R中基于时间的复杂子集划分

我有很多时间数据（YYYY/MM/DD HH:MM:SS.SSS）以不规则的数千秒间隔存储。在每个时间段有十个空间测量值（X、Y和Z值）

我希望得到的是数据的一个子集，例如每半秒（或几分之一秒）进行一组十次空间测量

我是R的新手，所以非常感谢您的帮助

以下是2次测量的示例：

2012/09/21 14:59:07:712,A,0.036,0.224,0.814
2012/09/21 14:59:07:712,B,0.042,0.057,0.934
2012/09/21 14:59:07:712,C,-0.104,0.008,0.930
2012/09/21 14:59:07:712,D,0.158,0.001,0.914
2012/09/21 14:59:07:712,E,-0.208,-0.168,0.778
2012/09/21 14:59:07:712,F,-0.185,0.087,0.748
2012/09/21 14:59:07:712,G,-0.176,0.155,0.738
2012/09/21 14:59:07:712,H,0.236,-0.171,0.790
2012/09/21 14:59:07:712,I,0.244,0.076,0.732
2012/09/21 14:59:07:712,J,0.248,0.137, 0.722
2012/09/21 14:59:07:848,A,0.036,0.224,0.814
2012/09/21 14:59:07:848,B,0.042,0.057,0.934
2012/09/21 14:59:07:848,C,-0.104,0.008,0.930
2012/09/21 14:59:07:848,D,0.158,0.001,0.914
2012/09/21 14:59:07:848,E,-0.208,-0.168,0.778
2012/09/21 14:59:07:848,F,-0.185,0.087,0.748
2012/09/21 14:59:07:848,G,-0.176,0.155,0.738
2012/09/21 14:59:07:848,H,0.236,-0.171,0.790
2012/09/21 14:59:07:848,I,0.244,0.076,0.732
2012/09/21 14:59:07:848,J,0.248,0.137, 0.722

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你想做什么还不清楚。您可以从读取数据开始。因为它是不规则的时间序列，并且包含一个因子变量（第一组），所以不能使用像

zoo

或

xts

这样方便的包，因为它们需要一个矩阵作为结构。但您可以从

数据中使用
fread
。表

包：

library(data.table)
dat <- fread('2012/09/21 14:59:07:712,A,0.036,0.224,0.814
2012/09/21 14:59:07:712,B,0.042,0.057,0.934
2012/09/21 14:59:07:712,C,-0.104,0.008,0.930
2012/09/21 14:59:07:712,D,0.158,0.001,0.914
2012/09/21 14:59:07:712,E,-0.208,-0.168,0.778
2012/09/21 14:59:07:712,F,-0.185,0.087,0.748
2012/09/21 14:59:07:712,G,-0.176,0.155,0.738
2012/09/21 14:59:07:712,H,0.236,-0.171,0.790
2012/09/21 14:59:07:712,I,0.244,0.076,0.732
2012/09/21 14:59:07:712,J,0.248,0.137, 0.722
2012/09/21 14:59:07:848,A,0.036,0.224,0.814
2012/09/21 14:59:07:848,B,0.042,0.057,0.934
2012/09/21 14:59:07:848,C,-0.104,0.008,0.930
2012/09/21 14:59:07:848,D,0.158,0.001,0.914
2012/09/21 14:59:07:848,E,-0.208,-0.168,0.778
2012/09/21 14:59:07:848,F,-0.185,0.087,0.748
2012/09/21 14:59:07:848,G,-0.176,0.155,0.738
2012/09/21 14:59:07:848,H,0.236,-0.171,0.790
2012/09/21 14:59:07:848,I,0.244,0.076,0.732
2012/09/21 14:59:07:848,J,0.248,0.137, 0.722',header=FALSE)

---

下面是我提出的解决方案，它能够解决这个问题（唯一的缺点是它无法以1秒的间隔创建移动平均线）：

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数据\u ID\u P001欢迎使用堆栈溢出！请提供更好的样本数据或可复制的示例，以便这里的优秀人员能够更好地帮助您。请参阅，其中还包括您迄今为止尝试过的内容。这不是免费完成工作的地方。又来了。让我试着更好地解释一下我想做什么。因此，我可以毫无问题地将数据加载到R中（我使用的是推断器）。我还可以将第一列从字符转换为时间。我遇到的问题是如何根据时间选择数据的子集。例如，假设我想每十分之一秒制作每组10个样本的第一个实例的子集（即，a-J，10行都是同时测量的）。这更有意义吗？我猜我需要使用某种类型的循环结构。或者是韦翰的plyr套餐。我知道subset函数允许您根据行中的特定值创建子集，但我不知道如何调整条件语句，以便它可以选择移动时间窗口中的第一个匹配项（例如，半秒内的第一个匹配项）。

 dat[V2 %in% LETTERS[1:5],]
                         V1 V2     V3     V4    V5
 1: 2012/09/21 14:59:07:712  A  0.036  0.224 0.814
 2: 2012/09/21 14:59:07:712  B  0.042  0.057 0.934
 3: 2012/09/21 14:59:07:712  C -0.104  0.008 0.930
 4: 2012/09/21 14:59:07:712  D  0.158  0.001 0.914
 5: 2012/09/21 14:59:07:712  E -0.208 -0.168 0.778
 6: 2012/09/21 14:59:07:848  A  0.036  0.224 0.814
 7: 2012/09/21 14:59:07:848  B  0.042  0.057 0.934
 8: 2012/09/21 14:59:07:848  C -0.104  0.008 0.930
 9: 2012/09/21 14:59:07:848  D  0.158  0.001 0.914
10: 2012/09/21 14:59:07:848  E -0.208 -0.168 0.778