反向地理编码（无谷歌API）；邮政编码

Question

将美国地区的地理代码“纬度”和“经度”反译为邮政编码的代码；最初用于确定纽约市枪击事件的邮政编码

Answer 1

下面是另一个解决方案（包括注释代码）：https://medium.com/@moritz.kittler/ever-struggled-with-reverse-geo-coding-36fe948ad5a3

Answer 2

示例输出：

    lat         lon         zipcode
0   40.896504   -73.859042  10470
1   40.732804   -74.005666  10014
2   40.674142   -73.936206  11213
3   40.648025   -73.904011  11236
4   40.764694   -73.914348  11103
... ... ... ...
20654   40.710989   -73.942949  11211
20655   40.682398   -73.840079  11416
20656   40.651014   -73.945707  11226
20657   40.835990   -73.916276  10452
20658   40.857771   -73.894606  10458

加载数据集（非必需）：

#load used dataset
df_shooting = pd.read_csv('Shooting_NY.csv',sep=';',low_memory=False)

反向地理编码代码：

pip install uszipcode

# Import packages
from uszipcode import SearchEngine
search = SearchEngine(simple_zipcode=True)
from uszipcode import Zipcode
import numpy as np

#define zipcode search function
def get_zipcode(lat, lon):
    result = search.by_coordinates(lat = lat, lng = lon, returns = 1)
    return result[0].zipcode

#load columns from dataframe
lat = df_shooting['Latitude']
lon = df_shooting['Longitude']

#define latitude/longitude for function
df = pd.DataFrame({'lat':lat, 'lon':lon})

#add new column with generated zip-code
df['zipcode'] = df.apply(lambda x: get_zipcode(x.lat,x.lon), axis=1)

#print result
print(df)

#(optional) save as csv
#df.to_csv(r'zip_codes.csv')

注意长时间运行（20k行=5-7分钟）。然而，我们在不利用（付费）谷歌API的情况下设法找到了最有效的代码

Answer 3

这是我的代码，我认为它更容易一点：

# !pip install uszipcode

# Import packages
from uszipcode import SearchEngine
search = SearchEngine(simple_zipcode=True)
from uszipcode import Zipcode

# Define zipcode search function
for index, row in df.iterrows():
    result = search.by_coordinates(lat = row[df lat column number], lng = row[df lon column number], returns = 1)
    zip = result[0].zipcode

# Add zipcode to the dataframe
df["Zipcode"] = zip

# Save dataframe to csv file (specify path)
df.to_csv("Resouces/df.csv", index=False)


# You can also use itertuples().  It is really faster than iterrows()
# Your for loop may change like the following

for row in df.itertuples(index = False):
     # follow remaining code explained above