Questing regarding sum_of_costs for PELT

[YanniPapandreou]

Hi,

I am new to change point detection and I have been greatly enjoying trying out your package, so thanks for creating this!

I have a quick question regarding the method sum_of_costs when applied to the PELT search method. I include some code below for reference:

algo = rpt.Pelt(custom_cost=c).fit(signal)
my_bkps = algo.predict(pen=3)
my_cost = algo.cost.sum_of_costs(my_bkps)

Here I have used PELT to find changepoints in a signal using a custom cost. I then use sum_of_costs to get the sum of costs for this segmentation. My question is the following: does sum_of_costs return the costs of the segments without the penalty term, or is this included as well? Sorry if this question is trivial, but I wasn't sure from the source code.

Best wishes,
Yanni

[deepcharles]

Thanks for using ruptures, glad to hear it is useful.

does sum_of_costs return the costs of the segments without the penalty term

sum_of_costs only returns the sum of costs without the penalty term. This is a bit hidden in the code (it is here if you are curious:

ruptures/src/ruptures/base.py

Line 53 in b84df2c

def sum_of_costs(self, bkps):

)

Cheers

Charles

[YanniPapandreou]

Thanks for clarifying!

Cheers,
Yanni

[frankcharltonsonos]

@deepcharles - I'm a little confused by the explanation here. I seem to get different sum_of_costs responses after adjusting the penalty for the same dataset. It appears that choice of penalty does affect the sum_of_costs return. I'm hoping this means it genuinely calculates the value, based on the penalty.

import numpy as np
import matplotlib.pylab as plt
import ruptures as rpt

# creation of data
n, dim = 52, 3
n_bkps, sigma = 3, 1
signal, bkps = rpt.pw_constant(n, dim, n_bkps, noise_std=sigma)

# change point detection
model = "l2"  # "l2", "rbf"
algo = rpt.Pelt(model=model, jump=1).fit(signal)

my_bkps_0 = algo.predict(pen=0)
algo.cost.sum_of_costs(my_bkps_0)

my_bkps_1 = algo.predict(pen=1)
algo.cost.sum_of_costs(my_bkps_1)

my_bkps_10 = algo.predict(pen=10)
algo.cost.sum_of_costs(my_bkps_10)

my_bkps_20 = algo.predict(pen=20)
algo.cost.sum_of_costs(my_bkps_20)

Each of these 4 penalty / sum_of_costs runs return a different value.

[deepcharles]

Hi,

The method .sum_of_costs(bkps) returns the sum of costs for the given set of change-points bkps, i.e. c(0..t_1) + c(t_1..t_2) + c(t_2..t_3) + ... where t_1, t_2,... are the change-point indexes contained in bkps.

sum_of_costs only depends on the cost function and the argument bkps, i.e. the set of changes (and the signal). It is independent of the value of pen.

In your example, algo.cost.sum_of_costs(my_bkps) changes value because the argument (my_bkps) changes too.

Hope this helps,

Charles

[frankcharltonsonos]

Thank you for your response - I'm hoping this means that they are still related, even if not directly passed into the function.

Each penalty run produces a different set of bkps...because the penalty is different. By passing an array of different penalties, I'd think that I'd be able to find the elbow for sum_of_costs associated with the 'ideal' number of changepoints. This just happens to also show me the penalty range associated with that 'ideal' number of changepoints, thereby showing me what penalty to use.

This seems to be working. I'm just hoping to confirm that this is working properly, or if using sum_of_costs with PELT (which requires a penalty value) is returning something unusable.

[frankcharltonsonos]

Realizing now that a plot of penalty values by number of changepoints is more helpful here. Sum of costs is not as helpful.

[deepcharles]

Here is an example where the sum of costs is plotted to find the number of changes : https://centre-borelli.github.io/ruptures-docs/examples/music-segmentation/.