在 R 中的 data.frames 列表中设置特定于集群的值

Question

受this answer 的启发，我的目标是在m 的data.frames 集群中找到仅特定于一个m（例如m[[15]]）而不是其他ms 的变量。

例如，我知道变量genre == 4 仅特定于m[[15]]（“Fazio”即names(m)[15]），而genre == 4 不会出现在任何其他m 集群中（由@ 确认987654331@).

因此，我希望我的输出给我命名 "Fazio" 和 genre == 4 。

我想对mods 中显示的所有变量重复此过程，而不仅仅是genre？

我尝试了以下方法但没有成功：

d <- read.csv("https://raw.githubusercontent.com/rnorouzian/m/master/v.csv", h = T) # DATA

mods <- c("genre","cont.type","time","cf.timely","ssci","setting","ed.level",  # mods
          "Age","profic","motivation","Ss.aware","random.grp","equiv.grp",
          "rel.inter","rel.intra","sourced","timed","Location",
          "cf.scope","cf.type","error.key","cf.provider","cf.revision","cf.oral",
          "Length","instruction","graded","acc.measure","cf.training","error.type")

m <- split(d, d$study.name) # `m` clusters of data.frames

# SOLUTION TRIED:

tmp = do.call(rbind, lapply(mods, function(x){
  d = unique(d[c("study.name", x)])
  names(d) = c("study.name", "val")
  transform(d, nm = x)
}))

# this logic may need to change:
tmp = tmp[ave(as.numeric(as.factor(tmp$val)), tmp$val, FUN = length) == 1,] 

lapply(split(tmp, tmp$study.name), function(a){
 setNames(a$val, a$nm)
})                               # doesn't return anything

Answer 1

我们也可以通过在ave 中添加“nm”来进行分组

tmp1 <- tmp[with(tmp, ave(val, val, nm, FUN = length)==1),]

现在执行split

tmp2 <- lapply(split(tmp1, tmp1$study.name, drop = TRUE), `row.names<-`, NULL)
rm.df <- data.frame(study.name = c(rep("Bitc_Knch_c", 3),
  rep("Sun", 3)), code = c(88,88,88,7,4,0), 

 mod.name = c("error.type","cf.scope","cf.type","error.type",
     "cf.type","error.key"))
rm.these <- split(rm.df, rm.df$study.name)

tmp2[names(rm.these)] <- Map(function(x, y) {
     subset(x, !(nm %in% y$mod.name & val %in% y$code))}, 
     tmp2[names(rm.these)], rm.these)
Filter(nrow, tmp2)