I have some JSON data coming from the API as a list of dictionaries, such as:
entities = [
{'name': 'McDonalds', 'city': 'New York', 'gross': 250000000, 'id': '000001'},
{'name': 'McDonalds', 'city': 'Philadelphia', 'gross': 190000000, 'id': '000002'},
{'name': 'Shake Shack', 'city': 'Los Angeles', 'gross': 17000000, 'id': '000003'},
{'name': 'In-N-Out Burger', 'city': 'Houston', 'gross': 23000000, 'id': '000004'},
{'name': 'In-N-Out Burger', 'city': 'Atlanta', 'gross': 12000000, 'id': '000005'},
{'name': 'In-N-Out Burger', 'city': 'Dallas', 'gross': 950000, 'id': '000006'},
]
I'm trying to group all the entries with the same name into another list of dictionaries named for whatever business it is.
def group_entities(entities):
entity_groups = []
# Establish a blank list for each unique name
for entity in entities:
entity['name'] = []
entity_groups.append(entity['name'])
# Within each business's list, add separate dictionaries with details
for entity in entities:
entity['name'].append({
'name':entity['name'],
'city':entity['city'],
'gross':entity['gross'],
'id':entity['id']
})
entity_groups.extend(entity['name'])
return entity_groups
I can't use entity['name'] as a variable name because it just changes the original value nor can I use a string version of the name. I want to end up with data I can iterate and display like:
Business
• All City 1 Dictionary Values
• All City 2 Dictionary Values, etc
Business
• All City 1 Dictionary Values
• All City 2 Dictionary Values, etc
I'm at a loss as to how to even do further research on this because I don't know proper 'googleable' terms to describe what I am trying to do.
If your data is ordered by name:
from itertools import groupby
from operator import itemgetter
entities = [
{'name': 'McDonalds', 'city': 'New York', 'gross': 250000000, 'id': '000001'},
{'name': 'McDonalds', 'city': 'Philadelphia', 'gross': 190000000, 'id': '000002'},
{'name': 'Shake Shack', 'city': 'Los Angeles', 'gross': 17000000, 'id': '000003'},
{'name': 'In-N-Out Burger', 'city': 'Houston', 'gross': 23000000, 'id': '000004'},
{'name': 'In-N-Out Burger', 'city': 'Atlanta', 'gross': 12000000, 'id': '000005'},
{'name': 'In-N-Out Burger', 'city': 'Dallas', 'gross': 950000, 'id': '000006'},
]
data = [{k: list(v)} for k, v in groupby(entities, itemgetter("name"))]
Which would give you:
[{'McDonalds': [{'id': '000001', 'city': 'New York', 'name': 'McDonalds', 'gross': 250000000}, {'id': '000002', 'city': 'Philadelphia', 'name': 'McDonalds', 'gross': 190000000}]}, {'Shake Shack': [{'id': '000003', 'city': 'Los Angeles', 'name': 'Shake Shack', 'gross': 17000000}]}, {'In-N-Out Burger': [{'id': '000004', 'city': 'Houston', 'name': 'In-N-Out Burger', 'gross': 23000000}, {'id': '000005', 'city': 'Atlanta', 'name': 'In-N-Out Burger', 'gross': 12000000}, {'id': '000006', 'city': 'Dallas', 'name': 'In-N-Out Burger', 'gross': 950000}]}]
Or if you don't want the name:
keys = ("id","gross", "city")
data = [{k: [dict(zip(keys, itemgetter(*keys)(dct))) for dct in v]} for k, v in groupby(entities, itemgetter("name"))]
If the data is not ordered you can use a defaultdict:
from collections import defaultdict
d = defaultdict(list)
for entity in entities:
d[entity["name"]].append(dict(entity))
print([{k: v} for k,v in d.items()])
Again you cab remove the name or maybe you wan to use the original dicts and you don't mind mutating them :
from collections import defaultdict
d = defaultdict(list)
for entity in entities:
d[entity.pop("name")].append(entity)
print([{k: v} for k,v in d.items()])
That will give you:
[{'Shake Shack': [{'id': '000003', 'city': 'Los Angeles', 'gross': 17000000}]}, {'McDonalds': [{'id': '000001', 'city': 'New York', 'gross': 250000000}, {'id': '000002', 'city': 'Philadelphia', 'gross': 190000000}]}, {'In-N-Out Burger': [{'id': '000004', 'city': 'Houston', 'gross': 23000000}, {'id': '000005', 'city': 'Atlanta', 'gross': 12000000}, {'id': '000006', 'city': 'Dallas', 'gross': 950000}]}]
It all depends on whether you want to use the original dicts again and/or if you want the names kept in the dicts. You can combine parts of the logic to get whatever format you like.
This should work:
def group_entities(entities):
entity_groups = {}
# Within each business's list, add separate dictionaries with details
for entity in entities:
name = entity['name'] # name is the key for entity_groups
del entity['name'] # remove it from each entity
# add the entity to the entity_groups with the key (name)
entity_groups[name] = entity_groups.get(name, []) + [entity]
return entity_groups
If you want to keep the entity name in each entity, remove the del statement.
bycompany = {}
for ent in entities:
if not ent['name'] in bycompany:
# if there is no location list for this company name,
# then start a new list for this company.
bycompany[ent['name']] = []
# Add the dict to the list of locations for this company.
bycompany[ent['name']].append(ent)
Related
I am using Python and I am trying to access the result of function PlacesAPI where I can see supermarkets around me and create a dataframe with few parts of each dictionary inside the main dictionary using a for loop, however I am getting the same information for different rows.
Can you please help me to put each different parts of dictionary in a different row?
Here is my code and the result (now reproducible):
from herepy import PlacesApi
import pandas as pd
def dataframe():
a = {'items': [{'title': 'Marcos Francisco dos Santos Padaria e Mercearia',
'id': 'here:pds:place:07675crc-dfd72cbf57bd45cc9277ed8530ffd61b',
'ontologyId': 'here:cm:ontology:supermarket',
'resultType': 'place',
'address': {'label': 'Marcos Francisco dos Santos Padaria e Mercearia, Rua Oswero Carmo Vilaça, 33, Petrópolis - RJ, 25635-101, Brazil',
'countryCode': 'BRA',
'countryName': 'Brazil',
'stateCode': 'RJ',
'state': 'Rio de Janeiro',
'city': 'Petrópolis',
'district': 'Petrópolis',
'street': 'Rua Oswero Carmo Vilaça',
'postalCode': '25635-101',
'houseNumber': '33'},
'position': {'lat': -22.5315, 'lng': -43.16904},
'access': [{'lat': -22.5314, 'lng': -43.16914}],
'distance': 134,
'categories': [{'id': '600-6300-0066', 'name': 'Grocery', 'primary': True},
{'id': '600-6300-0244', 'name': 'Bakery & Baked Goods Store'}],
'contacts': [{'phone': [{'value': '+552422312493'}]}]},
{'title': 'Mr. Frango',
'id': 'here:pds:place:076jx7ps-7c214f50052f0c23c9e5422ebde7d3cd',
'ontologyId': 'here:cm:ontology:supermarket',
'resultType': 'place',
'address': {'label': 'Mr. Frango, Rua Teresa, Petrópolis - RJ, 25635-530, Brazil',
'countryCode': 'BRA',
'countryName': 'Brazil',
'stateCode': 'RJ',
'state': 'Rio de Janeiro',
'city': 'Petrópolis',
'district': 'Petrópolis',
'street': 'Rua Teresa',
'postalCode': '25635-530'},
'position': {'lat': -22.52924, 'lng': -43.17222},
'access': [{'lat': -22.52925, 'lng': -43.1722}],
'distance': 545,
'categories': [{'id': '600-6300-0066', 'name': 'Grocery', 'primary': True},
{'id': '600-6000-0061', 'name': 'Convenience Store'}],
'references': [{'supplier': {'id': 'core'}, 'id': '1159487213'}],
'contacts': [{'phone': [{'value': '+552422201010'},
{'value': '+552422315720', 'categories': [{'id': '600-6000-0061'}]}]}]},
{'title': 'Mercadinho Flor de Petrópolis',
'id': 'here:pds:place:07675crc-6b03dfbac65a45c0bfc52ab9a3f04556',
'ontologyId': 'here:cm:ontology:supermarket',
'resultType': 'place',
'address': {'label': 'Mercadinho Flor de Petrópolis, Rua Teresa, 2060, Petrópolis - RJ, 25635-530, Brazil',
'countryCode': 'BRA',
'countryName': 'Brazil',
'stateCode': 'RJ',
'state': 'Rio de Janeiro',
'city': 'Petrópolis',
'district': 'Petrópolis',
'street': 'Rua Teresa',
'postalCode': '25635-530',
'houseNumber': '2060'},
'position': {'lat': -22.52895, 'lng': -43.17233},
'access': [{'lat': -22.52895, 'lng': -43.17219}],
'distance': 574,
'categories': [{'id': '600-6300-0066',
'name': 'Grocery',
'primary': True}]}]}
value = []
address = []
latlong = []
teste = pd.DataFrame(columns = ['nome','endereco','rua','numero',
'cidade','estado','cep','lat','long','raio'])
teste['nome'] = []
teste['endereco'] = []
teste['rua'] = []
teste['numero'] =[]
teste['cidade'] = []
teste['estado'] = []
teste['cep'] = []
teste['lat'] = []
teste['long'] = []
teste['raio'] = []
g = pd.DataFrame.from_dict(a.values())
h =[]
for i in range(3):
v = g[i].values[0]
h = v.items()
for k, l in h:
value.append(l)
for c, d in value[4].items():
address.append(d)
for la, lo in value[5].items():
latlong.append(lo)
novo_concorrente = {'nome': value[0], 'endereco':address[0],
'rua':address[7], 'numero':address[9],
'cidade':address[5], 'estado':address[3],
'cep':address[8],'lat':latlong[0],
'long':latlong[1],'raio':value[7]}
teste = teste.append(novo_concorrente, ignore_index=True)
return teste
You should focus on your for loop. I would suggest you to create a dictionary for each row you want to define in your final DataFrame, and then create a list to append those dictionaries to.
In example:
rows = []
for item in a["items"]:
row = {
"Latitude": item["position"]["lat"],
"Postal code": item["address"]["postalCode"],
}
rows.append(row)
result = DataFrame(rows)
result
Hope it helps as a starting point.
I have the following as part of a function to extract some info from a json response (at the bottom you can find a portion of it), and is working fine. But......
venues_list=[]
url = 'https://api.foursquare.com/v2/venues/search?client_id={}&client_secret={}&ll={},{}&v={}&categoryId={}&radius={}&limit={}'.format(CLIENT_ID, CLIENT_SECRET, lat, lng, VERSION, search_categoryId, radius, LIMIT)
results = requests.get(url).json()["response"]['venues']
venues_list.append([(
v['name'],
v['location']['lat'],
v['location']['lng'],
v['categories'][0]['name'],
v['id']) for v in results])
To the mix of objects I'm retrieving from the json file, i want to add:
v['location']['postalCode']
But is not working. I get a KeyError: 'postalCode'
If I do:
v['location']['distance']
v['location']['country']
v['location']['formattedAddress']
It works. I get no error.
These don't work either:
v['location']['cc']
v['location']['city']
I get the same KeyError: 'cc' , KeyError: 'city'
Is there something I'm missing? Can you help me understand why it behaves this way?
[{'id': '4ad4c061f964a52099f720e3',
'name': 'Live Organic Food Bar',
'location': {'address': '264 Dupont Street',
'lat': 43.67505287052667,
'lng': -79.40671518307245,
'labeledLatLngs': [{'label': 'display',
'lat': 43.67505287052667,
'lng': -79.40671518307245}],
'distance': 273,
'postalCode': 'M5R 1V7',
'cc': 'CA',
'city': 'Toronto',
'state': 'ON',
'country': 'Canada',
'formattedAddress': ['264 Dupont Street', 'Toronto ON M5R 1V7', 'Canada']},
'categories': [{'id': '4bf58dd8d48988d1d3941735',
'name': 'Vegetarian / Vegan Restaurant',
'pluralName': 'Vegetarian / Vegan Restaurants',
'shortName': 'Vegetarian / Vegan',
'icon': {'prefix': 'https://ss3.4sqi.net/img/categories_v2/food/vegetarian_',
'suffix': '.png'},
'primary': True}],
'referralId': 'v-1591407049',
'hasPerk': False},
fork_fig = {'categories': [{'alias': 'burgers', 'title': 'Burgers'},
{'alias': 'sandwiches', 'title': 'Sandwiches'},
{'alias': 'salad', 'title': 'Salad'}],
'coordinates': {'latitude': 35.10871, 'longitude': -106.56739},
'display_phone': '(505) 881-5293',
'distance': 3571.724649307866,
'id': 'fork-and-fig-albuquerque',
'image_url': 'https://s3-media1.fl.yelpcdn.com/bphoto/_-DpXKfS3jv6DyA47g6Fxg/o.jpg',
'is_closed': False,
'location': {'address1': '6904 Menaul Blvd NE',
'address2': 'Ste C',
'address3': '',
'city': 'Albuquerque',
'country': 'US',
'display_address': ['6904 Menaul Blvd NE', 'Ste C', 'Albuquerque, NM 87110'],
'state': 'NM',
'zip_code': '87110'},
'name': 'Fork & Fig',
'phone': '+15058815293',
'price': '$$',
'rating': 4.5,
'review_count': 604}
frontier_restaurant = {'categories': [{'alias': 'mexican', 'title': 'Mexican'},
{'alias': 'diners', 'title': 'Diners'},
{'alias': 'tradamerican', 'title': 'American (Traditional)'}],
'coordinates': {'latitude': 35.0808088832532, 'longitude': -106.619402244687},
'display_phone': '(505) 266-0550',
'distance': 4033.6583235266075,
'id': 'frontier-restaurant-albuquerque-2',
'image_url': 'https://s3-media4.fl.yelpcdn.com/bphoto/M9L2z6-G0NobuDJ6YTh6VA/o.jpg',
'is_closed': True,
'location': {'address1': '2400 Central Ave SE',
'address2': '',
'address3': '',
'city': 'Albuquerque',
'country': 'US',
'display_address': ['2400 Central Ave SE', 'Albuquerque, NM 87106'],
'state': 'NM',
'zip_code': '87106'},
'name': 'Frontier Restaurant',
'phone': '+15052660550',
'price': '$',
'rating': 4.0,
'review_count': 1369}
I have two restaurant list as above and i want to make function that returns a list of only the restaurants that are not closed by using conditional loops.
restaurants = [fork_fig, frontier_restaurant]
open_restaurants(restaurants)[0]['name'] ###I want restaurant name to be appear
Below is the code that I've been working on and cannot quite sure how I can fix this to get the value that I want to return.
def open_restaurants(restaurants):
selected = []
for i in restaurants:
if fork_fig['is_closed']:
selected = restaurants[1]
else:
selected = restaurants[0]
return selected
Continuing from the comment:
fork_fig = {'categories': [{'alias': 'burgers', 'title': 'Burgers'},
{'alias': 'sandwiches', 'title': 'Sandwiches'},
{'alias': 'salad', 'title': 'Salad'}],
'coordinates': {'latitude': 35.10871, 'longitude': -106.56739},
'display_phone': '(505) 881-5293',
'distance': 3571.724649307866,
'id': 'fork-and-fig-albuquerque',
'image_url': 'https://s3-media1.fl.yelpcdn.com/bphoto/_-DpXKfS3jv6DyA47g6Fxg/o.jpg',
'is_closed': False,
'location': {'address1': '6904 Menaul Blvd NE',
'address2': 'Ste C',
'address3': '',
'city': 'Albuquerque',
'country': 'US',
'display_address': ['6904 Menaul Blvd NE', 'Ste C', 'Albuquerque, NM 87110'],
'state': 'NM',
'zip_code': '87110'},
'name': 'Fork & Fig',
'phone': '+15058815293',
'price': '$$',
'rating': 4.5,
'review_count': 604}
frontier_restaurant = {'categories': [{'alias': 'mexican', 'title': 'Mexican'},
{'alias': 'diners', 'title': 'Diners'},
{'alias': 'tradamerican', 'title': 'American (Traditional)'}],
'coordinates': {'latitude': 35.0808088832532, 'longitude': -106.619402244687},
'display_phone': '(505) 266-0550',
'distance': 4033.6583235266075,
'id': 'frontier-restaurant-albuquerque-2',
'image_url': 'https://s3-media4.fl.yelpcdn.com/bphoto/M9L2z6-G0NobuDJ6YTh6VA/o.jpg',
'is_closed': True,
'location': {'address1': '2400 Central Ave SE',
'address2': '',
'address3': '',
'city': 'Albuquerque',
'country': 'US',
'display_address': ['2400 Central Ave SE', 'Albuquerque, NM 87106'],
'state': 'NM',
'zip_code': '87106'},
'name': 'Frontier Restaurant',
'phone': '+15052660550',
'price': '$',
'rating': 4.0,
'review_count': 1369}
restaurants = [fork_fig, frontier_restaurant]
def open_restaurants(restaurants):
selected = []
for i in restaurants:
if 'is_closed' in i:
if not i['is_closed']:
selected.append(i['name'])
return selected
print(open_restaurants(restaurants))
OUTPUT:
['Fork & Fig']
Shorter-version:
Using list-comprehension:
def open_restaurants(restaurants):
return [x['name'] for x in restaurants if 'is_closed' in x and not x["is_closed"]]
print(open_restaurants(restaurants))
Using get() instead of indexing:
def open_restaurants(restaurants):
return [x['name'] for x in restaurants if 'name' in x and not x.get('is_closed', True)]
print(open_restaurants(restaurants))
Use the filter function:
def open_restaurants(restaurants):
return filter(lambda r: not r['is_closed'], restaurants)
Use List Comprehension here. Its one line and easy to read
open_restaurants = [restaurant.get("name") for restaurant in restaurants if not restaurant.get("is_closed")]
Output: ["Fork & Fig"]
There is a simple way to append all your open restaurants to a new list and print their names.
You will need two loops to accomplish this.
Here is how to do it:
restaurants = [fork_fig, frontier_restaurant] # a list of restaurants
def open_restaurants(restaurants):
selected = []
for i in restaurants: # append all open restaurants to a new list
if not i['is_closed']:
selected.append(i)
return selected
my_open_restaurants = open_restaurants(restaurants) # call the function
for rest in my_open_restaurants: # print all the names
print(rest['name'])
In your example the output will be: Fork & Fig
Here‘s a simplified example of some data I have:
{"id": "1234565", "fields": {"name": "john", "email":"john#example.com", "country": "uk"}}
The wholeo nested dictionary is a bigger list of address data. The goal is to create pairs of people from the list with randomized partners where partners from the same country should be preferd. So my first real issue is to find a good way to group them by that country value.
I‘m sure there‘s a smarter way to do this than iterating through the dict and writing all records out to some new list/dict?
I think this is close to what you need:
result = {key:[i for i in value] for key, value in itertools.groupby(people, lambda item: item["fields"]["country"])}
What this does is use itertools.groupby to group all people in the people list by their specified country. The resulting dictionary has countries as keys, and the unpacked groupings (matching people) as values. Input is expected as a list of dictionaries like the one in your example:
people = [{"id": "1234565", "fields": {"name": "john", "email":"john#example.com", "country": "uk"}},
{"id": "654321", "fields": {"name": "sam", "email":"sam#example.com", "country": "uk"}}]
Sample output:
>>> print(result)
>>> {'uk': [{'fields': {'name': 'john', 'email': 'john#example.com', 'country': 'uk'}, 'id': '1234565'}, {'fields': {'name': 'sam', 'email': 'sam#example.com', 'country': 'uk'}, 'id': '654321'}]}
For a cleaner result, the looping construct can be tweaked so that only the ID of each person is included in the result dict:
result = {key:[i["id"] for i in value] for key, value in itertools.groupby(people, lambda item: item["fields"]["country"])}
>>> print(result)
>>> {'uk': ['1234565', '654321']}
EDIT: Sorry, I forgot about the sorting. Simply sort the list of people by country before putting it through groupby. It should now work properly:
sort = sorted(people, key=lambda item: item["fields"]["country"])
Here is another one that uses defaultdict:
import collections
def make_groups(nested_dicts, nested_key):
default = collections.defaultdict(list)
for nested_dict in nested_dicts:
for value in nested_dict.values():
try:
default[value[nested_key]].append(nested_dict)
except TypeError:
pass
return default
To test the results:
import random
COUNTRY = {'af', 'br', 'fr', 'mx', 'uk'}
people = [{'id': i, 'fields': {
'name': 'name'+str(i),
'email': str(i)+'#email',
'country': random.sample(COUNTRY, 1)[0]}}
for i in range(10)]
country_groups = make_groups(people, 'country')
for country, persons in country_groups.items():
print(country, persons)
Random output:
fr [{'id': 0, 'fields': {'name': 'name0', 'email': '0#email', 'country': 'fr'}}, {'id': 1, 'fields': {'name': 'name1', 'email': '1#email', 'country': 'fr'}}, {'id': 4, 'fields': {'name': 'name4', 'email': '4#email', 'country': 'fr'}}]
br [{'id': 2, 'fields': {'name': 'name2', 'email': '2#email', 'country': 'br'}}, {'id': 8, 'fields': {'name': 'name8', 'email': '8#email', 'country': 'br'}}]
uk [{'id': 3, 'fields': {'name': 'name3', 'email': '3#email', 'country': 'uk'}}, {'id': 7, 'fields': {'name': 'name7', 'email': '7#email', 'country': 'uk'}}]
af [{'id': 5, 'fields': {'name': 'name5', 'email': '5#email', 'country': 'af'}}, {'id': 9, 'fields': {'name': 'name9', 'email': '9#email', 'country': 'af'}}]
mx [{'id': 6, 'fields': {'name': 'name6', 'email': '6#email', 'country': 'mx'}}]
I have a list of dict what some data, and I would like to extract certain key:value pairs into a new list of dicts. I know one way that I could do this would be to use del i['unwantedKey'], however, I would rather not delete any data but instead create a new dict with the needed data.
The column order might change, so I need something to extract the two key:value pairs from the larger dict into a new dict.
Current Data Format
[{'Speciality': 'Math', 'Name': 'Matt', 'Location': 'Miami'},
{'Speciality': 'Science', 'Name': 'Ben', 'Location': 'Las Vegas'},
{'Speciality': 'Language Arts', 'Name': 'Sarah', 'Location': 'Washington DC'},
{'Speciality': 'Spanish', 'Name': 'Tom', 'Location': 'Denver'},
{'Speciality': 'Chemistry', 'Name': 'Jim', 'Location': 'Dallas'}]
Code to delete key:value from dict
import csv
data= []
for line in csv.DictReader(open('data.csv')):
data.append(line)
for i in data:
del i['Speciality']
print data
Desired Data Format without using del i['Speciality']
[{'Name': 'Matt', 'Location': 'Miami'},
{'Name': 'Ben', 'Location': 'Las Vegas'},
{'Name': 'Sarah', 'Location': 'Washington DC'},
{'Name': 'Tom', 'Location': 'Denver'},
{'Name': 'Jim', 'Location': 'Dallas'}]
If you want to give a positive list of keys to copy over into the new dictionaries:
import csv
with open('data.csv', 'rb') as csv_file:
data = list(csv.DictReader(csv_file))
keys = ['Name', 'Location']
new_data = [dict((k, d[k]) for k in keys) for d in data]
print new_data
suppose we have,
l1 = [{'Location': 'Miami', 'Name': 'Matt', 'Speciality': 'Math'},
{'Location': 'Las Vegas', 'Name': 'Ben', 'Speciality': 'Science'},
{'Location': 'Washington DC', 'Name': 'Sarah', 'Speciality': 'Language Arts'},
{'Location': 'Denver', 'Name': 'Tom', 'Speciality': 'Spanish'},
{'Location': 'Dallas', 'Name': 'Jim', 'Speciality': 'Chemistry'}]
to create a new list of dictionaries that do not contain the keys 'Speciality' we can do,
l2 = []
for oldd in l1:
newd = {}
for k,v in oldd.items():
if k != 'Speciality':
newd[k] = v
l2.append(newd)
and now l2 will be your desired output. In general you can exclude an arbitrary list of keys like so
exclude_keys = ['Speciality', 'Name']
l2 = []
for oldd in l1:
newd = {}
for k,v in oldd.items():
if k not in exclude_keys:
newd[k] = v
l2.append(newd)
the same can be done with an include_keys variable
include_keys = ['Name', 'Location']
l2 = []
for oldd in l1:
newd = {}
for k,v in oldd.items():
if k in include_keys:
newd[k] = v
l2.append(newd)
You can create a new list of dicts limited to the keys you want with one line of code (Python 2.6+):
NLoD=[{k:d[k] for k in ('Name', 'Location')} for d in LoD]
Try it:
>>> LoD=[{'Speciality': 'Math', 'Name': 'Matt', 'Location': 'Miami'},
{'Speciality': 'Science', 'Name': 'Ben', 'Location': 'Las Vegas'},
{'Speciality': 'Language Arts', 'Name': 'Sarah', 'Location': 'Washington DC'},
{'Speciality': 'Spanish', 'Name': 'Tom', 'Location': 'Denver'},
{'Speciality': 'Chemistry', 'Name': 'Jim', 'Location': 'Dallas'}]
>>> [{k:d[k] for k in ('Name', 'Location')} for d in LoD]
[{'Name': 'Matt', 'Location': 'Miami'}, {'Name': 'Ben', 'Location': 'Las Vegas'}, {'Name': 'Sarah', 'Location': 'Washington DC'}, {'Name': 'Tom', 'Location': 'Denver'}, {'Name': 'Jim', 'Location': 'Dallas'}]
Since you are using csv, you can limit the columns that you read in the first place to the desired columns so you do not need to delete the undesired data:
dc=('Name', 'Location')
with open(fn) as f:
reader=csv.DictReader(f)
LoD=[{k:row[k] for k in dc} for row in reader]
keys_lst = ['Name', 'Location']
new_data={key:val for key,val in event.items() if key in keys_lst}
print(new_data)