我想问一下:怎么在pycharm中实现，用点获取距离最近的建筑物面，然后根据建筑物面的最长的边的角度，生成一个5米*5米的正方形面
def euclidean_distance_sqr(point1, point2):
"""
>>> euclidean_distance_sqr([1,2],[2,4])
5
"""
return (point1[0] - point2[0]) ** 2 + (point1[1] - point2[1]) ** 2
def column_based_sort(array, column=0):
"""
>>> column_based_sort([(5, 1), (4, 2), (3, 0)], 1)
[(3, 0), (5, 1), (4, 2)]
"""
return sorted(array, key=lambda x: x[column])
def dis_between_closest_pair(points, points_counts, min_dis=float("inf")):
for i in range(points_counts - 1):
for j in range(i + 1, points_counts):
current_dis = euclidean_distance_sqr(points[i], points[j])
if current_dis < min_dis:
min_dis = current_dis
return min_dis
def dis_between_closest_in_strip(points, points_counts, min_dis=float("inf")):
for i in range(min(6, points_counts - 1), points_counts):
for j in range(max(0, i - 6), i):
current_dis = euclidean_distance_sqr(points[i], points[j])
if current_dis < min_dis:
min_dis = current_dis
return min_dis
def closest_pair_of_points_sqr(points_sorted_on_x, points_sorted_on_y, points_counts):
# base case
if points_counts <= 3:
return dis_between_closest_pair(points_sorted_on_x, points_counts)
# recursion
mid = points_counts // 2
closest_in_left = closest_pair_of_points_sqr(
points_sorted_on_x, points_sorted_on_y[:mid], mid
)
closest_in_right = closest_pair_of_points_sqr(
points_sorted_on_y, points_sorted_on_y[mid:], points_counts - mid
)
closest_pair_dis = min(closest_in_left, closest_in_right)
"""
cross_strip contains the points, whose Xcoords are at a
distance(< closest_pair_dis) from mid's Xcoord
"""
cross_strip = []
for point in points_sorted_on_x:
if abs(point[0] - points_sorted_on_x[mid][0]) < closest_pair_dis:
cross_strip.append(point)
closest_in_strip = dis_between_closest_in_strip(
cross_strip, len(cross_strip), closest_pair_dis
)
return min(closest_pair_dis, closest_in_strip)
def closest_pair_of_points(points, points_counts):
"""
>>> closest_pair_of_points([(2, 3), (12, 30)], len([(2, 3), (12, 30)]))
28.792360097775937
"""
points_sorted_on_x = column_based_sort(points, column=0)
points_sorted_on_y = column_based_sort(points, column=1)
return (
closest_pair_of_points_sqr(
points_sorted_on_x, points_sorted_on_y, points_counts
)
) ** 0.5
if name == "main":
points = [(2, 3), (12, 30), (40, 50), (5, 1), (12, 10), (3, 4)]
print("Distance:", closest_pair_of_points(points, len(points)))