根据groupby筛选天数_R_Filter - Fatal编程技术网

根据groupby筛选天数

r filter

根据groupby筛选天数,r,filter,R,Filter,我有一个df，我想根据分组筛选出一列。我想按组合保留分组（cc，odd，tree1，和tree2），如果第4天，则保留，否则放弃 df <- data_frame( cc = c('BB', 'BB', 'BB', 'BB','BB', 'BB','BB', 'BB', 'DD', 'DD', 'DD', 'DD', 'DD', 'DD', 'DD', 'DD', 'ZZ', 'ZZ', 'ZZ', 'ZZ', 'ZZ', 'ZZ', 'ZZ', 'ZZ'), odd = c(

我有一个df，我想根据分组筛选出一列。我想按组合保留分组（

cc

，

odd

，

tree1

，和

tree2

），如果第4天，则保留，否则放弃

df <- data_frame(
cc = c('BB', 'BB', 'BB', 'BB','BB', 'BB','BB', 'BB', 'DD', 'DD', 'DD', 'DD', 'DD', 'DD', 'DD', 'DD', 
     'ZZ', 'ZZ', 'ZZ', 'ZZ', 'ZZ', 'ZZ', 'ZZ', 'ZZ'),
odd = c(3434, 3434, 3434, 3434, 3435, 3435, 3435, 3435, 3434, 3434, 3434, 3434, 3435, 3435, 3435, 3435, 3434, 3434, 3434, 3434, 3435, 3435, 3435, 3435),
tree1 = c('ASP', 'ASP', 'ASP', 'ASP', 'SAP', 'SAP', 'SAP', 'SAP', 'ASP', 'ASP', 'ASP', 'ASP', 'SAP', 'SAP', 'SAP', 'SAP', 'ASP', 'ASP', 'ASP', 'ASP', 'SAP', 'SAP', 'SAP', 'SAP'),
tree2 = c('ATK', 'ATK','ATK','ATK','ATK','ATK','ATK','ATK', 'ATK', 'ATK','ATK','ATK','ATK','ATK','ATK','ATK', 'ATK', 'ATK','ATK','ATK','ATK','ATK','ATK','ATK'),
day = c(1, 2, 3, 4, 3, 4, 5, 6, 2, 3, 4, 5, 1, 3, 5, 7, 1, 2, 6, 8, 2, 4, 6, 8)
)

我想得到一个df如下

df2 <- data_frame(
  cc = c('BB', 'BB','BB', 'BB', 'DD', 'DD', 'DD', 'DD', 'DD', 'DD', 'DD', 'DD', 
     'ZZ', 'ZZ', 'ZZ', 'ZZ', 'ZZ', 'ZZ', 'ZZ', 'ZZ'),
  odd = c(3435, 3435, 3435, 3435, 3434, 3434, 3434, 3434, 3435, 3435, 3435, 3435, 3434, 3434, 3434, 3434, 3435, 3435, 3435, 3435),
  tree1 = c('SAP', 'SAP', 'SAP', 'SAP', 'ASP', 'ASP', 'ASP', 'ASP', 'SAP', 'SAP', 'SAP', 'SAP', 'ASP', 'ASP', 'ASP', 'ASP', 'SAP', 'SAP', 'SAP', 'SAP'),
  tree2 = c('ATK','ATK','ATK','ATK', 'ATK', 'ATK','ATK','ATK','ATK','ATK','ATK','ATK', 'ATK', 'ATK','ATK','ATK','ATK','ATK','ATK','ATK'),
  day = c(3, 4, 5, 6, 2, 3, 4, 5, 1, 3, 5, 7, 1, 2, 6, 8, 2, 4, 6, 8)
)

df2你可以试试
df %>%
  group_by(cc, odd, tree1, tree2) %>% 
  filter(any(day > 4)) 
# A tibble: 20 x 5
   cc      odd tree1 tree2   day
   <chr> <dbl> <chr> <chr> <dbl>
 1 BB     3435 SAP   ATK       3
 2 BB     3435 SAP   ATK       4
 3 BB     3435 SAP   ATK       5
 4 BB     3435 SAP   ATK       6
 5 DD     3434 ASP   ATK       2
 6 DD     3434 ASP   ATK       3
 7 DD     3434 ASP   ATK       4
 8 DD     3434 ASP   ATK       5
 9 DD     3435 SAP   ATK       1
10 DD     3435 SAP   ATK       3
11 DD     3435 SAP   ATK       5
12 DD     3435 SAP   ATK       7
13 ZZ     3434 ASP   ATK       1
14 ZZ     3434 ASP   ATK       2
15 ZZ     3434 ASP   ATK       6
16 ZZ     3434 ASP   ATK       8
17 ZZ     3435 SAP   ATK       2
18 ZZ     3435 SAP   ATK       4
19 ZZ     3435 SAP   ATK       6
20 ZZ     3435 SAP   ATK       8

df%>%
分组依据（cc，奇数，tree1，tree2）%>%
过滤器（任何（第4天以上））
#一个tibble:20x5
cc奇数树1树2天
1 BB 3435 SAP ATK 3
2 BB 3435 SAP ATK 4
3 BB 3435 SAP ATK 5
4 BB 3435 SAP ATK 6
5 DD 3434 ASP ATK 2
6 DD 3434 ASP ATK 3
7 DD 3434 ASP ATK 4
8 DD 3434 ASP ATK 5
9 DD 3435 SAP ATK 1
10 DD 3435 SAP ATK 3
11 DD 3435 SAP ATK 5
12 DD 3435 SAP ATK 7
13 ZZ 3434 ASP ATK 1
14 ZZ 3434 ASP ATK 2
15 ZZ 3434 ASP ATK 6
16 ZZ 3434 ASP ATK 8
17 ZZ 3435 SAP ATK 2
18 ZZ 3435 SAP ATK 4
19 ZZ 3435 SAP ATK 6
20 ZZ 3435 SAP ATK 8

df %>%
  group_by(cc, odd, tree1, tree2) %>% 
  filter(any(day > 4)) 
# A tibble: 20 x 5
   cc      odd tree1 tree2   day
   <chr> <dbl> <chr> <chr> <dbl>
 1 BB     3435 SAP   ATK       3
 2 BB     3435 SAP   ATK       4
 3 BB     3435 SAP   ATK       5
 4 BB     3435 SAP   ATK       6
 5 DD     3434 ASP   ATK       2
 6 DD     3434 ASP   ATK       3
 7 DD     3434 ASP   ATK       4
 8 DD     3434 ASP   ATK       5
 9 DD     3435 SAP   ATK       1
10 DD     3435 SAP   ATK       3
11 DD     3435 SAP   ATK       5
12 DD     3435 SAP   ATK       7
13 ZZ     3434 ASP   ATK       1
14 ZZ     3434 ASP   ATK       2
15 ZZ     3434 ASP   ATK       6
16 ZZ     3434 ASP   ATK       8
17 ZZ     3435 SAP   ATK       2
18 ZZ     3435 SAP   ATK       4
19 ZZ     3435 SAP   ATK       6
20 ZZ     3435 SAP   ATK       8