使用这种基本方法有两个大思路：

1. 由于我们需要比较所有值，因此我们应该将所有内容重新组合为一个data.frame。
2. 将未拆分的data.frame设置得较长会节省一些额外的步骤。

#https://stackoverflow.com/questions/58786052/find-variables-that-occur-only-once-across-a-split-data-frame-in-r/58788854#58788854
f <- data.frame(id = c(rep("AA",4),rep("BB",2),rep("CC",2)),X = c(1,2,3,1,4,3),Y = c(99,7,8,6,7))
m <- split(f,f$id) # Here is `m`

unsplit <- do.call(rbind,c(m,make.row.names = F))
molten <- data.frame(unsplit[,drop = F],stack(unsplit[,-1]))

# res <- subset(molten,!duplicated(values) & !duplicated(values,fromLast = T))
res <- molten[as.logical(ave(molten[['values']],molten[['ind']],FUN = function(x) !duplicated(x) & !duplicated(x,fromLast = T))),]
#I would stop here
res
#>    id values ind
#> 6  BB      4   X
#> 9  AA     99   Y
#> 11 AA      8   Y
#> 13 BB      6   Y

#to get exact output
res_vector <- res$values
names(res_vector) <- res$ind

split(res_vector,as.character(res$id))
#> $AA
#>  Y  Y 
#> 99  8 
#> 
#> $BB
#> X Y 
#> 4 6

^{由reprex package（v0.3.0）于2019-11-10创建}

这是另一种可能不太复杂的基本方法：

####Way 1 with rapply
vec <- rapply(lapply(m,'[',mods),I)
unique_vec <- vec[!duplicated(vec) & !duplicated(vec,fromLast = T)]

vec_names <- do.call(rbind,strsplit(names(unique_vec),'.',fixed = T))

names(unique_vec) <- substr(vec_names[,2],1) #turns Y1 into Y
split(unique_vec,vec_names[,1])

###Way 2 with data.frame already do.call(rbind,m)
vec <-   unlist(
  lapply(f[,-1],function(x){
           ind <- !duplicated(x) & !duplicated(x,fromLast = T)
           ret <- x[ind]
           names(ret) <- f[ind,1]
           ret
         } 
  )
)

#this is likely overly simplified:
split(vec,sub('.*\\.','',names(vec)))

#this leads to exact result
vec_names <- do.call(rbind,strsplit(names(vec),fixed = T))
names(vec) <- vec_names[,1]

split(vec,2])

$AA
 Y  Y 
99  8 

$BB
X Y 
4 6 

OP在提示中使用table()出现。 duplicated()表现出色：

unlist(lapply(f[mods],function(y) names(which(table(y) == 1))))
#   X   Y1   Y2   Y3 
# "4"  "6"  "8" "99"

vec
#X.BB Y.AA Y.AA Y.BB 
#   4   99    8    6 

# A tibble: 2 x 13
  expression   min median `itr/sec` mem_alloc
  <bch:expr> <bch> <bch:>     <dbl> <bch:byt>
1 table_meth 321us  336us     2794.    10.3KB
2 dup_meth   132us  136us     7105.    31.7KB

bench::mark(
  table_meth = {unlist(lapply(f[mods],function(y) names(which(table(y) == 1))))},dup_meth = {
  #could get slight performance boost with
    #f_id <- f[['id']]
  unlist(
    lapply(f[,function(x){
             ind <- !duplicated(x) & !duplicated(x,fromLast = T)
             ret <- x[ind]
             names(ret) <- f[ind,1]
             #names(ret) <- f_id[ind] 
             ret
           } 
    )
  )},check = F
)

还有data.table中的类似想法：

library(data.table)

molten_dt <- melt(rbindlist(m),id.vars = 'id')
molten_dt[!duplicated(value,by = variable) &
             !duplicated(value,by = variable,fromLast = T)]

还有dplyr中的类似想法：

library(dplyr)
library(tidyr)

m%>%
  bind_rows()%>%
  pivot_longer(cols = -id)%>%
  group_by(name)%>%
  filter(!duplicated(value) & !duplicated(value,fromLast = T))%>%
  group_by(id)%>%
  group_split()

这不是纯函数式编程，而是基于R的语言：

lapply(split(df,df$id),function(z){

  X <- z$X[which(!(z$X %in% df$X[duplicated(df$X)]))]

  Y <- z$Y[which(!(z$Y %in% df$Y[duplicated(df$Y)]))]

  cbind(X,Y)

  }

)

数据：

    df <-
  structure(list(
    id = structure(
      c(1L,1L,2L,3L,3L),.Label = c("AA","BB","CC"),class = "factor"
    ),7)
  ),class = "data.frame",row.names = c(NA,-8L))