to.period 不能与 xts 格式结合使用答案

【问题标题】：to.period doesn't work combined with xts formatto.period 不能与 xts 格式结合使用
【发布时间】：2021-06-03 13:57:57
【问题描述】：

让我们考虑以下数据框以作为可重现的示例：

df <- data.frame(
  "Date" =
    c(
      "2009-11-02", "2009-11-03", "2009-11-04", "2009-11-05", "2009-11-06",
      "2009-11-09", "2009-11-10", "2009-11-12", "2009-11-13", "2009-11-16",
      "2009-11-17", "2009-11-18", "2009-11-19", "2009-11-20"
    ),
  "Open" = c(
    64.97971, 64.64817, 63.88567, 64.34973, 67.16770, 67.63186,
    69.48868, 68.95794, 70.08527, 72.47256,
    72.53886, 72.73724, 71.07980, 69.75345
  ),
  "High" = c(
    65.47689, 65.14544, 65.44378, 66.96887, 68.75883, 69.62065, 70.81439, 73.26807,
    71.07980, 73.13536,
    73.26807, 72.93625, 71.87532, 72.27345
  ),
  "Low" = c(
    63.98508, 62.75843, 63.71976, 64.34973, 65.47689, 66.96887, 68.36125, 68.95794,
    69.28966, 72.00803,
    72.00803, 71.14620, 69.68705, 69.75345
  ),
  "Close" = c(
    64.64817, 62.85784, 65.21174, 66.96887, 65.70910, 69.62065, 70.81439, 71.94172, 70.61537, 72.53886,
    72.80355, 71.60999, 69.68705, 70.94709
  )
)

这是一些 OHLC 格式的数据框（开盘价、最高价、收盘价、最低价）。

现在让我们把这个数据框改成 xts 对象：

df <- as.xts(df, order.by = as.Date(df[, 1]))

现在我想申请.period函数，例如：

to.period(df, period = "days", k =3)

我得到错误：

'to.period(df, period = "days", k = 3)':unsupported type

我读到了这个错误，它的根源在于xts 对象的定义。因为xts 对象是一个矩阵，每个变量都应该具有相同的类型。问题就在这里，因为例如“打开”列是由数值创建的，并且第一列填充了日期格式的值。这就是as.xts() 将所有内容都转换为字符串的原因，这是最常见的数据格式。但是，即使我知道它为什么不起作用的理由——我也不知道如何让to.period 起作用。你知道如何解决吗？

【问题讨论】：

标签： r dataframe xts

【解决方案1】：

问题是在构造xts时没有删除第一列

df <- as.xts(df[-1], order.by = as.Date(df[, 1]))
to.period(df, period = "days", k =3) 
#            df.Open  df.High   df.Low df.Close
#2009-11-03 64.97971 65.47689 62.75843 62.85784
#2009-11-06 63.88567 68.75883 63.71976 65.70910
#2009-11-09 67.63186 69.62065 66.96887 69.62065
#2009-11-12 69.48868 73.26807 68.36125 71.94172
#2009-11-13 70.08527 71.07980 69.28966 70.61537
#2009-11-18 72.47256 73.26807 71.14620 71.60999
#2009-11-20 71.07980 72.27345 69.68705 70.94709

在不删除第一列即character 类列（'Date'）的情况下，xts 将整个数据转换为character 类，因为它也是matrix 和matrix 只能有单课

str(as.xts(df, order.by = as.Date(df[, 1])))
An ‘xts’ object on 2009-11-02/2009-11-20 containing:
  Data: chr [1:14, 1:5] "2009-11-02" "2009-11-03" "2009-11-04"  "2009-11-05" "2009-11-06" "2009-11-09" "2009-11-10" "2009-11-12" ...
 - attr(*, "dimnames")=List of 2
  ..$ : NULL
  ..$ : chr [1:5] "Date" "Open" "High" "Low" ...
  Indexed by objects of class: [Date] TZ: UTC
  xts Attributes:  
 NULL

【讨论】：