基于R中的两个标准聚合数据的有效方式

Question

我从大数据框中取样.我想前n从因素每个类别的每个ID.我已经尝试过使用嵌套for循环,但我认为它并不是有效的方式.请在下面找到数据和我的尝试.

id = c(1,1,1,1,1,1,2,2,2,2,2,2,2) 
category = c("A","A","A","B","B","B","A","A","A","A","B","B","B") 
factor= c("A1","A2","A3","B1","B2","B3","A2","A1","A3","A4","B2","B1","B3") 
rank = c(2,1,5,2,1,6,12,10,8,9,23,12,10) 

df = data.frame(id,category,factor,rank)

df$rank <- as.integer(df$rank)
cat <- as.data.frame(table(df$category))
list1 <- list()
list2 <- list()

for(i in 1:length(unique(df$id))){

  for(j in 1:nrow(cat))
  {
    list1[[j]] <- df[order(rank),][id==df[,id[i]]&category==cat[[1]][j] ,][1:2] 
  }
  rbind_list1 <- rbindlist(list1)
  list2[[i]] <- rbind_list1[rowSums(is.na(rbind_list1)) != ncol(rbind_list1),]

}
final <- do.call(rbind, list2)
df <- final[rowSums(is.na(final)) != ncol(final),]



DESIRED OUTPUT:

id  category factor rank
1     A        A2    1
1     A        A1    2
1     B        B1    2
1     B        B2    1
2     A        A3    8
2     A        A4    9
2     B        B3    10
2     B        B1    12

Answer 1

使用基数R,您可以这样做:

topn <- function(df, n) do.call(rbind, by(df, list(id, category), function(x) x[order(x$rank), ][1:n, ]))

topn(df, 2)  

   id category factor rank
2   1        A     A2    1
1   1        A     A1    2
9   2        A     A3    8
10  2        A     A4    9
5   1        B     B2    1
4   1        B     B1    2
13  2        B     B3   10
12  2        B     B1   12