问题描述
我有一个带有位置坐标的地理数据框，另一个带有表示洪水区域的多边形。对于每个位置点，我需要找到到最近洪水区的距离，但我无法这样做。这就是数据的样子和我所做的。
points.head()
Latitude Longitude geometry
0 29.873542 -95.300914 POINT (-95.30091 29.87354)
1 29.689952 -95.528468 POINT (-95.52847 29.68995)
2 29.789688 -95.090620 POINT (-95.09062 29.78969)
3 29.559194 -95.145382 POINT (-95.14538 29.55919)
4 29.817562 -95.720350 POINT (-95.72035 29.81756)
shape3.head()
geometry flood2
2 polyGON ((-96.17839 29.80032,-96.17845 29.800... AV
3 polyGON ((-96.14694 30.05772,-96.14695 30.057... AV
5 polyGON ((-96.09303 29.65069,-96.09323 29.650... AV
8 polyGON ((-96.46769 29.86118,-96.46754 29.861... AV
10 polyGON ((-96.42751 29.84184,-96.42716 29.841... AV
编辑：没有截断，多边形看起来像这样
geometry flood2
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检查它们是否具有相同的 CRS
points.crs
<Geographic 2D CRS: epsg:4269>
Name: NAD83
Axis Info [ellipsoidal]:
- Lat[north]: Geodetic latitude (degree)
- Lon[east]: Geodetic longitude (degree)
Area of Use:
- name: north America - NAD83
- bounds: (167.65,14.92,-47.74,86.46)
Datum: north American Datum 1983
- Ellipsoid: GRS 1980
- Prime Meridian: Greenwich
shape3.crs
<Geographic 2D CRS: epsg:4269>
Name: NAD83
Axis Info [ellipsoidal]:
- Lat[north]: Geodetic latitude (degree)
- Lon[east]: Geodetic longitude (degree)
Area of Use:
- name: north America - NAD83
- bounds: (167.65,86.46)
Datum: north American Datum 1983
- Ellipsoid: GRS 1980
- Prime Meridian: Greenwich
计算距离：
dist_point_pol = points.boundary.distance(shape3.geometry)
UserWarning: Geometry is in a geographic CRS. Results from 'distance' are likely incorrect. Use 'GeoSeries.to_crs()' to re-project geometries to a projected CRS before this operation.
我重新投影使用
points2 = points.to_crs(epsg=3857)
shape4 = shape3.to_crs(epsg=3857)
dist_point_pol2 = points2.boundary.distance(shape4.geometry)
UserWarning: The indices of the two GeoSeries are different.
warn("The indices of the two GeoSeries are different.")
我认为 .reset_index() 之类的东西可以解决这个问题，但事实并非如此。最后，我要么得到没有意义的东西，要么只是NaN。
我可能弄错了函数。我实际上需要一些东西来计算从每个点到最近的洪水区的距离（因此，不像 .distance 假设的那样对齐，但是添加 align=False 会给我一条错误消息，好像 align 不是预期的输入），从 this 的答案我认为 .boundary.distance() 是正确的，但似乎并非如此。
解决方法
为了回答您的问题，我们使用了 nearest_points 的 shapely 函数。
- 准备一些数据：
df1 = points.to_crs(epsg=3857)
df2 = shape3.to_crs(epsg=3857)
# Extract point from you GeoDataFrame
point = df1.geometry
# Compute centroid of your area
centroid = df2.centroid
# Create unary union from centroid of area
centroids = unary_union(centroid)
>>> point.to_wkt()
0 POINT (-10608848.77333559 3487304.924649447)
1 POINT (-10634180.63666051 3463757.741176611)
2 POINT (-10585439.39761668 3476544.997885468)
3 POINT (-10591535.25293251 3447012.74414724)
4 POINT (-10655540.62055392 3480120.372628092)
dtype: object
>>> centroid.to_wkt()
2 POINT (-10706645.37489606 3477794.283230069)
3 POINT (-10703147.61430024 3510692.1568683)
5 POINT (-10697256.38062128 3458615.305393973)
8 POINT (-10737620.52149788 3485241.842837139)
10 POINT (-10734474.301016 3483131.295266777)
dtype: object
>>> centroids.to_wkt()
'MULTIPOINT (-10737620.5214978829026222 3485241.8428371394984424,-10734474.3010160010308027 3483131.2952667772769928,-10706645.3748960644006729 3477794.2832300686277449,-10703147.6143002416938543 3510692.1568683004006743,-10697256.3806212786585093 3458615.3053939728997648)'
- 计算每个
point和centroids之间的距离
out = point.apply(lambda p: centroid == nearest_points(p,centroids)[1])
nearest_points 返回一个元组，在本例中：
-
[0]→POINTp -
[1]→POINT中最近的MULTIPOINT
只剩下在质心列表中找到POINT[1]，所以我们有以下交叉表：
>>> out
2 3 5 8 10
0 False False True False False
1 False False True False False
2 False False True False False
3 False False True False False
4 False False True False False
>>> out.idxmax(axis="columns")
0 5
1 5
2 5
3 5
4 5
dtype: int64
单线
>>> df1.apply(lambda p: df2.centroid == \
nearest_points(p.geometry,df2.centroid.unary_union)[1],axis="columns").idxmax(axis="columns")
0 5
1 5
2 5
3 5
4 5
dtype: int64
编辑：另一种方法
df1 = points.to_crs(epsg=3857)
df2 = shape3.to_crs(epsg=3857)
point = df1.geometry
polygon = df2.geometry
out = point.apply(lambda p: polygon.distance(p))
>>> out
2 3 5 8 10
0 98131.524964 96824.778253 92693.462319 127194.722351 125043.444702
1 73719.669946 83033.617116 63047.492025 104004.590418 101372.084859
2 121097.679806 122221.621814 113001.243242 150810.634442 148482.376394
3 119072.273823 128120.650093 106140.417246 149361.810890 146630.045936
4 51036.728734 56191.958501 46675.287835 80619.485414 78297.679404
>>> out.idxmin(axis="columns")
0 5
1 5
2 5
3 5
4 5
dtype: int64