添加人工断点

我有以下的时间序列：添加人工断点

Lines <- "Hour,PF 
0,14/01/2015 00:00,0.305 
1,14/01/2015 01:00,0.306 
2,14/01/2015 02:00,0.307 
3,14/01/2015 03:00,0.3081 
4,14/01/2015 04:00,0.3091 
5,14/01/2015 05:00,0.3101 
6,14/01/2015 06:00,0.3111 
7,14/01/2015 07:00,0.3122 
8,14/01/2015 08:00,0.455 
9,14/01/2015 09:00,0.7103 
10,14/01/2015 10:00,0.9656 
11,14/01/2015 11:00,1 
12,14/01/2015 12:00,0.9738 
13,14/01/2015 13:00,0.9476 
14,14/01/2015 14:00,0.9213 
15,14/01/2015 15:00,0.8951 
16,14/01/2015 16:00,0.8689 
17,14/01/2015 17:00,0.8427 
18,14/01/2015 18:00,0.6956 
19,14/01/2015 19:00,0.6006 
20,14/01/2015 20:00,0.5056 
21,14/01/2015 21:00,0.4106 
22,14/01/2015 22:00,0.3157 
23,14/01/2015 23:00,0.3157" 

library (zoo) 
library (strucchange) 

z <- read.zoo(text = Lines, tz = "", format = "%d/%m/%Y %H:%M", sep = ",") 

bp <- breakpoints(z ~ 1, h = 2) 

plot(z) 
abline(v = time(z)[bp$breakpoints]) 

The Breakpoints are at observation number: 
8 10 18 21

现在假设我有相同的时间序列，但在观察数11 - 24小时营业，下落不明。我想定义Gap（在这种情况下= 24小时），并找到与前面示例中相关的断点，并在间隙开始和结束时添加2个附加断点。对于下面的时间序列中的断点将是：

8 10 12 13 18 21

这里是时间序列的差距：

Lines <- "Hour,PF 
    0,14/01/2015 00:00,0.305 
    1,14/01/2015 01:00,0.306 
    2,14/01/2015 02:00,0.307 
    3,14/01/2015 03:00,0.3081 
    4,14/01/2015 04:00,0.3091 
    5,14/01/2015 05:00,0.3101 
    6,14/01/2015 06:00,0.3111 
    7,14/01/2015 07:00,0.3122 
    8,14/01/2015 08:00,0.455 
    9,14/01/2015 09:00,0.7103 
    10,14/01/2015 10:00,0.9656 
    11,14/01/2015 11:00,1   <--- 
    12,15/01/2015 12:00,0.9738  <--- GAP of 24 hours 
    13,15/01/2015 13:00,0.9476 
    14,15/01/2015 14:00,0.9213 
    15,15/01/2015 15:00,0.8951 
    16,15/01/2015 16:00,0.8689 
    17,15/01/2015 17:00,0.8427 
    18,15/01/2015 18:00,0.6956 
    19,15/01/2015 19:00,0.6006 
    20,15/01/2015 20:00,0.5056 
    21,15/01/2015 21:00,0.4106 
    22,15/01/2015 22:00,0.3157 
    23,15/01/2015 23:00,0.3157"

来源

2016-02-04 Avi

你的差距实际上是一整月（14/01至14/02）。

c(0,1)+which.max(diff(time(z))) # 12,13

将返回您可以追加到您的折点的最大差距的指数。

注意：所述间隙的开始和结束是不12,13 11,12自中的R指数从1

来源

2016-02-04 19:19:24 fishtank

由于@fishtank启动时，该间隙仅仅是一个例子可能有一些多个间隙更大或小。我想获得每个差距的折点，然后我想找到每个段的拟合（线性回归）并跳过这些差距。找到最大差异将不是解决方案。 – Avi

我不认为'断点'是用来识别“空白”的。所以这个例子的要点是你可以使用'diff（time（z））'来查找和评估（根据一些标准，例如超过24小时）你认为的“差距”。 – fishtank

非常感谢。如果我想为缺口添加轮廓（在跳动和结束时）有没有办法做到这一点？我使用abline（v = time（z）[which.max（diff（time（z）））]）但我不想使用which.max？我该怎么做？ – Avi

添加人工断点

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