有没有一种方法可以用这个代码在列表中循环

Question

基本上，我有一个大约5000个纬度和经度的列表，我想循环并分别运行这个查询。我一直在手工操作并单独运行查询，但我认为必须有一种方法来标准化流程。非常感谢您的帮助或想法，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)

Answer 1

places_result = gmaps.places(query=query, location=(latitude1, longitude1), radius=32186.9)
def get_field_value(result, field):
try:
for f in field.split('/'):
result = result[f]
return result
except KeyError:
return 'na'
col_field = [ ('name', 'name'),
('rating', 'rating')
('address', 'formatted_address')
('website', 'website')
('phone', 'formatted_phone_number')
('status', 'business_status')
('hours', 'opening_hours')
('lat', 'geometry/location/lat')
('lng', 'geometry/location/lng')
]
fields = [e[1] for e in col_field]
for place in places_result['results']:
place_details_result = gmaps.place(place_id=place['place_id'], fields=fields)['result']
df1 = { column: [get_field_value(place_details_result, field)] for column, field in col_field }

Answer 2

首先，关于您的尝试，除了块，您可以用重写那些

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"。

你可以用字典压缩：来进一步精简它

keys = ["name", "rating", "address", "website", "phone",
"status", "hours", "lat", "lng"]
entry = {k: result.get(k, 'na') for k in keys}

假设你有两个列表中的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