按索引合并多个 XTS 对象答案

【问题标题】：Merging multiple XTS objects by index按索引合并多个 XTS 对象
【发布时间】：2017-12-08 05:47:38
【问题描述】：

有多个 XTS 对象。每一个仅包含一列（价格）。对象可能具有不同的长度。我想按索引合并它们。我可以将两个 XTS 对象与merge.xts 合并。但我想合并多个（一个功能，无需多次重复合并过程）。

所以我采用了推荐的方法here。有用。只有一个问题：它将所有数据合并到一列，而不是将每个 XTS 对象合并到自己的列中。例如在可重现的示例中，期望的最终结果是 XTSM 有 3 列。

如何实现？谢谢。

library('xts')

#Data
XTS1 <- structure(c(125.26, 125.14, 125.21, 125.26, 125.21, 125.26, 125.24, 125.22, 125.3, 125.32, 125.38, 125.38, 125.29, 125.26, 125.3, 125.25, 125.29, 125.37, 125.21, 125.3, 125.25, 125.24, 125.21, 125.24, 125.24, 125.26, 125.23, 125.23), class = c("xts", "zoo"), .indexCLASS = c("POSIXct", "POSIXt"), tclass = c("POSIXct", "POSIXt"), .indexTZ = "", tzone = "", .CLASS = "double", index = structure(c(1403537400, 1403538300, 1403539200, 1403540100, 1403541000, 1403541900, 1403542800, 1403543700, 1403544600, 1403545500, 1403546400, 1403547300, 1403548200, 1403549100, 1403550000, 1403550900, 1403551800, 1403552700, 1403553600, 1403554500, 1403555400, 1403556300, 1403557200, 1403560800, 1403561700, 1403562600, 1403563500, 1403564400), tzone = "", tclass = c("POSIXct", "POSIXt")), .Dim = c(28L, 1L))
XTS2 <- structure(c(1867, 1867, 1868.5, 1868.75, 1868.75, 1868.75, 1868.25, 1867.25, 1867.5, 1867.75, 1868, 1868.25, 1868.25, 1868.75, 1868, 1868.5, 1869, 1869.75, 1868.75, 1868.5, 1868.25, 1867.5, 1867, 1866.75, 1866.75, 1867, 1867.5), class = c("xts", "zoo"), .indexCLASS = c("POSIXct", "POSIXt"), tclass = c("POSIXct", "POSIXt"), .indexTZ = "", tzone = "", .CLASS = "double", index = structure(c(1403537400, 1403538300, 1403539200, 1403540100, 1403541000, 1403541900, 1403542800, 1403543700, 1403544600, 1403545500, 1403546400, 1403547300, 1403548200, 1403549100, 1403550000, 1403550900, 1403551800, 1403552700, 1403553600, 1403555400, 1403556300, 1403557200, 1403560800, 1403561700, 1403562600, 1403563500, 1403564400), tzone = "", tclass = c("POSIXct", "POSIXt")), .Dim = c(27L, 1L))
XTS3 <- structure(c(1867, 1867, 1868.5, 1868.75, 1868.75, 1868.75, 1868.25, 1867.25, 1867.5, 1867.75, 1868, 1868.25, 1868.25, 1868.75, 1868, 1868.5, 1869, 1869.75, 1868.75, 1868.5, 1868.25, 1867.5, 1867, 1866.75, 1866.75, 1867, 1867.5), class = c("xts", "zoo"), .indexCLASS = c("POSIXct", "POSIXt"), tclass = c("POSIXct", "POSIXt"), .indexTZ = "", tzone = "", .CLASS = "double", index = structure(c(1403537400, 1403538300, 1403539200, 1403540100, 1403541000, 1403541900, 1403542800, 1403543700, 1403544600, 1403545500, 1403546400, 1403547300, 1403548200, 1403549100, 1403550000, 1403550900, 1403551800, 1403552700, 1403553600, 1403555400, 1403556300, 1403557200, 1403560800, 1403561700, 1403562600, 1403563500, 1403564400), tzone = "", tclass = c("POSIXct", "POSIXt")), .Dim = c(27L, 1L))

#Following code merges 2 xts objects into a 2 column xts
XTSM <- merge(XTS1,XTS2, all=TRUE)

#Following code merges multiple xts objects into a 1 column xts
XTSlist <- list(XTS1, XTS2, XTS3)

do.call.rbind <- function(lst) {
  while(length(lst) > 1) {
    idxlst <- seq(from=1, to=length(lst), by=2)
    lst <- lapply(idxlst, function(i) {
      if(i==length(lst)) { return(lst[[i]]) }
      return(rbind(lst[[i]], lst[[i+1]]))
    })
  }
  lst[[1]]
}

XTSM <- do.call.rbind (XTSlist)

【问题讨论】：

merge.xts 接受超过 2 个 xts 对象。见?merge.xts
确实如此。这解决了这个问题。出于某种奇怪的原因，当我在收到错误消息之前尝试使用 merge.xts 尝试多个对象时。谢谢。
请注意，如果join 不是"all"，merge.xts 只接受两个参数。另请注意，您只需致电merge()。直接调用方法不是一个好习惯。没有什么可以阻止您拨打电话，例如merge.xts(data.frame(...))，这可能会产生意想不到的结果。
mege.zoo 在所有情况下都接受多个参数。

标签： r list merge xts

【解决方案1】：

约书亚回答 cmets 中的问题。为了回答 Q 而发布解决方案。就这么简单：

XTSM.T <- merge(XTS1,XTS2,XTS3, all=TRUE)

or

XTSM.F <- merge(XTS1,XTS2,XTS3, all=FALSE)

【讨论】：