估计面板中​​组合过去时期的公共集合元素的百分比

Question

我有一个时间序列面板数据集，其结构如下：有 2 只基金在每个时间段各自拥有不同的股票。

df <- data.frame(
  fund_id = c(1,1,1,1,1,1,1,1, 1, 2,2,2,2),
  time_Q = c(1,1,1,2,2,2,2,3, 3, 1,1,2,2),
  stock_id = c("A", "B", "C", "A", "C", "D", "E", "D", "E", "A", "B", "B", "C")

)

> df
   fund_id time_Q stock_id
1        1      1        A
2        1      1        B
3        1      1        C
4        1      2        A
5        1      2        C
6        1      2        D
7        1      2        E
8        1      3        D
9        1      3        E
10       2      1        A
11       2      1        B
12       2      2        B
13       2      2        C

对于每只基金，我想计算当前 time_Q 持有的股票在前一到 2 年中曾经 持有的股票的百分比宿舍。所以基本上对于每个基金和每个 time_Q，我想有 2 列过去 1 time_Q，过去 1-2 time_Q 显示当时持有的股票百分比也是出现在过去的任何 time_Q 中。 结果应该是这样的：

result <- data.frame(
  fund_id = c(1,1,1,2,2),
  time_Q = c(1,2,3,1,2),
  past_1Q = c("NA",0.5,1,"NA",0.5),
  past_2Q = c("NA",0.5,1,"NA",0.5)
)

>  result
  fund_id time_Q past_1Q past_1_2Q
1       1      1      NA      NA
2       1      2     0.5     0.5
3       1      3       1       1
4       2      1      NA      NA
5       2      2     0.5     0.5

我已经问过类似的问题 ，但现在我正在寻找过去任何滞后时期的共同元素。我正在寻找一个 dplyr 或 data.table 可扩展的解决方案，我可以在过去的 12 个季度左右处理多个基金、股票和时间段。

提前致谢！

Answer 1

我的解决方案

# dummy data
df <- data.table(fund_id = c(1,1,1,1,1,1,1,1, 1, 2,2,2,2)
                 , time_Q = c(1,1,1,2,2,2,2,3, 3, 1,1,2,2)
                 , stock_id = c("A", "B", "C", "A", "C", "D", "E", "D", "E", "A", "B", "B", "C")
                 ); df

# lower case col names
names(df) <- tolower(names(df))

# unique grouping
x <- df[, .(dummy =.N), .(fund_id, time_q)][, dummy := NULL]

# initialise empty table
y <- NULL

# loop
for(i in 1:nrow(x))
{
  # current quarter & before
  z <- df[fund_id == x[i, fund_id]
          & time_q %between% c( x[i, time_q] - 12, x[i, time_q])
          ]
  
  # current quarter
  a <- z[fund_id == x[i, fund_id]
         & time_q == x[i, time_q]
         , unique(stock_id)
         ]
  
  # minus 1 to minus 12 quarter (lapply)
  b <- lapply(1:12, \(j) z[fund_id == x[i, fund_id]
                           & time_q %between% c( x[i, time_q] - j, x[i, time_q] - 1)
                           , unique(stock_id)
                           ]
              )
  
  # results
  c <- data.table(fund_id = x[i, fund_id]
                  , current_q = x[i, time_q]
                  )
  
  # no. of stocks in current quarter
  d <- length(a)
  
  # calculate % for the 12 periods
  c[, paste0('past_1_to_', 1:12, '_q') := lapply(1:12, \(j) length(intersect(a,b[[j]])) / d) ]
  
  # collect results
  y <- rbind(y, c)
}

基准

x <- 1e3
df <- data.table(fund_id = rep(1:x, each = x/10)
                 , time_Q = rep(1:4, each = x/4)
                 , stock_id = sample(letters[1:26], size=20, replace=T)
                 )

在上面 df 上用了 20 秒，有 100k 行和 1,200 个组（fund_id，time_q）