How do I draw the line for change-point detection using/after doing a Pettitt test.

[Ses4short]

This is my plot

This is my code so far for the plot
#From 2004 to 2019, Canada, Buildings--#
x = (2004:2019)
y = integer(19)
y = as.double(y)
len = dim(Buildings)[1]
rng = (1:len)
for (i in rng) {
idx = strtoi(Buildings[i,2]) - 2000
y[idx] = y[idx] + as.double(Buildings[i,77])
}
y=y[04:19]

matplot(x, y, type="b", lty=4, pch = 18, col="blue",xlab="Year",
ylab="Total Emission",main = "Building Sector 2004-2019", lwd = '3')

pettittTest <- trend::pettitt.test(x = Buildings[['TotalEmi']])
print(pettittTest)

print(Buildings[['Year']][pettittTest$estimate])

Thanks in advance

[jrkrideau]

We need a reproducible example (reprex)

FAQ: How to do a minimal reproducible example ( reprex ) for beginners

A minimal reproducible example consists of the following items: A minimal dataset, necessary to reproduce the issue The minimal runnable code necessary to reproduce the issue, which can be run on the given dataset, and including the necessary information on the used packages. Let's quickly go over each one of these with examples: Minimal Dataset (Sample Data) You need to provide a data frame that is small enough to be (reasonably) pasted on a post, but big enough to reproduce your issue. Let's say, as an example, that you are working with the iris data frame head(iris) #> Sepal.Length Sepal.Width Petal.Length Petal.Width Species #> 1 5.1 3.5 1.4 0.…

Any particular reason for using matplot?

When I try this

x = (2004:2019)

y = integer(19)

y = as.double(y)

y=y[04:19]

matplot(x, y, type="b", lty=4, pch = 18, col="blue",xlab="Year",
        
        ylab="Total Emission",main = "Building Sector 2004-2019", lwd = '3')

all I get is a flat line with y == 0