将字典的字典拼合为python中的列表

Question

I have a dictionary of dictionaries and need to flatten it into a list. My goal here is to have it in a form that I can send to a MySQL database. The dictionary is in the form

{
 'Customer 1': 
   {'Vitamin A': 0, 'Calcium': 1, 'Protein': 1, 'B vitamins': 0, 'HPMR': 0},
 'Customer 2': 
   {'Vitamin A': 0, 'Calcium': 0, 'Protein': 1, 'B vitamins': 1, 'HPMR': 0}
 }

I want it to end up like:

('Customer 1', 'Vitamin A', 0), ('Customer 1', 'Calcium', 1)
...
('Customer 2', 'Vitamin A', 0), ('Customer 2', 'Calcium', 0)`

Thanks for your help

Answer 1

customers = {'Customer 1': {'Vitamin A': 0, 
                            'Calcium': 1, 
                            'Protein': 1, 
                            'B vitamins': 0, 
                            'HPMR': 0}, 
             'Customer 2': {'Vitamin A': 0, 
                            'Calcium': 0, 
                            'Protein': 1, 
                            'B vitamins': 1, 
                            'HPMR': 0}}
all_orders = []
for customer, order in customers.items():
    customer_orders = [(customer, vitamin, quantity) for vitamin, quantity in order.items()]
    all_orders.append(customer_orders)

all_orders = [[('Customer 2', 'Calcium', 0),
               ('Customer 2', 'Vitamin A', 0),
               ('Customer 2', 'HPMR', 0),
               ('Customer 2', 'Protein', 1),
               ('Customer 2', 'B vitamins', 1)],
              [('Customer 1', 'Calcium', 1),
               ('Customer 1', 'Vitamin A', 0),
               ('Customer 1', 'HPMR', 0),
               ('Customer 1', 'Protein', 1),
               ('Customer 1', 'B vitamins', 0)]]

Answer 2

In [6]: dic={'Customer 1': {'Vitamin A': 0, 'Calcium': 1, 'Protein': 1, 'B vitamins': 0, 'HPMR': 0}, 'Customer 2': {'Vitamin A': 0, 'Calcium': 0, 'Protein': 1, 'B vitamins': 1, 'HPMR': 0}}

In [7]: [tuple([x]+list(y)) for x in dic for y in dic[x].items() ]

Out[7]: 
[('Customer 2', 'Calcium', 0),
 ('Customer 2', 'Vitamin A', 0),
 ('Customer 2', 'HPMR', 0),
 ('Customer 2', 'Protein', 1),
 ('Customer 2', 'B vitamins', 1),
 ('Customer 1', 'Calcium', 1),
 ('Customer 1', 'Vitamin A', 0),
 ('Customer 1', 'HPMR', 0),
 ('Customer 1', 'Protein', 1),
 ('Customer 1', 'B vitamins', 0)]

or using itertools.chain() :

In [8]: from itertools import chain

In [9]: [tuple(chain(*([x],y))) for x in dic for y in dic[x].items() ]
Out[9]: 
[('Customer 2', 'Calcium', 0),
 ('Customer 2', 'Vitamin A', 0),
 ('Customer 2', 'HPMR', 0),
 ('Customer 2', 'Protein', 1),
 ('Customer 2', 'B vitamins', 1),
 ('Customer 1', 'Calcium', 1),
 ('Customer 1', 'Vitamin A', 0),
 ('Customer 1', 'HPMR', 0),
 ('Customer 1', 'Protein', 1),
 ('Customer 1', 'B vitamins', 0)]

Answer 3

[(k, k2, v) for k in dct for k2, v in k.items()]

Answer 4

dictio = {'Customer 1': {'Vitamin A': 0, 'Calcium': 1, 'Protein': 1, 'B vitamins': 0, 'HPMR': 0}, 'Customer 2': {'Vitamin A': 0, 'Calcium': 0, 'Protein': 1, 'B vitamins': 1, 'HPMR': 0}}
a = []
for key in dictio.keys():
    for key_ in dictio[key].keys():
        a.append((key,key_,dictio[key][key_]))

print a

Output: [('Customer 2', 'Calcium', 0), ('Customer 2', 'Vitamin A', 0), ('Customer 2', 'HPMR', 0), ('Customer 2', 'Protein', 1), ('Customer 2', 'B vitamins', 1), ('Customer 1', 'Calcium', 1), ('Customer 1', 'Vitamin A', 0), ('Customer 1', 'HPMR', 0), ('Customer 1', 'Protein', 1), ('Customer 1', 'B vitamins', 0)]

Also what you are inserting are tuples to the list, not lists.

Answer 5

I too thought of a list comprehension like others:

cust = {'Customer 1': {'Vitamin A': 0, 'Calcium': 1, 'Protein': 1, 'B vitamins': 0, 'HPMR': 0}, 'Customer 2': {'Vitamin A': 0, 'Calcium': 0, 'Protein': 1, 'B vitamins': 1, 'HPMR': 0}}
cust_flat = [ (k, k2, v2) for k2, v2 in v.iteritems() for k, v in cust.iteritems() ]

Result looks like:

[('Customer 2', 'Calcium', 1),
 ('Customer 1', 'Calcium', 1),
 ('Customer 2', 'Vitamin A', 0),
 ('Customer 1', 'Vitamin A', 0),
 ('Customer 2', 'HPMR', 0),
 ('Customer 1', 'HPMR', 0),
 ('Customer 2', 'Protein', 1),
 ('Customer 1', 'Protein', 1),
 ('Customer 2', 'B vitamins', 0),
 ('Customer 1', 'B vitamins', 0)]