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xts 对象中是否有一种方法可以与以下相同，但对于具有多天盘中数据的 xts 对象？下面的工作就像一个时钟，但一天的数据。如果我将 xts 从 22 日传递到 26 日，则不会。似乎不可能一次性在 xts 中对多天的日内数据进行子集化，而是需要首先每天拆分数据，然后使用此 xts 功能。这个对吗？

indexTZ(tdata) = "GMT"
plotdata= tdata['20110822 10:00:00/20110822 12:00:00']

> plotdata= tdata['10:00:00/12:00:00'] works like swiss clock


> tail(plotdata)
                SYMBOL EX  PRICE              SIZE    COND CORR G127
2011-08-22 11:59:36 "BHP"  "N" "38370"            "185"   "1"  "0"  "0" 
2011-08-22 11:59:37 "BHP"  "N" "38380"            "314"   "1"  "0"  "0" 
2011-08-22 11:59:40 "BHP"  "N" "38390"            "675"   "1"  "0"  "0" 
2011-08-22 11:59:42 "BHP"  "N" "38390"            "119"   "1"  "0"  "0" 
2011-08-22 11:59:43 "BHP"  "N" "38390"            "359"   "1"  "0"  "0" 
2011-08-22 11:59:44 "BHP"  "N" "38400.7702920749" "42763" "1"  "0"  "0" 

 plotdata= tdata['10:00:00/12:00:00'] or 
 plotdata= tdata['20110822 10:00:00/20110826 12:00:00'] does not return expected xts 

tail(plotdata)
                SYMBOL EX  PRICE              SIZE   COND CORR G127
2011-08-26 15:59:53 "BHP"  "N" "38710"            "119"  "1"  "0"  "0" 
2011-08-26 15:59:55 "BHP"  "N" "38700"            "1004" "1"  "0"  "0" 
2011-08-26 15:59:56 "BHP"  "N" "38700.4838709677" "310"  "1"  "0"  "0" 
2011-08-26 15:59:57 "BHP"  "N" "38710"            "6"    "1"  "0"  "0" 
2011-08-26 15:59:58 "BHP"  "N" "38710"            "2"    "1"  "0"  "0" 
2011-08-26 15:59:59 "BHP"  "N" "38710"            "121"  "9"  "0"  "0"
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1 回答 1

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xts有时间子集：xts 刻度数据滚动子集

tdata['T10:00:00/T12:00:00']应该做。

于 2011-08-30T06:04:15.133 回答