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I'd like to know if there is a way to create a polygon in shapely from two not connected linestrings. I've been struggling with this question from some hours now and I can't find a way to deal with it.

Figure

For example, I'd like to have the polygon with those linestrings as its exterior coords.

from shapely.geometry import Polygon

p1 = Polygon([1.32, 7.66], [1.4, 5.56], [2.98, 4.5], [5.86, 4], [12.16, 4.88])
p2 = Polygon([1.62, 0], [2, 1], [3.16, 1.68], [5, 2], [8.44, 2.14], [9.92, 0.58], [12, 0], [13.62, 0.8])
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    By definition, a polygon is composed of the vertices which describe a closed ring. Your vertices are not closed rings. They are therefore not polygons. There are other requirements besides closed (not crossing, not touching other parts,...) but you need to work on that first. – Vince Nov 9 '20 at 14:06
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You can get the lines' coordinates and use them to make a polygon. But it depends. In many cases, you will probably have invalid polygon.

  • Let's make two lines using the coordinates in your question:

    from shapely.geometry import Polygon, LineString
l1 = LineString([[1.32, 7.66], [1.4, 5.56], [2.98, 4.5], [5.86, 4], [12.16, 4.88]])
l2 = LineString([[1.62, 0], [2, 1], [3.16, 1.68], [5, 2], [8.44, 2.14], [9.92, 0.58], [12, 0], [13.62, 0.8]])

p = Polygon([*list(l1.coords), *list(l2.coords)])

    enter image description here

    As you see, there are many self-intersecting edges. So the result is an invalid polygon.

  • Let's reverse the l2's coordinates:

    p = Polygon([*list(l1.coords), *list(l2.coords)[::-1]])

    enter image description here

    Again, self-intersecting edges.

  • Let's change l2's location. (added 10 to y-coordinate):

    l2 = LineString([[1.62, 10], [2, 11], [3.16, 11.68], [5, 12], [8.44, 12.14], [9.92, 10.58], [12, 10], [13.62, 10.8]])
p = Polygon([*list(l1.coords), *list(l2.coords)])

    enter image description here

  • Let's reverse the l2's coordinates:

    p = Polygon([*list(l1.coords), *list(l2.coords)[::-1]])

    enter image description here

    Now, this is a valid polygon.

Briefly, you cannot be sure if the result polygon will be valid. It depends.

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    Oh nice, thank you for your answer !! To manage the situation, I've made a list comprehensive loop with all the combinations of coordinates using itertools.product() and I'm only keeping the valid ones with shapely.obj.is_valid. – Titusium Nov 9 '20 at 15:39

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