Time Series change point detection

I have a time series data which looks like the figure below.

I want to detect the change points in the times series such that I get the following points as output.

Any help would be appreciated.

Thanks, Irtaza

Answer 1

[Kaiguang]

For those who may need a Bayesian alternative for time series changepoint detection, one such Matlab implemenation is available here from this FileExchange entry, which is developed and maintained by me. The algorithm is called BEAST (Bayesian estimator of Abrupt Change/changepoint, Seasonality, and Trend). It can be instantly installed by running eval(webread('http://b.link/beast',weboptions('cert',''))). Below is a quick example using a simulated time series:

% Quick installation of BEAST to a temporary path on your local drive
eval(webread('http://b.link/beast',weboptions('cert','')))  

% A simulated time series from another quesiton asked in this forum
y = [zeros(1,100) 1:100 99:-1:50  50*ones(1,250)] + 10*rand(1,500); 

% Apply beast to y. Here season='none' indicates that y has no periodic/seasonal component
o =  beast(y, 'season','none')    

printbeast(o)
plotbeast(o)

Below is a summary of the number and locations of the changepoints detected:

#####################################################################
#                      Trend  Changepoints                          #
#####################################################################
.-------------------------------------------------------------------.
| Ascii plot of probability distribution for number of chgpts (ncp) |
.-------------------------------------------------------------------.
|Pr(ncp = 0 )=0.000|*                                               |
|Pr(ncp = 1 )=0.000|*                                               |
|Pr(ncp = 2 )=0.000|*                                               |
|Pr(ncp = 3 )=0.914|*********************************************** |
|Pr(ncp = 4 )=0.083|*****                                           |
|Pr(ncp = 5 )=0.002|*                                               |
|Pr(ncp = 6 )=0.000|*                                               |
|Pr(ncp = 7 )=0.000|*                                               |
|Pr(ncp = 8 )=0.000|*                                               |
|Pr(ncp = 9 )=0.000|*                                               |
|Pr(ncp = 10)=0.000|*                                               |
.-------------------------------------------------------------------.
|    Summary for number of Trend ChangePoints (tcp)                 |
.-------------------------------------------------------------------.
|ncp_max    = 10   | MaxTrendKnotNum: A parameter you set           |
|ncp_mode   = 3    | Pr(ncp= 3)=0.91: There is a 91.4% probability  |
|                  | that the trend component has  3 changepoint(s).|
|ncp_mean   = 3.09 | Sum{ncp*Pr(ncp)} for ncp = 0,...,10            |
|ncp_pct10  = 3.00 | 10% percentile for number of changepoints      |
|ncp_median = 3.00 | 50% percentile: Median number of changepoints  |
|ncp_pct90  = 3.00 | 90% percentile for number of changepoints      |
.-------------------------------------------------------------------.
| List of probable trend changepoints ranked by probability of      |
| occurrence: Please combine the ncp reported above to determine    |
| which changepoints below are  practically meaningful              |
'-------------------------------------------------------------------'
|tcp#              |time (cp)                  |prob(cpPr)          |
|------------------|---------------------------|--------------------|
|1                 |199.000000                 |1.00000             |
|2                 |252.000000                 |0.92867             |
|3                 |96.000000                  |0.89042             |
|4                 |471.000000                 |0.01800             |
|5                 |413.000000                 |0.00733             |
|6                 |435.000000                 |0.00692             |
|7                 |483.000000                 |0.00679             |
|8                 |448.000000                 |0.00579             |
|9                 |343.000000                 |0.00204             |
|10                |63.000000                  |0.00154             |
.-------------------------------------------------------------------.

Below is the plot. The trend is fitted using a piecewise polynomial model. Again, as a Bayesian method, BEAST assumes the order of the polynomials for individual segments as uknowns. The orders of the polynomial needed to adequately fit the trend are estimated over time, as depicted iin the tOrder subplot below. The 1st and 4th segments are flat lines, so their estimated polynomial orders are close to zeros.

Answer 2

[Akira Agata]

ischange or findchangepts function will do that. If you could upload your data, I would be happy to check it.

https://jp.mathworks.com/help/matlab/ref/ischange.html?lang=en https://jp.mathworks.com/help/signal/ref/findchangepts.html?lang=en

[Surendar Babu B]

Hello,

I have applied both the functions in 52 year temperature data. Using ischangepts function, I found 1 changing point and 4 changing points obseved by using ischange function.

Which function I should use to detect the change point in the time series?

The temperature data:

-2.28 -1.01 -0.93 -1.16 -1.28 -0.86 -1.48 -2.38 -1.73 -0.93 -1.73 -2.03 -0.68 -1.25 -2.43 -2.40 -1.46 -0.85 -1.63 -1.18 -0.66 -2.06 -1.68 -1.78 -1.48 -1.43 -0.78 -1.71 -0.61 -1.56 -1.88 -0.65 -0.71 -0.43 -0.41 -0.66 -0.05 -0.86 -0.36 -0.36 -0.73 0.21 0.48 -0.88 -1.06 -1.23 -1.23 -0.63 0.43 0.40 0.63 -0.90

Thank you.

[Akira Agata]

Since change and findchangepts functions are using different algorithms (in default operation), detected changing points will be different.

Looking at your temperature data, there seems to be no clear changing point(s). Could you tell us which point(s) you would like to detect as a "changing point" ?

[Surendar Babu B]

These data set are from 1968 to 2019. As temeprerature is rising in recent decades, my study is focused on recent changes in the temperature. It would be great if I see any changes from 1999/2000 to 2019.

[Kaiguang]

Surendar Babu, not sure if you are still looking for an alternative solution to your problem. To add to Akira Agata's excellent explnations, please see below for another answer specificially for your tempeature time series.

Answer 3

[Kaiguang]

To Surendar Babu:

For your time series data, ineed, it is hard to see a signficant abrupt change. Similar to my answer to the oringal quesiton, I used the BEAST tool as another example to explain its relevance.

TLDR: The Bayesian changepoint detection method mentioned here (aka BEAST) is a FUZZY changepointe detection method. It tells not just when and how many changepoints exist but also the probability of having changepoints occurring over time.

Also, 'changepoint' is a misnomer. Regarding changepoint detection, here I borrow from the headline of a blog post from Dr. Andrew Gelman (https://statmodeling.stat.columbia.edu/2016/03/18/i-definitely-wouldnt-frame-it-as-to-determine-if-the-time-series-has-a-change-point-or-not-the-time-series-whatever-it-is-has-a-change-point-at-every-time-the-question/):

I definitely wouldn’t frame it as “To determine if the time series has a change-point or not.” The time series, whatever it is, has a change point AT EVERY TIME. The question might be, “Is a change point necessary to model these data?” That’s a question I could get behind.

With that said, here is the code snippet to apply BEAST to your data

% Your annual temprature data since 1968
Y=[-2.28	-1.01	-0.93	-1.16	-1.28	-0.86	-1.48	-2.38	-1.73	-0.93	-1.73	-2.03	-0.68	-1.25	-2.43	-2.40	-1.46	-0.85	-1.63	-1.18	-0.66	-2.06	-1.68	-1.78	-1.48	-1.43	-0.78	-1.71	-0.61	-1.56	-1.88	-0.65	-0.71	-0.43	-0.41	-0.66	-0.05	-0.86	-0.36	-0.36	-0.73	0.21 	0.48 	-0.88	-1.06	-1.23	-1.23	-0.63	0.43 	0.40 	0.63 	-0.90];

eval(webread('http://b.link/beast',weboptions('cert','')))  %First, install BEAST to a temporay folder on your local drive
o=beast(Y,'season','none','start',1968)                     % 'season'='none': no periodic variation in Y given your data is annual
                                                            % start=1968: the start year of your data
printbeast(o)   %print a summary of changepoints detected
plotbeast(o)    %plot the result 

BEAST also allows specifying the max and min orders of the polynomials allowed to fit individual trend segments. By default, it is set to 0 (const term) and 1 (linear term). Below run BEAST again to your Y but fix the min and max orders both to 0; that is, flat segments only

o = beast(Y,'season','none','start',1968, 'torder.minmax',[0,0]) % minorder=maxorder=0 (i.e., const/flat lines)
plotbeast(o)    % plot the result 

Below is the output. In particular, the Pr(tcp) subplot shows the probability of changepoint occurance over time. Aparently, peaks correspond to hihger pobabilities of changepoinits occuring there.

uninstallbeast % If BEAST is not needed, uninstall it from your machine

Answer 4

[shahina Joomun]

Hello, please can you please give the implementation of the function findchangepts, because I need to write the code in VB.net.

Your help would be much appreciated. Thank you