将坐标列表转换为点几何

Question

我的地质公园df看起来是这样的：

   events   datum   pz  ...  dst        Aktenzahl              geometry
0      22  2015.0  204  ...  BSm  BSm/204/30/2015  [16.34203, 48.21717]
1      22  2015.0  202  ...  BSm   BSm/202/8/2015  [16.34189, 48.21719]

如您所见，几何学列是一个包含经度和纬度值的列表。有没有一种有效的方法将它们转换成点？

这是可行的，但效率很低：

pts = gpd.GeoDataFrame(geometry=gpd.points_from_xy(
[df['geometry'][0][0], df['geometry'][1][0]], 
[df['geometry'][0][1], df['geometry'][1][1]]))

df["geometry"] = pts
df = gpd.GeoDataFrame(df, crs='epsg:4326').set_geometry('geometry')
df

   events   datum   pz  ...  dst        Aktenzahl                   geometry
0      22  2015.0  204  ...  BSm  BSm/204/30/2015  POINT (16.34203 48.21717)
1      22  2015.0  202  ...  BSm   BSm/202/8/2015  POINT (16.34189 48.21719)

当我把它放在循环中时，我得到了TypeError: len() of unsized object

for i in range(len(df['geometry'])):
  print(df['geometry'][i][0], df['geometry'][i][1]) # works
  print(gpd.GeoDataFrame(geometry=gpd.points_from_xy(df['geometry'][i][0], df['geometry'][i][1]))) # TypeError: len() of unsized object

基本上，我需要this问题的逆。

可复制的数据如下：

df={'events': {0: 22, 1: 22}, 'datum': {0: 2015.0, 1: 2015.0}, 'pz': {0: 204, 1: 202}, 'strasse': {0: 'Hernalser Gürtel', 1: 'Hernalser Gürtel'}, 'hnr': {0: 'unter U6', 1: 'unter U6'}, 'dst': {0: 'BSm', 1: 'BSm'}, 'Aktenzahl': {0: 'BSm/204/30/2015', 1: 'BSm/202/8/2015'}, 'geometry': {0: [16.34203, 48.21717], 1: [16.34189, 48.21719]}}

Answer 1

我不知道你为什么要做个循环。

df={'events': {0: 22, 1: 22}, 'datum': {0: 2015.0, 1: 2015.0}, 'pz': {0: 204, 1: 202}, 'strasse': {0: 'Hernalser Gürtel', 1: 'Hernalser Gürtel'}, 'hnr': {0: 'unter U6', 1: 'unter U6'}, 'dst': {0: 'BSm', 1: 'BSm'}, 'Aktenzahl': {0: 'BSm/204/30/2015', 1: 'BSm/202/8/2015'}, 'geometry': {0: [16.34203, 48.21717], 1: [16.34189, 48.21719]}}
df = pd.DataFrame(df)
df.geometry = gpd.points_from_xy(df.geometry[0], df.geomety[1])
gdf = gpd.GeoDataFrame(df, geometry='geometry', crs = 'epsg:4326)

所有坐标都更改为点。

你认为这真的是低效的吗？

更新

对不起，我搞错了。我觉得你想要的解决办法就是这个。

df.geometry = gpd.points_from_xy(pd.DataFrame(df.geometry.to_list())[0], pd.DataFrame(df.geometry.to_list())[1])