Telling two paths apart in a road network graph?-CodePudding

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Here is an example of the data. We split the data into groups that represent real car trips. So sightings of the same license plate by different cameras sufficiently close in time (on the same day of course). Here is an example of one group (unrelated to picture above), like which there are about 10M others.

index camera_ID encoded_plate date time_seconds velocity 9200 480301111 660.0 2021-03-11 62000.0 54 9201 480321111 660.0 2021-03-11 62135.0 28 9202 480331111 660.0 2021-03-11 62235.0 5 9203 480341112 660.0 2021-03-11 62302.0 42 9204 450371112 660.0 2021-03-11 62648.0 32

CodePudding user response：

Here's a possible approach: sort the recorded journey types by number of camera points, descending (so e.g. ABCD has four points, it should appear before EFG which has three). Then for each journey type, estimate how many spurious observations of each subsequence you would expect to see, and subtract that number from the actual observations of each of those subsequences (e.g. if we think there were 100 jouneys for ABCD, then we'd expect 100 * p(1-p)^3 spurious observations of each of ABC, ABD, ACD and BCD, and so on for shorter subsequences).

Also keep track of the uncertainty in each count, which is initially zero, but each time you subtract an estimate from the count of some journey type, add on the variance of that estimate to the uncertainty for that journey type. The variance can then be used to estimate the likelihood that a measurement should really be zero; if so, just ignore that journey type when you come to it, and move onto the next one.

At the end, the "real" journey types are those where a count of zero isn't likely based on the estimated count and variance. Don't forget to also adjust the counts of "real" journeys upwards, e.g. a path ABCD has a (1-p)^4 chance of not being recorded correctly so you should divide the estimated count by that factor to compensate. This adjustment should be made before estimating how many spurious observations to deduct from the subsequences.

CodePudding user response：

if there is just one way to get from A->C, say via B, then you will still see a lot of cars at A then directly at C because camera B just happened not to pick them up, and so AC will appear to be a popular path as well as ABC, when really it is just ABC

I do not see the problem. Why can you not just immediately replace all occurrences of AC with ABC?