add function to compute (signed) area of path

Author: brunobeltran

doc/users/next_whats_new/path-size-methods.rst

@@ -0,0 +1,6 @@
+Path area
+~~~~~~~~~
+
+A `~.path.Path.signed_area` method was added to compute the signed filled area
+of a Path object analytically (i.e. without integration). This should be useful
+for constructing Paths of a desired area.

lib/matplotlib/bezier.py

@@ -2,7 +2,6 @@
 A module providing some utility functions regarding Bézier path manipulation.
 """
 
-from functools import lru_cache
 import math
 import warnings
 
@@ -11,15 +10,7 @@
 from matplotlib import _api
 
 
-# same algorithm as 3.8's math.comb
-@np.vectorize
-@lru_cache(maxsize=128)
-def _comb(n, k):
-    if k > n:
-        return 0
-    k = min(k, n - k)
-    i = np.arange(1, k + 1)
-    return np.prod((n + 1 - i)/i).astype(int)
+_comb = np.vectorize(math.comb, otypes=[int])
 
 
 class NonIntersectingPathException(ValueError):
@@ -229,6 +220,121 @@ def point_at_t(self, t):
         """
         return tuple(self(t))
 
+    @property
+    def arc_area(self):
+        r"""
+        Signed area swept out by ray from origin to curve.
+
+        Counterclockwise area is counted as positive, and clockwise area as
+        negative.
+
+        The sum of this function for each Bezier curve in a Path will give the
+        signed area enclosed by the Path.
+
+        Returns
+        -------
+        float
+            The signed area of the arc swept out by the curve.
+
+        Notes
+        -----
+        An analytical formula is possible for arbitrary bezier curves.
+        The formulas can be found in computer graphics references [1]_ and
+        an example derivation is given below.
+
+        For a bezier curve :math:`\vec{B}(t)`, to calculate the area of the arc
+        swept out by the ray from the origin to the curve, we need to compute
+        :math:`\frac{1}{2}\int_0^1 \vec{B}(t) \cdot \vec{n}(t) dt`, where
+        :math:`\vec{n}(t) = u^{(1)}(t)\hat{x}^{(0)} - u^{(0)}(t)\hat{x}^{(1)}`
+        is the normal vector oriented away from the origin and
+        :math:`u^{(i)}(t) = \frac{d}{dt} B^{(i)}(t)` is the :math:`i`\th
+        component of the curve's tangent vector.  (This formula can be found by
+        applying the divergence theorem to :math:`F(x,y) = [x, y]/2`, and
+        calculates the *signed* area for a counter-clockwise curve, by the
+        right hand rule).
+
+        The control points of the curve are its coefficients in a Bernstein
+        expansion, so if we let :math:`P_i = [P^{(0)}_i, P^{(1)}_i]` be the
+        :math:`i`\th control point, then
+
+        .. math::
+
+            \frac{1}{2}\int_0^1 B(t) \cdot n(t) dt
+                    &= \frac{1}{2}\int_0^1 B^{(0)}(t) \frac{d}{dt} B^{(1)}(t)
+                                        - B^{(1)}(t) \frac{d}{dt} B^{(0)}(t)
+                                        dt \\
+                    &= \frac{1}{2}\int_0^1
+                        \left( \sum_{j=0}^n P_j^{(0)} b_{j,n} \right)
+                        \left( n \sum_{k=0}^{n-1} (P_{k+1}^{(1)} -
+                                P_{k}^{(1)}) b_{j,n} \right)
+                        \\
+                    &\hspace{1em} - \left( \sum_{j=0}^n P_j^{(1)} b_{j,n}
+                        \right) \left( n \sum_{k=0}^{n-1} (P_{k+1}^{(0)}
+                                        - P_{k}^{(0)}) b_{j,n} \right)
+                        dt,
+
+        where :math:`b_{\nu, n}(t) = {n \choose \nu} t^\nu {(1 - t)}^{n-\nu}`
+        is the :math:`\nu`\th Bernstein polynomial of degree :math:`n`.
+
+        Grouping :math:`t^l(1-t)^m` terms together for each :math:`l`,
+        :math:`m`, we get that the integrand becomes
+
+        .. math::
+
+            \sum_{j=0}^n \sum_{k=0}^{n-1}
+                {n \choose j} {{n - 1} \choose k}
+                &\left[P_j^{(0)} (P_{k+1}^{(1)} - P_{k}^{(1)})
+                    - P_j^{(1)} (P_{k+1}^{(0)} - P_{k}^{(0)})\right] \\
+                &\hspace{1em}\times{}t^{j + k} {(1 - t)}^{2n - 1 - j - k}
+
+        or more compactly,
+
+        .. math::
+
+            \sum_{j=0}^n \sum_{k=0}^{n-1}
+                \frac{{n \choose j} {{n - 1} \choose k}}
+                        {{{2n - 1} \choose {j+k}}}
+                [P_j^{(0)} (P_{k+1}^{(1)} - P_{k}^{(1)})
+                    - P_j^{(1)} (P_{k+1}^{(0)} - P_{k}^{(0)})]
+                b_{j+k,2n-1}(t).
+
+        Interchanging sum and integral, and using the fact that :math:`\int_0^1
+        b_{\nu, n}(t) dt = \frac{1}{n + 1}`, we conclude that the original
+        integral can be written as
+
+        .. math::
+
+            \frac{1}{2}&\int_0^1 B(t) \cdot n(t) dt
+            \\
+            &= \frac{1}{4}\sum_{j=0}^n \sum_{k=0}^{n-1}
+                \frac{{n \choose j} {{n - 1} \choose k}}
+                    {{{2n - 1} \choose {j+k}}}
+                [P_j^{(0)} (P_{k+1}^{(1)} - P_{k}^{(1)})
+                - P_j^{(1)} (P_{k+1}^{(0)} - P_{k}^{(0)})]
+
+        References
+        ----------
+        .. [1] Sederberg, Thomas W., "Computer Aided Geometric Design" (2012).
+          Faculty Publications. 1. https://scholarsarchive.byu.edu/facpub/1
+        """
+        n = self.degree
+        P = self.control_points
+        dP = np.diff(P, axis=0)
+        j = np.arange(n + 1)
+        k = np.arange(n)
+        return (1/4)*np.sum(
+            np.multiply.outer(_comb(n, j), _comb(n - 1, k))
+            / _comb(2*n - 1, np.add.outer(j, k))
+            * (np.multiply.outer(P[j, 0], dP[k, 1]) -
+               np.multiply.outer(P[j, 1], dP[k, 0]))
+        )
+
+    @classmethod
+    def differentiate(cls, B):
+        """Return the derivative of a BezierSegment, itself a BezierSegment"""
+        dcontrol_points = B.degree*np.diff(B.control_points, axis=0)
+        return cls(dcontrol_points)
+
     @property
     def control_points(self):
         """The control points of the curve."""

lib/matplotlib/bezier.pyi

@@ -36,6 +36,10 @@ class BezierSegment:
     def __call__(self, t: ArrayLike) -> np.ndarray: ...
     def point_at_t(self, t: float) -> tuple[float, ...]: ...
     @property
+    def arc_area(self) -> float: ...
+    @classmethod
+    def differentiate(cls, B: BezierSegment) -> BezierSegment: ...
+    @property
     def control_points(self) -> np.ndarray: ...
     @property
     def dimension(self) -> int: ...

lib/matplotlib/path.py

@@ -666,6 +666,62 @@ def intersects_bbox(self, bbox, filled=True):
         return _path.path_intersects_rectangle(
             self, bbox.x0, bbox.y0, bbox.x1, bbox.y1, filled)
 
+    def signed_area(self):
+        """
+        Get signed area of the filled path.
+
+        Area of a filled region is treated as positive if the path encloses it
+        in a counter-clockwise direction, but negative if the path encloses it
+        moving clockwise.
+
+        All sub paths are treated as if they had been closed. That is, if there
+        is a MOVETO without a preceding CLOSEPOLY, one is added.
+
+        If the path is made up of multiple components that overlap, the
+        overlapping area is multiply counted.
+
+        Returns
+        -------
+        float
+            The signed area enclosed by the path.
+
+        Notes
+        -----
+        If the Path is not self-intersecting and has no overlapping components,
+        then the absolute value of the signed area is equal to the actual
+        filled area when the Path is drawn (e.g. as a PathPatch).
+
+        Examples
+        --------
+        A symmetric figure eight, (where one loop is clockwise and
+        the other counterclockwise) would have a total *signed_area* of zero,
+        since the two loops would cancel each other out.
+        """
+        area = 0
+        prev_point = None
+        prev_code = None
+        start_point = None
+        for B, code in self.iter_bezier():
+            # MOVETO signals the start of a new connected component of the path
+            if code == Path.MOVETO:
+                # if the previous segment exists and it doesn't close the
+                # previous connected component of the path, do so manually to
+                # match Agg's convention of filling unclosed path segments
+                if prev_code not in (None, Path.CLOSEPOLY):
+                    Bclose = BezierSegment([prev_point, start_point])
+                    area += Bclose.arc_area
+                # to allow us to manually close this connected component later
+                start_point = B.control_points[0]
+            area += B.arc_area
+            prev_point = B.control_points[-1]
+            prev_code = code
+        # add final implied CLOSEPOLY, if necessary
+        if start_point is not None \
+                and not np.all(np.isclose(start_point, prev_point)):
+            B = BezierSegment([prev_point, start_point])
+            area += B.arc_area
+        return area
+
     def interpolated(self, steps):
         """
         Return a new path resampled to length N x *steps*.

lib/matplotlib/path.pyi

@@ -90,6 +90,7 @@ class Path:
     def get_extents(self, transform: Transform | None = ..., **kwargs) -> Bbox: ...
     def intersects_path(self, other: Path, filled: bool = ...) -> bool: ...
     def intersects_bbox(self, bbox: Bbox, filled: bool = ...) -> bool: ...
+    def signed_area(self) -> float: ...
     def interpolated(self, steps: int) -> Path: ...
     def to_polygons(
         self,

lib/matplotlib/tests/test_bezier.py

@@ -3,6 +3,10 @@
 """
 
 from matplotlib.bezier import inside_circle, split_bezier_intersecting_with_closedpath
+from matplotlib.tests.test_path import _test_curves
+
+import numpy as np
+import pytest
 
 
 def test_split_bezier_with_large_values():
@@ -15,3 +19,28 @@ def test_split_bezier_with_large_values():
     # All we are testing is that this completes
     # The failure case is an infinite loop resulting from floating point precision
     # pytest will timeout if that occurs
+
+
+# get several curves to test our code on by borrowing the tests cases used in
+# `~.tests.test_path`. get last path element ([-1]) and curve, not code ([0])
+_test_curves = [list(tc.path.iter_bezier())[-1][0] for tc in _test_curves]
+# np2+ uses trapezoid, but we need to fallback to trapz for np<2 since it isn't there
+_trapezoid = getattr(np, "trapezoid", np.trapz)  # type: ignore[attr-defined]
+
+
+def _integral_arc_area(B):
+    """(Signed) area swept out by ray from origin to curve."""
+    dB = B.differentiate(B)
+    def integrand(t):
+        x = B(t)
+        y = dB(t)
+        # np.cross for 2d input
+        return (x[:, 0] * y[:, 1] - x[:, 1] * y[:, 0]) / 2
+    x = np.linspace(0, 1, 1000)
+    y = integrand(x)
+    return _trapezoid(y, x)
+
+
+@pytest.mark.parametrize("B", _test_curves)
+def test_area_formula(B):
+    assert np.isclose(_integral_arc_area(B), B.arc_area)

lib/matplotlib/tests/test_path.py

@@ -1,8 +1,8 @@
 import platform
 import re
+from collections import namedtuple
 
 import numpy as np
-
 from numpy.testing import assert_array_equal
 import pytest
 
@@ -88,25 +88,29 @@ def test_contains_points_negative_radius():
     np.testing.assert_equal(result, [True, False, False])
 
 
-_test_paths = [
+_ExampleCurve = namedtuple('_ExampleCurve', ['path', 'extents', 'area'])
+_test_curves = [
     # interior extrema determine extents and degenerate derivative
-    Path([[0, 0], [1, 0], [1, 1], [0, 1]],
-           [Path.MOVETO, Path.CURVE4, Path.CURVE4, Path.CURVE4]),
-    # a quadratic curve
-    Path([[0, 0], [0, 1], [1, 0]], [Path.MOVETO, Path.CURVE3, Path.CURVE3]),
+    _ExampleCurve(Path([[0, 0], [1, 0], [1, 1], [0, 1]],
+                       [Path.MOVETO, Path.CURVE4, Path.CURVE4, Path.CURVE4]),
+                  extents=(0., 0., 0.75, 1.), area=0.6),
+    # a quadratic curve, clockwise
+    _ExampleCurve(Path([[0, 0], [0, 1], [1, 0]],
+                       [Path.MOVETO, Path.CURVE3, Path.CURVE3]),
+                  extents=(0., 0., 1., 0.5), area=-1/3),
     # a linear curve, degenerate vertically
-    Path([[0, 1], [1, 1]], [Path.MOVETO, Path.LINETO]),
+    _ExampleCurve(Path([[0, 1], [1, 1]], [Path.MOVETO, Path.LINETO]),
+                  extents=(0., 1., 1., 1.), area=0.),
     # a point
-    Path([[1, 2]], [Path.MOVETO]),
+    _ExampleCurve(Path([[1, 2]], [Path.MOVETO]), extents=(1., 2., 1., 2.),
+                  area=0.),
+    # non-curved triangle
+    _ExampleCurve(Path([(1, 1), (2, 1), (1.5, 2)]), extents=(1, 1, 2, 2), area=0.5),
 ]
 
 
-_test_path_extents = [(0., 0., 0.75, 1.), (0., 0., 1., 0.5), (0., 1., 1., 1.),
-                      (1., 2., 1., 2.)]
-
-
-@pytest.mark.parametrize('path, extents', zip(_test_paths, _test_path_extents))
-def test_exact_extents(path, extents):
+@pytest.mark.parametrize('precomputed_curve', _test_curves)
+def test_exact_extents(precomputed_curve):
     # notice that if we just looked at the control points to get the bounding
     # box of each curve, we would get the wrong answers. For example, for
     # hard_curve = Path([[0, 0], [1, 0], [1, 1], [0, 1]],
@@ -116,6 +120,7 @@ def test_exact_extents(path, extents):
     # the way out to the control points.
     # Note that counterintuitively, path.get_extents() returns a Bbox, so we
     # have to get that Bbox's `.extents`.
+    path, extents = precomputed_curve.path, precomputed_curve.extents
     assert np.all(path.get_extents().extents == extents)
 
 
@@ -129,6 +134,28 @@ def test_extents_with_ignored_codes(ignored_code):
     assert np.all(path.get_extents().extents == (0., 0., 1., 1.))
 
 
+@pytest.mark.parametrize('precomputed_curve', _test_curves)
+def test_signed_area(precomputed_curve):
+    path, area = precomputed_curve.path, precomputed_curve.area
+    np.testing.assert_allclose(path.signed_area(), area)
+    # now flip direction, sign of *signed_area* should flip
+    rcurve = Path(path.vertices[::-1], path.codes)
+    np.testing.assert_allclose(rcurve.signed_area(), -area)
+
+
+def test_signed_area_unit_rectangle():
+    rect = Path.unit_rectangle()
+    assert rect.signed_area() == 1
+
+
+def test_signed_area_unit_circle():
+    circ = Path.unit_circle()
+    # Not a "real" circle, just an approximation of a circle made out of bezier
+    # curves. The actual value is 3.1415935732517166, which is close enough to
+    # pass here.
+    assert np.isclose(circ.signed_area(), np.pi)
+
+
 def test_point_in_path_nan():
     box = np.array([[0, 0], [1, 0], [1, 1], [0, 1], [0, 0]])
     p = Path(box)