point-in-polygon | determine if a point is inside a polygon with a Sweep Line

I realized that your point data is an sf object since you have POINT (-82.34323174 29.67058748) as character. Hence, I reconstructed your data first. I assigned a projection here as well.

All TurfJS methods accept (well-formed) GeoJSON as input. There is no need to use integer values for the coordinates.

It returns false

Your call to boolanPointInPolygon returns false because that point is outside that polygon:

As you can clearly see from your console.log that everything outside of your query is running first.

This is what is called Asynchronous behavior. Which basically means you cannot get values outside from a function that runs Asynchronously because Javascript/Node.js is Synchronous languages (Single Thread) and it does not wait for you async function to finish and console log whatever value it currently holds which by default is undefined.

People get confused on this behaviour a lot and you might want to understand how Node.Js and async stuff is handled by Event Loop

When P is in the same plane as S is guaranteed, make (virtual) projection of polygon and point onto any coordinate plane (not perpendicular to polygon plane). For example, to make projection onto OXZ (suitable for the second example), just ignore Y-components.

Then use any algorithm intended for 2D (pnpoly)

Point-in-polygon already assumes a 2D restriction, and GPS coordinates are 3D. Right away, that gets you in trouble.

A simple workaround is to discard the GPS height, reducing it to 2D surface coordinates. Your next problem is that that your 2D surface is now a sphere. On a spherical surface, a polygon divides the surface in two parts, but there is no obvious "inside". There's a left-hand side and a right-hand side which follows from the order of points in the polygon, but neither side is the obvious "inside". Consider the equator as a trivial polygon - which hemisphere is "inside" the equator?

Next up is the issue of the polygon edges. By definition, these are straight, i.e. line segments. But lines on a spherical surface are weird - they're generally known as great circles. And any two great circles cross in exactly two points. That's not how cartesian lines behave. Worse, the equations for a great circle are not linear when expressed in GPS coordinates, because those are longitude/latitude pairs.

I can imagine that at this point you're feeling a bit confused. You might want to look at this from another side - we have a similar problem with maps. Globe maps are by definition attempts to flatten that non-flat surface. Since that's not exactly possible, you end up with map projections. You can also project the corner points of your polygons on such projections. And because the projections are flat, you can draw the edges on the projection. You now see the problem visually: On two different projections, identical polygons will contain different parts of the world!

So, since we agreed that in the real world, the edges of the polygons are great circles, we should really consider a projection that keeps the great circles straight. There's exactly one family of projections that has this property, and that's the Gnomonic projection. It's a family of projections because you can pick any point as the center.

As it happens, we have one natural point to consider here: the GPS point we're considering. If you put that in the center, draw a gnomonic projection around it, project the polygon edges, and then draw the polygon, you have an exact solution.

Except that the actual earth isn't spherical. Sorry. How exact did you need the test to be, anyway?

You should have a look at the following packages:

• salem and the region of interest example
• regionmask

Both may get you to what you want.

To test if a point P lies on an edge QR, consider the transformation that maps Q to the origin and R to the point (0, 1).

Using complex numbers, write this transformation as Z = (z - Q) / (R - Q) and transform P with (P - Q) / (R - Q). This number must be pure real and its real part be in [0, 1].

You can use this test to detect points on the outline. It is also possible to allow a tolerance, by allowing the imaginary part to be a small number. In this case, it is advisable to normalize the denominator R - Q, so that the imaginary part becomes the Euclidean distance to the segment.

If you are talking about the index, cause the pandas is index/column sensitive, which mean it will assign the new value base on the index matched.

So what you can do