根据另一列（R）的数量删除组

Question

我有一个数据框，我只想保留至少有2箱有汽车的组，而保留组的另一个输出将至少有2箱没有汽车的组：

Group = c('a','a','a','b','b','b','c','c','c','c')
Car = c(1,1,0,0,0,0,1,0,0,0) # 1 = Have car, 0 = No car
df = data.frame(Group,Car)
df$Group = factor(df$Group)
df$Car = factor(df$Car)

   Group Car
1      a   1
2      a   1
3      a   0
4      b   0
5      b   0
6      b   0
7      c   1
8      c   0
9      c   0
10     c   0

输出应为：

Group Car
a      1
a      1
a      0

第二输出：

Group Car
b      0
b      0
b      0
c      1
c      0
c      0
c      0

我有一个非常庞大的数据集。 请帮忙。 谢谢！

Answer 1

第一种情况：至少有2箱汽车的团体

library(dplyr)

df %>%
  group_by(Group) %>%
  filter(sum(Car) > 1)

#   Group   Car
#   <fct> <dbl>
#1  a         1
#2  a         1
#3  a         0

或基础R ave

subset(df, ave(Car, Group, FUN = sum) > 1)

和data.table

library(data.table)
setDT(df)[, if (sum(.SD) > 1) .SD, by = Group]

---

第2种情况：至少有2例无车的团体

df %>%
  group_by(Group) %>%
  filter(sum(Car == 0) > 1)


#  Group   Car
#  <fct>  <dbl>
#1  b       0
#2  b       0
#3  b       0
#4  c       1
#5  c       0
#6  c       0
#7  c       0

并具有基准R ave

subset(df, ave(Car == 0, Group, FUN = sum) > 1)

与data.table

setDT(df)[, if (sum(.SD == 0) > 1) .SD, by = Group]

数据

Group = c('a','a','a','b','b','b','c','c','c','c')
Car = c(1,1,0,0,0,0,1,0,0,0) 
df = data.frame(Group,Car)

Answer 2

我们可以在一个同时获得数据集list使用一步split

lst1 <- split(df, df$Group %in% names(which(rowsum(df$Car, df$Group)[,1] >= 2)))
lst1
#$`FALSE`
#   Group Car
#4      b   0
#5      b   0
#6      b   0
#7      c   1
#8      c   0
#9      c   0
#10     c   0

#$`TRUE`
#  Group Car
#1     a   1
#2     a   1
#3     a   0

如果我们需要提取list元素，请使用[[

lst1[[1]]
lst1[[2]]

数据

Group <- c('a','a','a','b','b','b','c','c','c','c')
Car <- c(1,1,0,0,0,0,1,0,0,0) 
df <- data.frame(Group,Car)