根据组和另一个数据帧删除R数据帧中的行

这可以通过 反连接 . 但是,我们需要确定哪些组ID group 必须从 dat1

library(data.table)
# count names per ID
setDT(dat2)[, n.id := .N, by = ID]
# identify groups to remove by joining and ... 
groups_to_remove <- dat2[setDT(dat1), on = "name", nomatch = 0L][
  # ... check which groups have a match with the complete set of names
  , which(n.id == .N), by = .(ID, group)]
# anti join
dat1[!groups_to_remove, on = "group"]

   group name
1:    12    C
2:    12    D
3:    14    G
4:    14    H
5:    15    I
6:    15    J
7:    19    A
8:    19    X

第19组是 不 dat2

更精简的方法使用 all()

library(data.table)
setDT(dat1)
setDT(dat2)
groups_to_remove <- dat1[dat2, on = "name"][, which(all(ID == ID[1])), by = group]
dat1[!groups_to_remove, on = "group"]

1: 12摄氏度
3: 14克
5: 15英寸
6: 15焦耳
8: 19倍

同上 dplyr 语法:

library(dplyr)
dat2 %>% 
  left_join(dat1, by = "name") %>% 
  group_by(group) %>% 
  summarise(all_have_same_id = all(ID == ID[1L])) %>% 
  filter(all_have_same_id) %>% 
  anti_join(dat1, ., by = "group")

  group name
1    12    C
2    12    D
3    14    G
4    14    H
5    15    I
6    15    J
7    19    A
8    19    X
Warning message:
Column `name` joining factors with different levels, coercing to character vector

数据

示例数据集 由OP提供的组由不包含名称的组组成 数据2 或者所有的名字都在一个ID中 数据2 数据2 . 因此,我添加了这个用例(作为组19):

dat1 <- data.frame(
  group= c(11,11,12,12,13,13,14,14,15,15,16,16,17,17,17,18,18,18,19,19),
  name= c("A","B","C","D","E","F","G","H","I","J","A","B","E","F","W","A","B","V","A","X"))

dat2 <- data.frame(ID=c(1,1,2,2,3,3),name =c("A","B","E","F","X","Y"))

像这样。。。?

dat1[dat1$name %in% setdiff(dat1$name, dat2$name), ]
3     12    C
4     12    D
7     14    G
8     14    H
9     15    I
10    15    J
15    17    W
18    18    V