我正在使用shapely进行地理信息系统的开发,但是当加载每个邮政编码的几何形状时,由于几何形状过于复杂和崎岖不平,导致内存错误。
我希望通过尽可能减少边界点的数量来缩小形状的内存占用,而又不会过度扭曲形状。使用凸包似乎是一个解决方案,还可以从边界中简单地丢弃很多点。我想知道是否已经有现成的方法解决这个问题。
这里有两个区域。
import matplotlib.pyplot as plt
import shapely.geometry
import cv2
import numpy as np
# gen. mask
mask=np.zeros((600,600),dtype=bool)
mask[300:500,300:500]=True
mask[:150,30:120]=True
mask[70:120,30:220]=True
mask[100:200,200:260]=True
# get contours == polygon
contours, _ = cv2.findContours(mask.astype(np.uint8), # cv2 requires special types
cv2.RETR_TREE,
cv2.CHAIN_APPROX_NONE)
contours = [i.reshape((-1, 2)) for i in contours]
def simplify(polygon, tolerance = .1):
""" Simplify a polygon with shapely.
Polygon: ndarray
ndarray of the polygon positions of N points with the shape (N,2)
tolerance: float
the tolerance
"""
poly = shapely.geometry.Polygon(i)
poly_s = poly.simplify(tolerance=tolerance)
# convert it back to numpy
return np.array(poly_s.boundary.coords[:])
# Simplify all contours
contours_s = []
for i in contours:
contours_s.append(simplify(i)
plt.figure(figsize=(4,4))
plt.imshow(mask, label='2D mask')
for i, c_i in enumerate(contours_s):
plt.plot(*c_i.T, '-', label=f'cv2 contours {i}')
for i, c_i in enumerate(contours_s):
plt.plot(*c_i.T, 'o', label=f'shapely simplify {i}')
plt.legend()
plt.tight_layout()
contours_s.append(simplify(i)
。应该是contours_s.append(simplify(i))
吧? - undefined