使用geopy pandas坐标的新列

Question

我有一个df:

import pandas as pd
import numpy as np
import datetime as DT
import hmac
from geopy.geocoders import Nominatim
from geopy.distance import vincenty

df


     city_name  state_name  county_name
0    WASHINGTON  DC  DIST OF COLUMBIA
1    WASHINGTON  DC  DIST OF COLUMBIA
2    WASHINGTON  DC  DIST OF COLUMBIA
3    WASHINGTON  DC  DIST OF COLUMBIA
4    WASHINGTON  DC  DIST OF COLUMBIA
5    WASHINGTON  DC  DIST OF COLUMBIA
6    WASHINGTON  DC  DIST OF COLUMBIA
7    WASHINGTON  DC  DIST OF COLUMBIA
8    WASHINGTON  DC  DIST OF COLUMBIA
9    WASHINGTON  DC  DIST OF COLUMBIA

我想得到下面数据框中任何一列的纬度和经度坐标.在处理各个位置的文档时,文档(http://geopy.readthedocs.org/en/latest/#data)非常简单.

>>> from geopy.geocoders import Nominatim
>>> geolocator = Nominatim()
>>> location = geolocator.geocode("175 5th Avenue NYC")
>>> print(location.address)
Flatiron Building, 175, 5th Avenue, Flatiron, New York, NYC, New York,     ...
>>> print((location.latitude, location.longitude))
(40.7410861, -73.9896297241625)
>>> print(location.raw)
{'place_id': '9167009604', 'type': 'attraction', ...}

但是我想将函数应用于df中的每一行并创建一个新列.我尝试了以下内容

df['city_coord'] = geolocator.geocode(lambda row: 'state_name' (row))

但我想我的代码中缺少一些东西,因为我得到以下内容:

    city_name   state_name  county_name coordinates
0    WASHINGTON  DC  DIST OF COLUMBIA    None
1    WASHINGTON  DC  DIST OF COLUMBIA    None
2    WASHINGTON  DC  DIST OF COLUMBIA    None
3    WASHINGTON  DC  DIST OF COLUMBIA    None
4    WASHINGTON  DC  DIST OF COLUMBIA    None
5    WASHINGTON  DC  DIST OF COLUMBIA    None
6    WASHINGTON  DC  DIST OF COLUMBIA    None
7    WASHINGTON  DC  DIST OF COLUMBIA    None
8    WASHINGTON  DC  DIST OF COLUMBIA    None
9    WASHINGTON  DC  DIST OF COLUMBIA    None

希望使用Lambda函数我希望这样的东西:

     city_name  state_name  county_name  city_coord
0    WASHINGTON  DC  DIST OF COLUMBIA    38.8949549, -77.0366456 
1    WASHINGTON  DC  DIST OF COLUMBIA    38.8949549, -77.0366456 
2    WASHINGTON  DC  DIST OF COLUMBIA    38.8949549, -77.0366456 
3    WASHINGTON  DC  DIST OF COLUMBIA    38.8949549, -77.0366456 
4    WASHINGTON  DC  DIST OF COLUMBIA    38.8949549, -77.0366456 
5    WASHINGTON  DC  DIST OF COLUMBIA    38.8949549, -77.0366456 
6    WASHINGTON  DC  DIST OF COLUMBIA    38.8949549, -77.0366456 
7    WASHINGTON  DC  DIST OF COLUMBIA    38.8949549, -77.0366456 
8    WASHINGTON  DC  DIST OF COLUMBIA    38.8949549, -77.0366456 
9    WASHINGTON  DC  DIST OF COLUMBIA    38.8949549, -77.0366456
10   GLYNCO      GA  GLYNN               31.2224512, -81.5101023

我感谢任何帮助.在我得到坐标后,我想要映射它们.任何推荐的映射坐标资源也非常受欢迎.谢谢

Answer 1

您可以调用apply并传递要在每行上执行的函数,如下所示:

In [9]:

geolocator = Nominatim()
df['city_coord'] = df['state_name'].apply(geolocator.geocode)
df
Out[9]:
    city_name state_name       county_name  \
0  WASHINGTON         DC  DIST OF COLUMBIA   
1  WASHINGTON         DC  DIST OF COLUMBIA   

                                          city_coord  
0  (District of Columbia, United States of Americ...  
1  (District of Columbia, United States of Americ...  

然后,您可以访问纬度和经度属性:

In [16]:

df['city_coord'] = df['city_coord'].apply(lambda x: (x.latitude, x.longitude))
df
Out[16]:
    city_name state_name       county_name                       city_coord
0  WASHINGTON         DC  DIST OF COLUMBIA  (38.8937154, -76.9877934586326)
1  WASHINGTON         DC  DIST OF COLUMBIA  (38.8937154, -76.9877934586326)

或者通过apply两次调用在一个班轮中完成:

In [17]:
df['city_coord'] = df['state_name'].apply(geolocator.geocode).apply(lambda x: (x.latitude, x.longitude))
df

Out[17]:
    city_name state_name       county_name                       city_coord
0  WASHINGTON         DC  DIST OF COLUMBIA  (38.8937154, -76.9877934586326)
1  WASHINGTON         DC  DIST OF COLUMBIA  (38.8937154, -76.9877934586326)

此外,您的尝试geolocator.geocode(lambda row: 'state_name' (row))没有做任何事情,因此为什么您有一个充满None值的列

编辑

@leb在这里提出了一个有趣的观点,如果你有许多重复值,那么对每个唯一值进行地理编码会更加高效,然后添加:

In [38]:
states = df['state_name'].unique()
d = dict(zip(states, pd.Series(states).apply(geolocator.geocode).apply(lambda x: (x.latitude, x.longitude))))
d

Out[38]:
{'DC': (38.8937154, -76.9877934586326)}

In [40]:    
df['city_coord'] = df['state_name'].map(d)
df

Out[40]:
    city_name state_name       county_name                       city_coord
0  WASHINGTON         DC  DIST OF COLUMBIA  (38.8937154, -76.9877934586326)
1  WASHINGTON         DC  DIST OF COLUMBIA  (38.8937154, -76.9877934586326)

所以上面得到了所有的唯一值unique,从它们构造一个dict,然后调用map执行查找并添加coords,这比尝试按行进行地理编码更有效

Answer 2

投票并接受@EdChum 的回答，我只是想对此进行补充。他的方法非常有效，但从个人经验来看，我想分享一些事情：

处理地理编码时，如果您有多个重复的城市/州组合，则仅发送 1 个进行地理编码，然后将其余的复制到下面的其他行会更快：

这对于大数据非常有帮助，可以通过两种方式完成：

1. 仅基于您的数据，因为行看起来完全重复，并且仅当您愿意时，才删除多余的行并对其中之一执行地理编码。这可以使用以下方法完成drop_duplicate
2. 如果您想保留所有行（group_by城市/州组合），请通过调用 对第一个行应用地理编码head(1)，然后复制到其余行。

原因是每次您致电 Nominatim 时，即使您在同一城市/州连续排队，也会出现小延迟问题。当您的数据变大时，这种小的延迟会变得更糟，导致响应的巨大延迟和可能的超时。

再说一遍，这都是个人处理的。如果现在对您没有好处，请记住以供将来使用。