0
$\begingroup$

I have a finite series of line segments, each continuous from the previous one[1]. I want to quantify how "straight" or "zig-zaggy" the run of line segments is.

My idea is the following:

1) Rotate all segments such that the first point of the first segment and the last point of the last segment are vertical with one another (ΔX = 0, ΔY > 0).

2) Walk each segment and separately sum the absolute value of both Δx and Δy for each segment. Call these absDx and absDy.

For a perfectly straight line, absDx = ΔX = 0 and absDy = ΔY.

For a series that deviates from straight, either absDx and/or absDy will be greater than 0. How much they're greater than 0 defines how zig-zaggy the lines are.

I am considering scaling the values based on ΔY such that a large ΔY requires larger absDx and absDy.

I have found other recommendations for quantifying straightness that fail to account for the idea that the line segments may double-back on themselves, or may not be of equal length and spacing.

My worry is that by simply summing values directly, I may be introducing bias in unintended ways.

[1] This represents gesture data captured from a touch screen.

$\endgroup$
0
$\begingroup$

I haven't read through your proposal, but have an alternative suggestion for the problem you are trying to solve. Since the end of each segment is the start of the next, what you really have is a sequence of points. You want to quantify the extent to which they lie on a single line.

That reformulation suggests simple linear regression. Then $R^2$ will measure the goodness of fit.

That might be a problem for nearly vertical lines, since linear regression works with the $y$-distances from the line. Perhaps the variant that finds the line that minimizes the orthogonal distances would be better. You can read about that at http://mathworld.wolfram.com/LeastSquaresFittingPerpendicularOffsets.html .

$\endgroup$
1
  • $\begingroup$ The problem is that this doesn't handle points which lie on the line, but reverse direction. I.e. if it has -Δy's. $\endgroup$ – dave mankoff Jun 7 '19 at 17:14

Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.