python - 有没有办法用这段代码遍历一个列表?
问题描述
基本上,我有一个大约 5,000 个纬度和经度的列表,我想分别循环并运行这个查询。我一直在手动执行此操作并单独运行查询,但我想必须有一种方法来标准化该过程。非常感谢任何帮助或想法,谢谢!
df = []
places_result = gmaps.places(query=query, location=(latitude1, longitude1), radius=32186.9, )
for place in places_result['results']:
my_place_id = place['place_id']
my_fields = ['name', 'business_status', 'formatted_address', 'opening_hours', 'rating', 'website',
'formatted_phone_number', 'geometry/location/lng', 'geometry/location/lat']
place_details = gmaps.place(place_id=my_place_id, fields=my_fields)
column_names = ['name', 'rating', 'address', 'website', 'phone', 'status', 'hours', 'lat', 'lng']
# df=pd.DataFrame(columns = column_names)
try:
rating = place_details['result']['rating']
except KeyError:
rating = 'na'
try:
name = place_details['result']['name']
except KeyError:
name = "na"
try:
address = place_details['result']['formatted_address']
except KeyError:
address = "na"
try:
website = place_details['result']['website']
except KeyError:
website = "na"
try:
phone = place_details['result']['formatted_phone_number']
except KeyError:
phone = "na"
try:
lat = place_details['result']['geometry']['location']['lat']
except KeyError:
lat = "na"
try:
lng = place_details['result']['geometry']['location']['lng']
except KeyError:
lng = "na"
try:
status = place_details['result']['business_status']
except KeyError:
status = "na"
try:
hours = place_details['result']['opening_hours']
except KeyError:
hours = 'na'
df1 = {"name": [name, ], "rating": [rating, ], "address": [address, ], "website": [website, ], "phone": [phone, ],
"status": [status, ], "hours": [hours, ], "lat": [lat, ], "lng": [lng, ]}
data: DataFrame = pd.DataFrame(data=df1)
df.append(data)
dfr = pd.concat(df)
解决方案
首先,关于你的尝试,除了块,你可以用以下方式重写那些:
result = place_details['result']
rating = result.get('rating', 'na')
name = result.get('name', 'na')
...
等等。dict.get(key, default=None) 方法将避免引发 KeyError 并且您可以在您的情况下提供默认值“na”。
您可以通过字典理解进一步缩小它:
keys = ["name", "rating", "address", "website", "phone",
"status", "hours", "lat", "lng"]
entry = {k: result.get(k, 'na') for k in keys}
假设您在两个列表中有您的 lat 和 lon 值:
entries = []
for lat, lon in zip(latitudes, longitudes):
q = gmaps.places(query=query, location=(lat, lon), radius=32186.9,)
result = q.get('result', {}) # returns an empty dict if the query fails
entry = {k: result.get(k, 'na') for k in keys}
entries.append(entry)
df = pd.DataFrame(entries) # create a pandas dataframe
推荐阅读
- python - 如何使用 Python 为 RL 生成所有可能的组合?
- javascript - 由于超时逻辑,分支覆盖失败
- arrays - 结构问题 C(声明)
- java - 在 Java 中使用 Sendgrid X-SMTPAPI
- solr - Solr API 创建 copyField: "copyField source :'[title, director... description]' 不是 glob 并且不匹配任何显式字段或 dynamicField
- powershell - PowerShell Select-Object:将 -Unique 与 First/Last/Skip/Index 结合使用
- node.js - 部署 firebase 功能 - 没有创建 Firebase 应用“[DEFAULT]”
- c# - Automapper - 在两个类之间创建一个简单的映射并对特定类型使用自定义转换器
- python - 如何计算嵌套列表中的元素
- c# - 如何从 ViewModel 访问/更改按钮的内容