您可以使用ave。

df[ave(df$value, df$group, FUN=\(x) diff(sort(c(-x, Inf)))[1]) <= 2,]
#   group value
#1      A     5
#2      A     1
#3      A     1
#4      A     5
#9      C    10
#10     C     1
#11     C     1
#12     C     8

如果你可以确定你有所有的时间，至少有两个值，你可以使用。

df[ave(df$value, df$group, FUN=\(x) diff(tail(sort(x), 2))) <= 2,]
df[ave(df$value, df$group, FUN=\(x) diff(sort(-x)[1:2])) <= 2,]

使用dplyr

library(dplyr)

df %>% 
  group_by(group) %>% 
  filter(abs(diff(sort(value, decreasing=T)[1:2])) <= 2) %>%
  ungroup()
# A tibble: 8 × 2
  group value
  <chr> <int>
1 A         5
2 A         1
3 A         1
4 A         5
5 C        10
6 C         1
7 C         1
8 C         8

A碱基R替代物

grp <- na.omit(aggregate(. ~ group, df, function(x) 
  abs(diff(sort(x, decreasing=T)[1:2])) <= 2))

do.call(rbind, c(mapply(function(g, v) 
  list(df[df$group == g & v,]), grp$group, grp$value), make.row.names=F))
  group value
1     A     5
2     A     1
3     A     1
4     A     5
5     C    10
6     C     1
7     C     1
8     C     8

我可能会先创建一个向量与组，以满足您的条件，然后在原始的data.frame过滤器。

library(dplyr)

group_to_keep <-
  df %>% 
  group_by(group) %>% 
  slice_max(n = 2,value) %>% 
  filter(abs(diff(value)) <= 2) %>% 
  pull(group) %>% 
  unique()

df %>% 
  filter(group %in% group_to_keep)