r - 如果存在于 =.>3 个数据框中(共 5 个)，如何删除公共(public)元素

Question

继 here 中的上一个问题之后, 并进行进一步的统计分析, 我想知道是否有可能去除存在于 >= 3 个数据帧中的常见峰值

a <- data.frame(ID = c("1", "2", "3", "4", "5"), peak = c("peak1", "peak2", "peak3", "peak4", "peak10"))
b <- data.frame(ID = c("1", "2", "3", "4"), peak = c("peak1","peak3", "peak20", "peak21"))
c <- data.frame(ID = c("1", "2", "3"), peak = c("peak1", "peak5", "peak3"))
d <- data.frame(ID = c("1", "2", "3", "4", "5", "6"),peak = c("peak1", "peak3", "peak7", "peak8", "peak11", "peak12"))
e <- data.frame(ID = c("1", "2", "3"), peak = c("peak1", "peak3",  "peak9"))

并且我想删除存在于 >= 3 个数据帧中的常见峰值，并具有所需的输出:

a <- data.frame(ID = c("1", "2", "3", "4", "5"), peak = c("peak2",  "peak4", "peak10"))
b <- data.frame(ID = c("1", "2", "3", "4"), peak = c("peak20", "peak21"))
c <- data.frame(ID = c("1", "2", "3"), peak = c( "peak5"))
d <- data.frame(ID = c("1", "2", "3", "4", "5", "6"),peak = c(  "peak7", "peak8", "peak11", "peak12"))
e <- data.frame(ID = c("1", "2", "3"), peak = c ("peak9"))

Answer 1

如果'peak'值在每个数据集中是唯一的，将数据集绑定(bind)在一起成为一个数据(bind_rows)，得到'peak'的count，filter 'n' 小于 3 的行并pull 那些 'peak' 元素

library(dplyr)
to_keep <- bind_rows(a, b, c, d, e, .id = 'grp') %>%
             count(peak) %>% 
             filter(n < 3) %>% 
             pull(peak)

现在我们更新全局环境中的对象(不推荐)，在根据“to_keep”的峰值对这些元素进行子集之后使用assign

for(obj in letters[1:5]) {
      assign(obj, subset(get(obj), peak %in% to_keep))
 }

或者将对象保存在列表中并从那里获取子集

library(purrr)
lst1 <- lst(a, b, c, d, e) %>%
           map(~ .x %>%
                  filter(peak %in% to_keep))