如何从5中删除=.>3数据帧中的公共元素

Question

根据here中先前的问题，并进行进一步的统计分析，我想知道是否可以删除>= 3数据帧中存在的公共峰值。

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"))

我希望删除>= 3数据帧中存在的公共峰值，并提供所需的输出：

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“峰值”值在每个数据集中是唯一的，则将数据集绑定到单个数据(bind_rows)中，获取“峰值”的count，将“n”小于3的行filter，并将这些“峰值”元素pull

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env中的对象(不推荐)，在基于‘assign’的峰值对这些元素进行subset之后使用subset

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

或者将对象保存在list和子集中。

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

Answer 2

在R基，你可以：

my_list <- list(a = a, b = b, c = c, d = d, e = e)
y <-table(do.call(rbind, my_list)) < 3
list2env(lapply(my_list, function(x) subset(x, y[peak])), .GlobalEnv)

现在调用a或任何数据格式。

将结果保存在一份清单中是明智的。ie lapply(my_list, function(x) subset(x, y[peak]))而不是将它们转换到环境中。

Answer 3

另一个基本R选项

lst <- list(a, b, c, d, e)
v <- names(Filter(function(x) x < 3, table(unlist(sapply(lst, `[[`, "peak")))))
lapply(
  lst,
  function(x) {
    subset(x, peak %in% v)
  }
)

这样的话

[[1]]
  ID   peak
2  2  peak2
4  4  peak4
5  5 peak10

[[2]]
  ID   peak
3  3 peak20
4  4 peak21

[[3]]
  ID  peak
2  2 peak5

[[4]]
  ID   peak
3  3  peak7
4  4  peak8
5  5 peak11
6  6 peak12

[[5]]
  ID  peak
3  3 peak9