Question

我想在o年发生事件r之后更新变量y的值。一个lapply()解决方案在行数超过500k时似乎很慢。因此，在寻找一个更快的解决方案时，我试图避免使用aggregate()进行子集设置，这会更快一些，但并没有达到预期的效果。在base R中（或者也许在data.table中，有更快的方法来做这些事情吗？

示例

# lapply
lapply(split(df2, df2$id), function(x) {
  ry <- x$y[which(x[, "r"] == 1)]
  x[x$y >= ry, "o"] <- 1
  x
})

# aggregate
df2 <- merge(df1, with(df1, aggregate(list(ry=r), by=list(id=id), 
                    function(x) y[which(x == 1)])))
lapply(split(df2, df2$id), function(x) {
  x$o[x$y >= unique(x$ry)] <- 1
  x
})

# Output for id `11`
  id   ry    y o r
1 11 2003 2005 1 0
2 11 2003 2004 1 0
3 11 2003 2003 1 1
4 11 2003 2002 0 0
5 11 2003 2001 0 0

基准

Unit: microseconds
      expr     min      lq     mean  median       uq      max neval cld
    lapply 915.181 929.724 988.2273 934.699 943.2465 5150.221   100   b
 aggregate 790.418 803.175 844.8039 810.192 817.4635 3474.984   100  a

数据

df1 <- structure(list(id = c(11, 11, 11, 11, 11, 22, 22, 22, 22, 22, 
33, 33, 33, 33, 33, 44, 44, 44, 44, 44, 55, 55, 55, 55, 55), 
    y = c(2005L, 2004L, 2003L, 2002L, 2001L, 2005L, 2004L, 2003L, 
    2002L, 2001L, 2005L, 2004L, 2003L, 2002L, 2001L, 2005L, 2004L, 
    2003L, 2002L, 2001L, 2005L, 2004L, 2003L, 2002L, 2001L), 
    o = c(0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 0, 
    0, 0, 0, 0, 0, 0, 0, 0), r = c(0, 0, 1, 0, 0, 0, 1, 0, 0, 
    0, 0, 0, 0, 0, 0, 1, 0, 0, 0, 0, 0, 0, 0, 0, 1)), out.attrs = list(
    dim = c(id = 5L, y = 5L, o = 1L, r = 1L), dimnames = list(
        id = c("id=11", "id=22", "id=33", "id=44", "id=55"), 
        y = c("y=2001", "y=2002", "y=2003", "y=2004", "y=2005"
        ), o = "o=0", r = "r=0")), row.names = c(NA, -25L), class = "data.frame")

如何在事件后快速更新变量值？

0 个答案: