无效语法- Shapely空间操作

Question

我正在尝试使用Shapely...just中的'.within'函数，这是一个简单的检查，但我无法绕过这个语法错误。

我只是想重新创建我在docs.中看到的示例

代码为：

>>> from shapely.geometry import Point, LineString

>>> LineString([(-9765787.9981184918000000 5488940.9749489054000000, -9748582.8016368076000000 5488402.1275707092000000)]).within(Point(-9763788.9782693591000000 5488878.3678984242000000))

返回：

 >> Traceback (  File "<interactive input>", line 1
    LineString([(-9765787.9981184918000000 5488940.9749489054000000, -9748582.8016368076000000 5488402.1275707092000000)]).within(Point(-9763788.9782693591000000 5488878.3678984242000000))
                                                                  ^
SyntaxError: invalid syntax

Answer 1

您有两个选择：

1)加载WKT字符串：

from shapely.wkt import loads as wkt_loads
line = wkt_loads('LINESTRING(-9765787.9981184918000000 5488940.9749489054000000, -9748582.8016368076000000 5488402.1275707092000000)')
point = wkt_loads('POINT(-9763788.9782693591000000 5488878.3678984242000000)')

2)将格式正确的坐标对列表格式化为常规Python浮点数：

from shapely.geometry import Point, LineString
line = LineString([(-9765787.9981184918000000, 5488940.9749489054000000), (-9748582.8016368076000000, 5488402.1275707092000000)])
point = Point(-9763788.9782693591000000, 5488878.3678984242000000)

接下来，您将在测试中遇到一些浮点精度错误：

line.within(point)  # False
line.distance(point)  # 7.765244949417793e-11
line.distance(point) < 1e-8  # True

我建议使用最后一种测试方法来确定该点是否处于危险状态。

Answer 2

明白了。

>> LineString([(-9765787.9981184918000000,5488940.9749489054000000), (-9748582.8016368076000000,5488402.1275707092000000)]).within(Point(-9765787.9981184918000000,5488940.9749489054000000))