0
$\begingroup$

Given the following ts:

ts <- c("08:43:48", "09:17:52", "12:56:22", "12:27:32", "10:59:23", 
"07:22:45", "11:13:59", "10:13:26", "10:07:01", "06:09:56", "12:43:17", 
"07:07:35", "09:36:44", "10:45:00", "08:27:36", "07:55:35", "11:32:56", 
"13:18:35", "11:09:51", "09:46:33", "06:59:12", "10:19:36", "09:39:47", 
"09:39:46", "18:23:54")

Converting to circular:

ts <- circular(ts, units = "hours", template = "clock24")

# Estimate the periodic mean from the von Mises distribution

estimates <- mle.vonmises(ts)

p_mean <- estimates$mu %% 24
concentration <- estimates$kappa

# Estimate densities of all 25 timestamps
densities <- dvonmises(ts, mu = p_mean, kappa = concentration)

Here is what I can't figure out, given alpha = 95%:

If I need a 95% CI why I need to tell qvonmises to calculate the percentile of (1-alpha)/2 - one number only. I have 2 tails clockwise and anti clockwise. Please clarify what I am missing here. Is it because this distribution is circular?

# Check if the densities are larger than the cutoff of 95%-CI
cutoff <- dvonmises(qvonmises((1 - alpha)/2, mu = p_mean, kappa = concentration), mu = p_mean, kappa = concentration)

# Define the variable time_feature
time_feature <- densities >= cutoff
$\endgroup$
  • $\begingroup$ even after loading the circular package, your estimates <- mle.vonmises(ts) is giving me: Error in x/12 : non-numeric argument to binary operator $\endgroup$ – Henry Jan 20 at 19:42

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Browse other questions tagged or ask your own question.