从 csv 创建嵌套字典

Question

I have some data in a csv that when opened looks like this:

Example Data
['', 'Name', 'Phone', 'Address', 'City', 'Country', 'Email']
['1', 'Bob Myers', '410-504-5887', '12334 Hamilton Way', 'Toronto', 'Canada', 'bob@gmail.com']
['2', 'Carlton James', '455-323-8479', '1234 James Rd', 'New York', 'USA', 'carlton@example.com']
['3', 'Frank Wright', '744-521-9874', '567 Travis St', 'Boston', 'USA', 'fw4322@yahoo.com']

I want to create a nested dictionary where I take the integer in '' and use it as a key for each nested dictionary. I found a few similar questions and tried:

import csv
f = csv.DictReader(open('data.csv'))

result = {}
for row in f:
    key = row.pop('')
    result[key] = row
print(result)

Which yielded something like:

{'1': {'Name': 'Bob Myers', 'Phone': '410-504-5887', 'Address': '12334 Hamilton Way', 'City': 'Toronto',
'Country': 'Canada', 'Email': 'bob@gmail.com'}...

What do I change in my code to get my data to look something like this (without pandas):

my_dict = {{'Name': {1: 'Bob Myers', 2: 'Carlton James', 3: 'Frank Wright'}}, 
 {'Phone': {1: '410-504-5887', 2: '455-323-8479', 3: '744-521-9874'}}, 
 {'Address': {1: '12334 Hamilton Way', 2: '1234 James Rd', 3: '567 Travis St'}},
 {'City': {1: 'Toronto', 2: 'New York', 3: 'Boston'}},
 {'Country': {1: 'Canada', 2: 'USA', 3: 'USA'}},
 {'Email': {1: 'bob@gmail.com', 2: 'carlton@example.com', 3: 'fw4322@yahoo.com'}}
}

Answer 1

Best way is to (ab)use Pandas ;-)

import pandas as pd
d = pd.DataFrame([result]).to_dict()
print(d)

For test case result :

{'1': {'Name': 'Bob Myers', 'Phone': '410-504-5887', 'Address': '12334 Hamilton Way', 'City': 'Toronto',
'Country': 'Canada', 'Email': 'bob@gmail.com'}

Output is like:

{'Name': {0: 'Bob Myers'}, 'Phone': {0: '410-504-5887'}, 'Address': {0: '12334 Hamilton Way'}, 'City': {0: 'Toronto'}, 'Country': {0: 'Canada'}, 'Email': {0: 'bob@gmail.com'}} 

Answer 2

Since you have a csv data file, it's better to use pandas .

import pandas as pd

a = pd.read_csv(file location)

nested_dict = a.todict()