Finding Anchors and slope for segmented time series

Question

I have the following time series:

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$breakpoints])
fit <- zoo(fitted(bp), time(z))
lines(fit, col = "blue", lty = 2, lwd = 2)
levs <- fit[bp$breakpoints + 0:1]
a<-diff(levs) / diff(as.numeric(time(levs)) / 3600)
DF <- fortify.zoo(a)

I get the following DF:

> DF
                Index             a
1 2015-01-14 10:00:00  2.061000e-01
2 2015-01-14 17:00:00 -9.516197e-17
3 2015-01-14 21:00:00 -1.448854e-01

I tried to change the formula in breakpoints to get a linear model with slope and intercept:

bp <- breakpoints(z ~ Lines$PF, h = 2)

with no success. The final result I would like to have is the beginning of the segment and the end of the segment, the slope (a) as is now and the Intercept , left point of a segment (Anchor) and the right point of the segment. As followed (only example with no connection to real numbers):

> DF
    Start Segment       End Segment             Slope          Intercept   Anchor Beginning Anchor End
1 2015-01-14 10:00:00  2015-01-14 08:00:00     2.061000e-01    8.123            0.50        0.30
2 2015-01-14 08:00:00  2015-01-14 17:00:00    -9.516197e-17    9.456            0.70        0.40
3 2015-01-14 17:00:00  2015-01-14 23:00:00    -1.448854e-01    2.9009           0.60        0.90

Note that strucchange's model does not constrain the segments to be connected. Also when posting to SO please reference prior questions if your question is a follow up. — G. Grothendieck, Jan 30 '16 at 20:02
Thanks, is there a way to constrain the segments to be connected? — Avi, Jan 30 '16 at 20:12

score 1 · Accepted Answer · answered Jan 30 '16 at 20:00

1

Well, you have it at your hands with your breakpoints, e.g.

(breaks <- data.frame(
  start = index(z[c(1, bp$breakpoints+1)]),
  end = c(index(z[bp$breakpoints]), index(z[length(z)]))
))
#                 start                 end
# 1 2015-01-14 00:00:00 2015-01-14 07:00:00
# 2 2015-01-14 08:00:00 2015-01-14 09:00:00
# 3 2015-01-14 10:00:00 2015-01-14 17:00:00
# 4 2015-01-14 18:00:00 2015-01-14 20:00:00
# 5 2015-01-14 21:00:00 2015-01-14 23:00:00
fits <- lapply(seq_len(nrow(breaks)), function(x) {
  idx <- index(z)>=breaks[x, 1] & index(z)<=breaks[x, 2]
  fit <- lm(z[idx]~index(z[idx]))
})
sapply(fits, coefficients)
#                        [,1]          [,2]          [,3]          [,4]          [,5]
# (Intercept)   -4.048094e+02 -1.007876e+05  8.358223e+03  3.750603e+04  1.873346e+04
# index(z[idx])  2.850529e-07  7.091667e-05 -5.880291e-06 -2.638889e-05 -1.318056e-05

The last step would be to merge together all data your need in the format you want.

answered Jan 30 '16 at 20:00

lukeA

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5
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Thanks a lot @lukeA! I would like to ask why in the breaks is there discontinuity between the end of the previous segment to the start of the next one? And additional question - Is index(z[idx]) is the left Anchor for the specific segment? – Avi Jan 30 '16 at 20:11
If you don't want the discontinuity, remove the `+1` from `bp$breakpoints+1`. `index(z[idx])` is the x value for the segment, for which the regression line is calculated. – lukeA Jan 30 '16 at 20:17
Thanks a lot, If I would like to find the real value of the left point of the segment (the real value) and the right as well. Is there a way to get it? – Avi Jan 30 '16 at 20:27
E.g. `z[breaks$start]` (values) and `index(z[breaks$start])` (indices) for the lefts – lukeA Jan 30 '16 at 20:29
Thanks again, so the right hand value is the next left hand value? And if I would like to plot the fitted line what do I have to change in lines(fit, col = "blue", lty = 2, lwd = 2)? – Avi Jan 30 '16 at 20:34
The right is `z[breaks$end]`, and plotting is `lapply(fits, abline, col="blue")` – lukeA Jan 30 '16 at 20:36
Thanks a again, lapply(fits, abline, col="blue") plots along the model I would to plot per segment? – Avi Jan 30 '16 at 20:40
And another question, how can I get the slope for each segment? – Avi Jan 30 '16 at 22:15
`index(z[idx])` is the slope. You can draw segments using the [`segments`](https://stat.ethz.ch/R-manual/R-devel/library/graphics/html/segments.html) function. You just need to arrange the values (from x,y and to x,y). – lukeA Jan 31 '16 at 16:06
Thank you so much, If I use real_slope = ((as.numeric(z[breaks$end]) - as.numeric(z[breaks$start])) ) / (as.numeric(time ((z[breaks$end])) - time ((z[breaks$start]))) ) to calculate the real slope why is there a difference at segment 3 (9:00 to 17:00) in the direction? (i.e -1.055556e-07 vs 0.016550000) – Avi Jan 31 '16 at 16:46

Finding Anchors and slope for segmented time series

1 Answers1