r 中包含 group_by 函数的子集数据

Question

这是对以下问题的后续问题给here

我有以下数据

数据：

df = structure(list(Org_ID = c(1L, 1L, 1L, 2L, 2L, 3L, 3L, 3L, 3L), 
    Market_volume = c(100L, 200L, 300L, 50L, 500L, 400L, 200L, 
    300L, 100L), Indicator_variable = c(1L, 0L, 0L, 1L, 1L, 0L, 
    0L, 0L, 0L),variable3=c(10L, 1L, 1L, 4L, 2L, 3L, 3L, 10L, 3L),variable4=c(2L, 1L, 1L, 7L, 2L, 3L, 3L, 8L, 3L)).Names = c("Org_ID", "Market_volume", "Indicator_variable","Var3","Var4"
), class = "data.frame", row.names = c(NA, -9L))

使用 (dplyr)，我通过以下函数按 Org_ID 计算了 NA 按市场交易量计算的百分比

df %>%
  group_by(Org_ID) %>%
  summarize(sum_market_vol = sum(Market_volume*!Indicator_variable),
            tot_market_vol = sum(Market_volume)) %>%
  transmute(Org_ID, Perc_Market_Vol = 100*sum_market_vol/tot_market_vol)

结果：

# A tibble: 3 x 2
  Org_ID Perc_Market_Vol
   <int>           <dbl>
1      1        83.33333
2      2         0.00000
3      3       100.00000

问题： 我想通过删除 Org_ID 的所有行（比如 2）# X if perc_market_vol

我希望新数据如下所示：

df1 = structure(list(Org_ID = c(1L, 1L, 1L, 3L, 3L, 3L, 3L), 
    Market_volume = c(100L, 200L, 300L, 400L, 200L, 
    300L, 100L), Indicator_variable = c(1L, 0L, 0L, 0L, 
    0L, 0L, 0L),variable3=c(10L, 1L, 1L, 3L, 3L, 10L, 3L),variable4=c(2L, 1L, 1L, 3L, 3L, 8L, 3L)).Names = c("Org_ID", "Market_volume", "Indicator_variable","Var3","Var4"
), class = "data.frame", row.names = c(NA, -7L))

Answer 1

您可以使用group_by %>% filter 在不具体化聚合数据框的情况下进行过滤，并且在过滤器中您可以计算每个组的聚合条件：

df %>% 
    group_by(Org_ID) %>% 
    filter(sum(Market_volume * !Indicator_variable)/sum(Market_volume) > 0.3)

# A tibble: 7 x 5
# Groups:   Org_ID [2]
#  Org_ID Market_volume Indicator_variable  Var3  Var4
#   <int>         <int>              <int> <int> <int>
#1      1           100                  1    10     2
#2      1           200                  0     1     1
#3      1           300                  0     1     1
#4      3           400                  0     3     3
#5      3           200                  0     3     3
#6      3           300                  0    10     8
#7      3           100                  0     3     3