%
  summarise(count1 = max(count)) %>%
  summarise(max(count1))

Output:

Source: local data frame [1 x 1]

  max(count1)
1          12

Answer 1

Try:

mtcars %>%
  count(cyl, gear, name = "count") %>%
  arrange(desc(count))

You will get:

#Source: local data frame [8 x 3]
#
#  cyl gear count
#1   8    3    12
#2   4    4     8
#3   6    4     4
#4   4    5     2
#5   6    3     2
#6   8    5     2
#7   4    3     1
#8   6    5     1