Calculate area under a polyline

Question

Consider the following code:

tmp = {{0, 0}, {1, 1}, {2, 1}, {3, 2}, {1, 0.5}};
ListLinePlot[tmp, Filling -> Axis]

Is there any easy way to compute filled area?

Edit: Actual data consists of thousands of points, so some analytical solution would be the best

Answer 1

tmp = {{0, 0}, {1, 1}, {2, 1}, {3, 2}, {1, 0.5}};
plot = ListLinePlot[tmp, Filling -> Axis];

Area@DiscretizeGraphics[plot]
(*  0.833333  *)

---

I figured out how to use Graphics`PolygonUtils`SimplePolygonPartition. It subdivides the polygon from the plot into non self-intersecting, possibly nonconvex polygons, but some of the polygons it creates lie outside the original polygon. One difficulty is finding a point inside a polygon. We do that by searching for a point, the average of three consecutive vertices, that is inside a convex angle. The subdivision also creates vertices along edges. Numerical round-off error makes it difficult to detect when point is inside or outside the polygon, so we skip those.

ClearAll[findPtInPoly];
SetAttributes[findPtInPoly, Listable];
findPtInPoly::nopt = "Warning: Could not find point inside polygon ``; returning a vertex";
tolerance = 1*^-10;
findPtInPoly[Polygon[poly_]] := 
 Module[{point}, 
  Do[With[{pts = poly[[t ;; t + 2]]}, 
    If[VectorAngle @@ Differences[pts] > tolerance && (* == not collinear *)
      Graphics`PolygonUtils`InPolygonQ[Polygon[poly], Mean[pts]],
     point = Mean[pts];
     Break[]]],
   {t, Length[poly] - 2}];
  If[VectorQ[point],
   point,
   Message[findPtInPoly::nopt, poly];
   First[poly]]]

SeedRandom[0];
npts = 300;
tmp = RandomReal[100, {npts, 2}];
plot = ListLinePlot[tmp, Filling -> Axis]
With[{poly = First@Cases[Normal@plot, _Polygon, Infinity]},
  Total[If[Graphics`PolygonUtils`InPolygonQ[poly, findPtInPoly[#]],
      Graphics`PolygonUtils`PolygonArea[#], 0] & /@
    Graphics`PolygonUtils`SimplePolygonPartition[poly]]
  ] // AbsoluteTiming
Area@DiscretizeGraphics[plot] // AbsoluteTiming
(*
  {2.931620, 4368.75}
  {32.840698, 4368.75}
*)

It seems to have better time complexity than Area @* DiscretizeGraphics. On inputs of size npts equal to 100, 200, 300, the timing ratios of the two methods are {1.8112, 4.16505, 11.2022}.

Answer 2

tmp = {{0, 0}, {1, 1}, {2, 1}, {3, 2}, {1, 0.5}}; 
Graphics@ Polygon[tmp]

yields this:

Its area is

Area[Polygon[tmp]]

(*  0.583333   *)

Have fun!