以分钟为单位的时滞(R)
我每5米采集一次数据,我想估计它们之间的共因失效系数, 为了做到这一点,我对时间序列x和y进行了预白处理,然后应用CCF 但是滞后与我的采样频率没有直接关系, 我怎么知道他们的关系?例如,滞后-1000分钟的关系是什么以分钟为单位的时滞(R),r,correlation,lag,R,Correlation,Lag,我每5米采集一次数据,我想估计它们之间的共因失效系数, 为了做到这一点,我对时间序列x和y进行了预白处理,然后应用CCF 但是滞后与我的采样频率没有直接关系, 我怎么知道他们的关系?例如,滞后-1000分钟的关系是什么 row.names fecha y x 1 53542 2012-10-20 00:00:00 1.1055 1.0219 2 53543 2012-10-20 00:05:00 1.1059 1.0164 3 53544 2012
row.names fecha y x
1 53542 2012-10-20 00:00:00 1.1055 1.0219
2 53543 2012-10-20 00:05:00 1.1059 1.0164
3 53544 2012-10-20 00:10:00 1.1003 1.0201
4 53545 2012-10-20 00:15:00 1.1015 1.0441
5 53546 2012-10-20 00:20:00 1.1090 1.0224
6 53547 2012-10-20 00:30:00 1.1136 0.9965
7 53548 2012-10-20 00:35:00 1.1140 1.0017
8 53549 2012-10-20 00:40:00 1.1118 1.0379
9 53550 2012-10-20 00:45:00 1.1141 1.0234
10 53551 2012-10-20 00:50:00 1.1128 1.0092
11 53552 2012-10-20 00:55:00 1.1123 1.0232
12 53553 2012-10-20 01:00:00 1.1167 1.0656
13 53554 2012-10-20 01:05:00 1.1195 1.0266
14 53555 2012-10-20 01:10:00 1.1205 1.0142
15 53556 2012-10-20 01:15:00 1.1122 1.0486
16 53557 2012-10-20 01:20:00 1.1168 1.0196
17 53558 2012-10-20 01:30:00 1.1116 1.0070
18 53559 2012-10-20 01:35:00 1.1117 0.9783
19 53560 2012-10-20 01:40:00 1.1170 0.9589
20 53561 2012-10-20 01:45:00 1.1198 1.0070
21 53562 2012-10-20 01:50:00 1.1126 0.9907
22 53563 2012-10-20 01:55:00 1.1190 1.0191
23 53564 2012-10-20 02:00:00 1.1162 0.9625
24 53565 2012-10-20 02:05:00 1.1206 0.9517
25 53566 2012-10-20 02:10:00 1.1237 0.9545
26 53567 2012-10-20 02:15:00 1.1168 0.9894
27 53568 2012-10-20 02:20:00 1.1225 0.9832
28 53569 2012-10-20 02:25:00 1.1239 1.0424
29 53570 2012-10-20 02:30:00 1.1272 1.0140
30 53571 2012-10-20 02:35:00 1.1304 1.0103
library(forecast)
fit <- Arima(x,order=c(2,0,2))
yfiltered <- residuals(Arima(y,model=fit))
ccf(residuals(fit),yfiltered)
row.name fecha y x
1 53542 2012-10-20 00:00:00 1.1055 1.0219
2 53543 2012-10-20 00:05:00 1.1059 1.0164
3 53544 2012-10-20 00:10:00 1.1003 1.0201
4 53545 2012-10-20 00:15:00 1.1015 1.0441
5 53546 2012-10-20 00:20:00 1.1090 1.0224
6 53547 2012-10-20 00:30:00 1.1136 0.9965
7 53548 2012-10-20 00:35:00 1.1140 1.0017
8 53549 2012-10-20 00:40:00 1.1118 1.0379
9 53550 2012-10-20 00:45:00 1.1141 1.0234
10 53551 2012-10-20 00:50:00 1.1128 1.0092
11 53552 2012-10-20 00:55:00 1.1123 1.0232
12 53553 2012-10-20 01:00:00 1.1167 1.0656
13 53554 2012-10-20 01:05:00 1.1195 1.0266
14 53555 2012-10-20 01:10:00 1.1205 1.0142
15 53556 2012-10-20 01:15:00 1.1122 1.0486
16 53557 2012-10-20 01:20:00 1.1168 1.0196
17 53558 2012-10-20 01:30:00 1.1116 1.0070
18 53559 2012-10-20 01:35:00 1.1117 0.9783
19 53560 2012-10-20 01:40:00 1.1170 0.9589
20 53561 2012-10-20 01:45:00 1.1198 1.0070
21 53562 2012-10-20 01:50:00 1.1126 0.9907
22 53563 2012-10-20 01:55:00 1.1190 1.0191
23 53564 2012-10-20 02:00:00 1.1162 0.9625
24 53565 2012-10-20 02:05:00 1.1206 0.9517
25 53566 2012-10-20 02:10:00 1.1237 0.9545
26 53567 2012-10-20 02:15:00 1.1168 0.9894
27 53568 2012-10-20 02:20:00 1.1225 0.9832
28 53569 2012-10-20 02:25:00 1.1239 1.0424
29 53570 2012-10-20 02:30:00 1.1272 1.0140
30 53571 2012-10-20 02:35:00 1.1304 1.0103
图书馆(预测)
fit滞后绝对以秒为单位,并且与您的观察结果一致。这只是个阴谋问题。可以在打印前将滞后转换为观察数:
ccf.out=ccf(residuals(fit),yfiltered, plot=F)
ccf.out$lag=ccf.out$lag*frequency(x)
plot(ccf.out)
滞后绝对以秒为单位,并与您的观察结果一致。这只是个阴谋问题。可以在打印前将滞后转换为观察数:
ccf.out=ccf(residuals(fit),yfiltered, plot=F)
ccf.out$lag=ccf.out$lag*frequency(x)
plot(ccf.out)
我认为滞后-3000是秒,因为频率是1/300=1/(5*60),所以我的滞后可能是秒,但我认为滞后-3000是秒,因为频率是1/300=1/(5*60),所以我的滞后可能是秒,但我认为滞后-3000是秒,因为频率是1/300=1/(5*60),所以我的滞后可能是秒,但