提问人:Fletch 提问时间:4/21/2021 更新时间:11/9/2023 访问量:859
Geopy Google v3 - 从location.raw中提取地址组件
Geopy Google v3 - Extracting address components from location.raw
问:
我有这个 python 脚本,我从 SQL 表中获取地址列表,然后使用 Geopy 将它们传递给 googles api,对它们进行地理编码,然后将数据写回不同的 SQL 表。
我目前被困在尝试从address_components中提取地址部分。
我已经尝试了很多事情,例如转换为 Json,使用其他地址解析器(但我不在美国),但我的 python 并不强大。我也不能直接引用列表部分,因为不同的地址会有不同的长度,所以当我稍后将其应用于数据帧时,它会失败,因为列表的长度并不完全相同。例如location.raw
loc_raw0.append(location.raw['address_components'][0]['long_name'])
我目前正在尝试使用嵌套的 For 循环来获取街道号码,然后复制其他部分。
发生的事情是相等的,但会相等,而不仅仅是.k
'address_components'
v
'{'long_name': '46', 'short_name': '46', 'types': ['street_number']}'
'street_number'
for k, v in location.raw.items():
if k == 'address_components' and 'street_number' in v:
loc_street_number.append(location.raw['long_name'])
print(loc_street_number)
df = pd.DataFrame(SQL_Query, columns=['address'])
loc_Inputaddress = []
loc_Longitude = []
loc_Latitude = []
loc_Matchedaddress = []
loc_subpremise = []
loc_street_number = []
loc_road = []
loc_locality = []
loc_AdminArea1 = []
loc_AdminArea2 = []
loc_postcode = []
loc_type = []
for address in df.address:
try:
inputAddress = address
location = g.geocode(inputAddress, timeout=15)
loc_Inputaddress.append(inputAddress)
loc_Longitude.append(location.longitude)
loc_Latitude.append(location.latitude)
loc_Matchedaddress.append(location.address)
loc_type.append(location.raw['types'][0])
#get address type
print(loc_type.append(location.raw['types'][0]))
#print(location.raw['address_components'])
for k, v in location.raw.items():
if k == 'address_components' and 'street_number' in v:
loc_street_number.append(location.raw['long_name'])
print(loc_street_number)
except Exception as e:
print('Error, skipping address...', e)
答:
0赞
Hernán Alarcón
4/21/2021
#1
在您的代码中,是一个字典列表,据我了解,您想要具有类型的字典。此示例应帮助您:v
long_name
street_number
v = [{
"long_name": "40",
"short_name": "40",
"types": ["subpremise"]
},
{
"long_name": "46",
"short_name": "46",
"types": ["street_number"]
},
{
"long_name": "Aongatete",
"short_name": "Aongatete",
"types": ["locality", "political"]
}]
# Iterate over v because it is a list
for address_component in v:
# Check if one of the address component types is "street_number"
if "street_number" in address_component["types"]:
print(address_component["long_name"])
break
# Output:
# 46
评论
0赞
Fletch
4/21/2021
非常感谢,一切都很好。我花了 1.5 天的时间尝试不同的东西,但我终于可以把这些数据拿出来了!
0赞
Fletch
4/21/2021
例如,在某些情况下,地址可能没有门牌号 - 如果没有要添加的数据,您有什么技巧可以保持数组的长度相同吗?
0赞
N8-
11/9/2023
#2
下面是一个函数,用于根据 geocodezip 的 js 示例提取地址组件,在这里找到:
def extract_address_details(address_components):
"""
extract_address_details extracts address parts from the details of the google maps api response
:param address_components: a dict representing the details['address_components'] response from the google maps api
:return: a dict of the address components
"""
# set up the loop parameters for each component
count = len(address_components)
looplist = range(0, count)
#loop through the indices of the address components
for i in looplist:
#set up the loop parameters for the component types
tcount = len(address_components[i]['types'])
tlooplist = range(0, tcount)
#loop through the indices of the address component types
for t in tlooplist:
#match the type, pull the short_name from the appropriate component as a string
match address_components[i]['types'][t]:
case "subpremise":
subpremise = str(address_components[i]['short_name'])
case "street_number":
street_number = str(address_components[i]['short_name'])
case "route":
route = str(address_components[i]['short_name'])
case "locality":
city = str(address_components[i]['short_name'])
case "administrative_area_level_1":
state = str(address_components[i]['short_name'])
case "postal_code":
postal_code = str(address_components[i]['short_name'])
#assemble the street address
address1 = street_number + " " + route + " " + subpremise
#populate the return values
data = {
'address1': address1,
'city': city,
'state': state,
'postal_code': postal_code
}
return data
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