本文介绍了如何使用geopy从坐标获取邮政编码?的处理方法,对大家解决问题具有一定的参考价值,需要的朋友们下面随着小编来一起学习吧!
问题描述
所以下面是我一直在使用的代码.我有点新手.由于使用该API的配额,我一直仅在数据的开头进行测试.下面是数据框的快照:
So below is the code I have been using. I'm a bit of a newb. I've been testing with just the head of the data because of the quota for using the API. Below is a snapshot of the dataframe:
latitude longitude
0 -73.99107 40.730054
1 -74.000193 40.718803
2 -73.983849 40.761728
3 -73.97499915 40.68086214
4 -73.89488591 40.66471445
这是我被绊倒的地方.
train['latlng'] = train.apply(lambda row: '{},{}'.format(row['latitude'],
row['longitude']), axis=1)
train['geocode_data'] = train['latlng'].map(reverse_geocode)
train['Zip'] =train['latlng'].apply(geolocator.reverse)
train['Zip'].apply(lambda x: pd.Series(x.split(',')))
foo = lambda x: pd.Series([i for i in reversed(x.split(','))])
train['Zip']=train['Zip'].apply(lambda x: str(x))
train['Zip']=train['Zip'].apply(foo)[1]
train
此刻,我遇到一个错误:
At the moment, I'm getting an error that:
AttributeError:位置"对象没有属性拆分"
我该如何拆分位置,以便我可以提取邮政编码?
How would I go about splitting up location so that I can just pick up the zip code?
推荐答案
将 addressdetails = True
传递给Nominatim以获得此信息( https://geopy.readthedocs.io/en/stable/#nominatim )
Pass addressdetails=True
to Nominatim to get this information (https://geopy.readthedocs.io/en/stable/#nominatim)
location = geolocator.geocode(query={'postalcode':'95119'}, addressdetails=True)
print(location)
location.raw
产生
San Jose, Santa Clara County, California, 95119, United States of America
{'place_id': 237756323,
'licence': 'Data © OpenStreetMap contributors, ODbL 1.0. https://osm.org/copyright',
'boundingbox': ['37.072728062914',
'37.392728062914',
'-121.94612695296',
'-121.62612695296'],
'lat': '37.23272806291426',
'lon': '-121.78612695295647',
'display_name': 'San Jose, Santa Clara County, California, 95119, United States of America',
'class': 'place',
'type': 'postcode',
'importance': 0.33499999999999996,
'address': {'city': 'San Jose',
'county': 'Santa Clara County',
'state': 'California',
'postcode': '95119',
'country': 'United States of America',
'country_code': 'us'}}
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