Reputation: 33
This is my first time with strucchange so bear with me. The problem I'm having seems to be that strucchange doesn't recognize my time series correctly but I can't figure out why and haven't found an answer on the boards that deals with this. Here's a reproducible example:
require(strucchange)
# time series
nmreprosuccess <- c(0,0.50,NA,0.,NA,0.5,NA,0.50,0.375,0.53,0.846,0.44,1.0,0.285,
0.75,1,0.4,0.916,1,0.769,0.357)
dat.ts <- ts(nmreprosuccess, frequency=1, start=c(1996,1))
str(dat.ts)
Time-Series [1:21] from 1996 to 2016: 0 0.5 NA 0 NA 0.5 NA 0.5 0.375 0.53 ...
To me this means that the time series looks OK to work with.
# obtain breakpoints
bp.NMSuccess <- breakpoints(dat.ts~1)
summary(bp.NMSuccess)
Which gives:
Optimal (m+1)-segment partition:
Call:
breakpoints.formula(formula = dat.ts ~ 1)
Breakpoints at observation number:
m = 1 6
m = 2 3 7
m = 3 3 14 16
m = 4 3 7 14 16
m = 5 3 7 10 14 16
m = 6 3 7 10 12 14 16
m = 7 3 5 7 10 12 14 16
Corresponding to breakdates:
m = 1 0.333333333333333
m = 2 0.166666666666667 0.388888888888889
m = 3 0.166666666666667
m = 4 0.166666666666667 0.388888888888889
m = 5 0.166666666666667 0.388888888888889 0.555555555555556
m = 6 0.166666666666667 0.388888888888889 0.555555555555556 0.666666666666667
m = 7 0.166666666666667 0.277777777777778 0.388888888888889 0.555555555555556 0.666666666666667
m = 1
m = 2
m = 3 0.777777777777778 0.888888888888889
m = 4 0.777777777777778 0.888888888888889
m = 5 0.777777777777778 0.888888888888889
m = 6 0.777777777777778 0.888888888888889
m = 7 0.777777777777778 0.888888888888889
Fit:
m 0 1 2 3 4 5 6 7
RSS 1.6986 1.1253 0.9733 0.8984 0.7984 0.7581 0.7248 0.7226
BIC 14.3728 12.7421 15.9099 20.2490 23.9062 28.7555 33.7276 39.4522
Here's where I start having the problem. Instead of reporting the actual breakdates it reports numbers which then makes it impossible to plot the break lines onto a graph because they're not at the breakdate (2002) but at 0.333.
plot.ts(dat.ts, main="Natural Mating")
lines(fitted(bp.NMSuccess, breaks = 1), col = 4, lwd = 1.5)
Nothing shows up for me in this graph (I think because it's so small for the scale of the graph).
In addition, when I try fixes that may possibly work around this problem,
fm1 <- lm(dat.ts ~ breakfactor(bp.NMSuccess, breaks = 1))
I get:
Error in model.frame.default(formula = dat.ts ~ breakfactor(bp.NMSuccess, :
variable lengths differ (found for 'breakfactor(bp.NMSuccess, breaks = 1)')
I get errors because of the NA values in the data so the length of dat.ts
is 21 and the length of breakfactor(bp.NMSuccess, breaks = 1)
18 (missing the 3 NAs).
Any suggestions?
Upvotes: 3
Views: 1402
Reputation: 17223
The problem occurs because breakpoints()
currently can only (a) cope with NA
s by omitting them, and (b) cope with times/date through the ts
class. This creates the conflict because when you omit internal NA
s from a ts
it loses its ts
property and hence breakpoints()
cannot infer the correct times.
The "obvious" way around this would be to use a time series class that can cope with this, namely zoo
. However, I just never got round to fully integrate zoo
support into breakpoints()
because it would likely break some of the current behavior.
To cut a long story short: Your best choice at the moment is to do the book-keeping about the times yourself and not expect breakpoints()
to do it for you. The additional work is not so huge. First, we create a time series with the response and the time vector and omit the NA
s:
d <- na.omit(data.frame(success = nmreprosuccess, time = 1996:2016))
d
## success time
## 1 0.000 1996
## 2 0.500 1997
## 4 0.000 1999
## 6 0.500 2001
## 8 0.500 2003
## 9 0.375 2004
## 10 0.530 2005
## 11 0.846 2006
## 12 0.440 2007
## 13 1.000 2008
## 14 0.285 2009
## 15 0.750 2010
## 16 1.000 2011
## 17 0.400 2012
## 18 0.916 2013
## 19 1.000 2014
## 20 0.769 2015
## 21 0.357 2016
Then we can estimate the breakpoint(s) and afterwards transform from the "number" of observations back to the time scale. Note that I'm setting the minimal segment size h
explicitly here because the default of 15% is probably somewhat small for this short series. 4 is still small but possibly enough for estimating of a constant mean.
bp <- breakpoints(success ~ 1, data = d, h = 4)
bp
## Optimal 2-segment partition:
##
## Call:
## breakpoints.formula(formula = success ~ 1, h = 4, data = d)
##
## Breakpoints at observation number:
## 6
##
## Corresponding to breakdates:
## 0.3333333
We ignore the break "date" at 1/3 of the observations but simply map back to the original time scale:
d$time[bp$breakpoints]
## [1] 2004
To re-estimate the model with nicely formatted factor levels, we could do:
lab <- c(
paste(d$time[c(1, bp$breakpoints)], collapse = "-"),
paste(d$time[c(bp$breakpoints + 1, nrow(d))], collapse = "-")
)
d$seg <- breakfactor(bp, labels = lab)
lm(success ~ 0 + seg, data = d)
## Call:
## lm(formula = success ~ 0 + seg, data = d)
##
## Coefficients:
## seg1996-2004 seg2005-2016
## 0.3125 0.6911
Or for visualization:
plot(success ~ time, data = d, type = "b")
lines(fitted(bp) ~ time, data = d, col = 4, lwd = 2)
abline(v = d$time[bp$breakpoints], lty = 2)
One final remark: For such short time series where just a simple shift in the mean is needed, one could also consider conditional inference (aka permutation tests) rather than the asymptotic inference employed in strucchange
. The coin
package provides the maxstat_test()
function exactly for this purpose (= short series where a single shift in the mean is tested).
library("coin")
maxstat_test(success ~ time, data = d, dist = approximate(99999))
## Approximative Generalized Maximally Selected Statistics
##
## data: success by time
## maxT = 2.3953, p-value = 0.09382
## alternative hypothesis: two.sided
## sample estimates:
## "best" cutpoint: <= 2004
This finds the same breakpoint and provides a permutation test p-value. If however, one has more data and needs multiple breakpoints and/or further regression coefficients, then strucchange
would be needed.
Upvotes: 5