按一天中的时间子集 xts 对象 [英] Subset xts object by time of day

问题描述

一个简单的问题:我知道如何从帮助中将 xts 中的时间序列子集化为年、月和日:x['2000-05/2001']等等.

A simple question: I know how to subset time series in xts for years, months and days from the help: x['2000-05/2001'] and so on.

但是我如何按一天中的小时数来划分我的数据?我想在 07:00 am 和 06:00 pm 之间获取所有数据.即，我想在工作时间提取数据 - 与当天无关(稍后我会照顾周末).帮助有一个表单示例:

But how can I subset my data by hours of the day? I would like to get all data between 07:00 am and 06:00 pm. I.e., I want to extract the data during business time - irrelevant of the day (I take care for weekends later on). Help has an example of the form:

.parseISO8601('T08:30/T15:00')

但这在我的情况下不起作用.有人知道吗?

But this does not work in my case. Does anybody have a clue?

推荐答案

如果你的 xts 对象被称为 x 那么类似于 y <- x[例如，T09:30/T11:00"] 对我有用，可以从早上的会议中分一杯羹.

If your xts object is called x then something like y <- x["T09:30/T11:00"] works for me to get a slice of the morning session, for example.

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