如何合并具有稍微不同列的 xts 对象？

Question

给定各种单行 xts 对象：

z1 = xts(t(c("9902"=0,"9903"=0,"9904"=0,"9905"=2,"9906"=2)),as.Date("2015-01-01"))
z2 = xts(t(c("9902"=3,"9903"=4,"9905"=6,"9906"=5,"9908"=8)),as.Date("2015-01-02"))
z3 = xts(t(c("9901"=1,"9903"=3,"9905"=5,"9906"=6,"9907"=7,"9909"=9)),as.Date("2015-01-03"))

我想将它们合并为一个 xts 对象。但是 cbind(z1,z2,z3) 给出：

           X9902 X9903 X9904 X9905 X9906 X9902.1 X9903.1 X9905.1 X9906.1 X9908 X9901 X9903.2 X9905.2 X9906.2 X9907 X9909
2015-01-01     0     0     0     2     2      NA      NA      NA      NA    NA    NA      NA      NA      NA    NA    NA
2015-01-02    NA    NA    NA    NA    NA       3       4       6       5     8    NA      NA      NA      NA    NA    NA
2015-01-03    NA    NA    NA    NA    NA      NA      NA      NA      NA    NA     1       3       5       6     7     9

而我期望的是：

           9901 9902 9903 9904 9905 9906 9907 9908 9909
2015-01-01    0    0    0    0    2    2    0    0    0
2015-01-02    0    3    4    0    6    5    0    8    0
2015-01-03    1    0    3    0    5    6    7    0    9

（我可以通过给出 fill=0 将 NA 更改为零，即 cbind(z1,z2,z3,fill=0)。）

rbind(z1,z2,z3) 抱怨行的列数不同。但是，我相信如果事先将缺失的列添加到每个 xts 对象中，这将是一个好方法吗？

真实数据可能有数千行和几百列（合并后），所以我只关注效率。

Answer 1

library(xts)
library(plyr)

z1df <- as.data.frame(z1)
z2df <- as.data.frame(z2)
z3df <- as.data.frame(z3)

res <- rbind.fill(z1df, z2df, z3df)
res[is.na(res)] <- 0
res

#  9902 9903 9904 9905 9906 9908 9901 9907 9909
#1    0    0    0    2    2    0    0    0    0
#2    3    4    0    6    5    8    0    0    0
#3    0    3    0    5    6    0    1    7    9

这类似于下面的Whosebug post

combining two data frames of different lengths

包括日期列

res$Date <- c("2015-01-01", "2015-01-02", "2015-01-03") # the appropriate values
res$Date <- as.Date(res$Date)

并转化为xts对象

xts(res[,-10], order.by=res[,10])

Answer 2

正如我在评论中提到的，merge.xts（和 merge.zoo）仅按索引合并，因此您无法使用 merge（或 cbind).所以看起来您确实需要 rbind，但是（如您所说）它将要求所有对象以相同的顺序具有相同数量的列。

我在下面创建了两个函数来帮助处理对象，因此您可以 rbind 它们来创建您想要的结果。

# put all xts objects in a list for easier processing
x <- list(z1, z2, z3)

# function to create template xts object
template <- function(xlist) {
  # find set of unique column names from all objects
  cn <- unique(unlist(lapply(xlist, colnames)))
  # create template xts object
  # using a date that doesn't occur in the actual data
  minIndex <- do.call(min, lapply(xlist, function(x) index(x[1L,])))
  # template object
  xts(matrix(0,1,length(cn)), minIndex-1, dimnames=list(NULL, sort(cn)))
}

# function to apply to each xts object
proc <- function(x, template) {
  # columns we need to add
  neededCols <- !(colnames(template) %in% colnames(x))
  # merge this object with template object, filling w/zeros
  out <- merge(x, template[,neededCols], fill=0)
  # reorder columns (NB: merge.xts always uses make.names)
  # and remove first row (from template)
  out <- out[-1L,make.names(colnames(template))]
  # set column names back to desired values
  # (using attr<- because dimnames<-.xts copies)
  attr(out, "dimnames") <- list(NULL, colnames(template))
  # return object
  out
}
(res <- do.call(rbind, lapply(x, proc, template=template(x))))
#            9901 9902 9903 9904 9905 9906 9907 9908 9909
# 2015-01-01    0    0    0    0    2    2    0    0    0
# 2015-01-02    0    3    4    0    6    5    0    8    0
# 2015-01-03    1    0    3    0    5    6    7    0    9

如何合并具有稍微不同列的 xts 对象？

How to merge xts objects with slightly different columns?

r

xts