feat: broadcasting solution exercise 79

source/exercises100.ktx

@@ -1118,6 +1118,32 @@ P1 = np.random.uniform(-10,10,(10,2))
 p = np.random.uniform(-10, 10, (10,2))
 print(np.array([distance(P0,P1,p_i) for p_i in p]))
 
+# Author: Yang Wu (Broadcasting)
+def distance_points_to_lines(p: np.ndarray, p_1: np.ndarray, p_2: np.ndarray) -> np.ndarray:
+    x_0, y_0 = p.T  # Shape -> (n points, )
+    x_1, y_1 = p_1.T  # Shape -> (n lines, )
+    x_2, y_2 = p_2.T  # Shape -> (n lines, )
+
+    # Displacement vector coordinates from p_1 -> p_2
+    dx = x_2 - x_1  # Shape -> (n lines, )
+    dy = y_2 - y_1  # Shape -> (n lines, )
+
+    # The 'cross product' term
+    cross_term = x_2 * y_1 - y_2 * x_1  # Shape -> (n lines, )
+
+    # Broadcast x_0, y_0 (n points, 1) and dx, dy, cross_term (1, n lines) -> (n points, n lines)
+    numerator = np.abs(
+        dy[np.newaxis, :] * x_0[:, np.newaxis]
+        - dx[np.newaxis, :] * y_0[:, np.newaxis]
+        + cross_term[np.newaxis, :]
+    )
+    denominator = np.sqrt(dx**2 + dy**2)  # Shape -> (n lines, )
+
+    # Shape (n points, n lines) / (1, n_lines) -> (n points, n lines)
+    return numerator / denominator[np.newaxis, :]
+
+distance_points_to_lines(p, P0, P1)
+
 < q80
 Consider an arbitrary array, write a function that extract a subpart with a fixed shape and centered on a given element (pad with a `fill` value when necessary) (★★★)