从xts对象中删除连续重复行

您可以使用 rle() 关于每行的总和。

x <- structure(
  c(10L, 10L, 11L, 12L, 10L, 10L, 10L, 11L, 11L, 12L, 13L, 11L, 11L, 11L),
  .Dim = c(7L, 2L), .Dimnames = list(NULL, c("Bid", "Ask")),
  index = structure(1:7, tzone = "", tclass = c("POSIXct", "POSIXt")),
  .indexCLASS = c("POSIXct", "POSIXt"), .indexTZ = "",
  tclass = c("POSIXct", "POSIXt"), tzone = "", class = c("xts", "zoo"))
r <- rle(rowSums(x))

如果你想要每组中的最后一个观察结果,你可以使用 cumsum(r$lengths)

R> x[cumsum(r$lengths),]
                    Bid Ask
1969-12-31 18:00:02  10  11
1969-12-31 18:00:03  11  12
1969-12-31 18:00:04  12  13
1969-12-31 18:00:07  10  11

由于您希望对每组进行第一次观察,因此需要预先准备 r$lengths 1 (您总是需要第一次观察),然后删除 r$长度 . 然后打电话 cumsum() 关于结果。

R> x[cumsum(c(1, head(r$lengths, -1))),]
                    Bid Ask
1969-12-31 18:00:01  10  11
1969-12-31 18:00:03  11  12
1969-12-31 18:00:04  12  13
1969-12-31 18:00:05  10  11

很好地抓住了 rowSums() . 稳健的解决方案是 diff() bids和ASK并选择其中一行不为零的行。

d <- diff(x) != 0           # rows with price changes
d[1,] <- TRUE               # always select first observation
g <- cumsum(d$Bid | d$Ask)  # groups of repeats
r <- rle(as.numeric(g))     # run length encoding on groups

# now use the solution above
x[cumsum(c(1, head(r$lengths, -1))),]