r - 将两个 xts 时间序列合并到一个流中

Question

给定这 2 个 xts 对象:

> X
2012-02-01 09:31:40 2012-02-01 09:31:50 2012-02-01 09:33:30 
         -3.8993007          -2.0860621          -0.6308867
> Y
2012-02-01 09:32:00 2012-02-01 09:32:10 2012-02-01 09:32:20
          1.1983066          -0.2071149           0.3887806

如何按时间连接它们并拥有索引为 9:31:40、9:31:50、9:32:00、9:32:10、9:32:20、9: 的单个列对象: 33:30？也就是说，我需要在单个序列中流式传输两个对象(而不是在组合矩阵中合并两列 X/Y)。

Answer 1

> X <- xts(rnorm(1:10), Sys.time() + 1:10)
> Y <- xts(rnorm(1:10), Sys.time() - 10:1)
> rbind(X, Y)
                          [,1]
2012-05-15 13:07:25  1.1022975
2012-05-15 13:07:26 -0.4755931
2012-05-15 13:07:27 -0.7094400
2012-05-15 13:07:28 -0.5012581
2012-05-15 13:07:29 -1.6290935
2012-05-15 13:07:30 -1.1676193
2012-05-15 13:07:31 -2.1800396
2012-05-15 13:07:32 -1.3409932
2012-05-15 13:07:33 -0.2942939
2012-05-15 13:07:34 -0.4658975
2012-05-15 13:07:36  0.1340882
2012-05-15 13:07:37 -0.4906859
2012-05-15 13:07:38 -0.4405479
2012-05-15 13:07:39  0.4595894
2012-05-15 13:07:40 -0.6937202
2012-05-15 13:07:41 -1.4482049
2012-05-15 13:07:42  0.5747557
2012-05-15 13:07:43 -1.0236557
2012-05-15 13:07:44 -0.0151383
2012-05-15 13:07:45 -0.9359486