为什么我不能直接打包xts？

Question

> x <- data.frame(a = rnorm(10), b = rnorm(10), c = rnorm(10))
> x
             a          b            c
1  -1.09651022 -0.7416278  0.209405373
2   1.53644398 -0.9463432  0.374955227
3   1.71132675 -0.3828052  2.024143398
4  -1.10622882 -0.3599187 -0.808780103
5  -0.49616562  0.7061180  0.644142118
6  -1.75452442  0.3890812 -0.623815889
7   0.06315648  0.5103820 -1.501873998
8   0.64856129 -1.0973679  1.432024595
9  -0.62828873 -0.3159317  0.183674189
10 -0.82657934  1.6376569 -0.003601196
> rownames(x) <- as.Date(Sys.Date() - 9:0)
> x
                     a          b            c
2013-06-07 -1.09651022 -0.7416278  0.209405373
2013-06-08  1.53644398 -0.9463432  0.374955227
2013-06-09  1.71132675 -0.3828052  2.024143398
2013-06-10 -1.10622882 -0.3599187 -0.808780103
2013-06-11 -0.49616562  0.7061180  0.644142118
2013-06-12 -1.75452442  0.3890812 -0.623815889
2013-06-13  0.06315648  0.5103820 -1.501873998
2013-06-14  0.64856129 -1.0973679  1.432024595
2013-06-15 -0.62828873 -0.3159317  0.183674189
2013-06-16 -0.82657934  1.6376569 -0.003601196
> class(x)
[1] "data.frame"
> boxplot(x)

> xx <- as.xts(x, order.by = as.Date(rownames(x)))
> xx
                     a          b            c
2013-06-07 -1.09651022 -0.7416278  0.209405373
2013-06-08  1.53644398 -0.9463432  0.374955227
2013-06-09  1.71132675 -0.3828052  2.024143398
2013-06-10 -1.10622882 -0.3599187 -0.808780103
2013-06-11 -0.49616562  0.7061180  0.644142118
2013-06-12 -1.75452442  0.3890812 -0.623815889
2013-06-13  0.06315648  0.5103820 -1.501873998
2013-06-14  0.64856129 -1.0973679  1.432024595
2013-06-15 -0.62828873 -0.3159317  0.183674189
2013-06-16 -0.82657934  1.6376569 -0.003601196
> class(xx)
[1] "xts" "zoo"
> boxplot(xx)
Error in try.xts(c(2.12199579096527e-314, 2.12199579096527e-314, 0, 2.12199579096527e-314,  : 
  Error in xts(coredata(x), order.by = index(x), .CLASS = "xts", ...) :   order.by requires an appropriate time-based object
> 

当然......

> boxplot(as.matrix(xx), col = "red")

这不是一个大问题，但我不想每次都强迫我的xts，或者更一般地说，我想知道这里发生了什么，以防有可能发现的其他问题。

Answer 1

xts对象是矩阵加索引，因此您不能将对象传递给boxplot()。如果你看str(xx)：

# An ‘xts’ object from 2013-06-07 to 2013-06-16 containing:
#   Data: num [1:10, 1:3] 0.321 -0.462 1.715 0.575 0.83 ...
# - attr(*, "dimnames")=List of 2
# ..$ : NULL
# ..$ : chr [1:3] "a" "b" "c"
# Indexed by objects of class: [Date] TZ: 
#   xts Attributes:  
#   NULL

您可以通过coredata()：

访问矩阵

coredata(xx)
#                a           b            c
# [1,]  0.32120813 -0.07370657 -0.601288169
# [2,] -0.46154742 -1.09475940  3.028664653
# [3,]  1.71515544 -0.31000454 -0.009281175
# [4,]  0.57482616 -0.06260991  1.198034802
# [5,]  0.83015688 -2.49614565 -1.689812377
# [6,]  0.01748081 -0.55332675  2.391426111
# [7,]  0.69852800 -0.10337251 -0.267939285
# [8,]  0.75139113 -0.17427970 -0.561434122
# [9,] -0.68942598  0.18685817 -1.508917140
# [10,] -0.76381007 -2.44387628  0.290524821

您可以直接使用，但是您可以使用矩阵：

boxplot(coredata(xx))

如果您想使用xts对象的索引，只需使用index()：

index(xx)
# [1] "2013-06-07" "2013-06-08" "2013-06-09" "2013-06-10" "2013-06-11" "2013-06-12" "2013-06-13" "2013-06-14" "2013-06-15"
# [10] "2013-06-16"

Answer 2

因为没有boxplot.xts功能。写它，神奇地你可以做boxplot（anxtsobject）。如果xts的作者喜欢它，它甚至可能进入包中。哦，如果xts是S3类，那可能只是真的......

你能做hist（anxtsobject）吗？

Answer 3

您可以使用PerformanceAnalytics这样的包，例如：

library(PerformanceAnalytics)
chart.Boxplot(xx)

在内部，xts对象被强制转换为data.frame，然后可以调用boxplot：

library(PerformanceAnalytics)
R <- checkData(xx, method = "data.frame")
boxplot(R)