trying to replace all elements named 'number' to 'numbr' in the data list but doesn't get it working.
Edit: So each key number should be renamed to numbr. Values stay as they are.
What am I doing wrong?
Thank you for your help!
data = [{'address': {
'city': 'city A',
'company_name': 'company A'},
'amount': 998,
'items': [{'description': 'desc A1','number': 'number A1'}],
'number': 'number of A',
'service_date': {
'type': 'DEFAULT',
'date': '2015-11-18'},
'vat_option': 123},
{'address': {
'city': 'city B',
'company_name': 'company B'},
'amount': 222,
'items': [{'description': 'desc B1','number': 'number B1'},
{'description': 'desc B2','number': 'number B2'}],
'number': 'number of B',
'service_date': {
'type': 'DEFAULT',
'date': '2015-11-18'},
'vat_option': 456}
]
def replace(l, X, Y):
for i,v in enumerate(l):
if v == X:
l.pop(i)
l.insert(i, Y)
replace(data, 'number', 'numbr')
print data
The following is a recursive replace implementation that replaces p1 by p2 in any string it encounters in the s object, recursing through lists, sets, tuples, dicts (both keys and values):
def nested_replace(s, p1, p2):
if isinstance(s, basestring): # Python2
# if isinstance(s, (str, bytes)): # Python3
return s.replace(p1, p2)
if isinstance(s, (list, tuple, set)):
return type(s)(nested_replace(x, p1, p2) for x in s)
if isinstance(s, dict):
return {nested_replace(k, p1, p2): nested_replace(v, p1, p2) for k, v in s.items()}
return s
>>> from pprint import pprint
>>> pprint(nested_replace(data, 'number', 'numbr'))
[{'address': {'city': 'city A', 'company_name': 'company A'},
'amount': 998,
'items': [{'description': 'desc A1', 'numbr': 'numbr A1'}],
'numbr': 'numbr of A',
'service_date': {'date': '2015-11-18', 'type': 'DEFAULT'},
'vat_option': 123},
{'address': {'city': 'city B', 'company_name': 'company B'},
'amount': 222,
'items': [{'description': 'desc B1', 'numbr': 'numbr B1'},
{'description': 'desc B2', 'numbr': 'numbr B2'}],
'numbr': 'numbr of B',
'service_date': {'date': '2015-11-18', 'type': 'DEFAULT'},
'vat_option': 456}]
eval function is anti pattern, but I think eval is best solution here
data1 = eval(repr(data).replace('number', 'numbr'))
If you are trying to replace both keys and values this will work.
from json import dumps, loads
data = [{'address': {
'city': 'city A',
'company_name': 'company A'},
'amount': 998,
'items': [{'description': 'desc A1','number': 'number A1'}],
'number': 'number of A',
'service_date': {
'type': 'DEFAULT',
'date': '2015-11-18'},
'vat_option': 123},
{'address': {
'city': 'city B',
'company_name': 'company B'},
'amount': 222,
'items': [{'description': 'desc B1','number': 'number B1'},
{'description': 'desc B2','number': 'number B2'}],
'number': 'number of B',
'service_date': {
'type': 'DEFAULT',
'date': '2015-11-18'},
'vat_option': 456}
]
data_string = dumps(data)
data = loads(data_string.replace('number', 'numbr')
Related
I'm trying to replace the value of a nested dictionary with a for loop and if statement, it kinda works but it replaces other keys as well. I tried solutions from other similar problems but no luck. Any help is appreciated, thanks!
d = {'Item 1': {'Name': 'Wireless Power Bank',
'Price': '$50',
'Status': 'sold-out'},
'Item 2': {'Name': 'Case - iPhone 12',
'Price': '$50',
'Status': 'not'},
'Item 3': {'Name': 'The Case - iPhone 12',
'Price': '$50',
'Status': 'not'},
'Item 4': {'Name': 'Case - iPhone 11',
'Price': '$50',
'Status': 'sold-out'},
'Item 5': {'Name': 'The Card Case - iPhone X/XS',
'Price': '$50',
'Status': 'not'}}
for k, v in d.items():
for inK, inV in v.items():
if "sold-out" in inV:
d[k][inK] = 'Out of Stock'
else:
d[k][inK] = 'In Stock'
Output:
{'Item 1': {'Name': 'In Stock', 'Price': 'In Stock', 'Status': 'Out of Stock'},
'Item 2': {'Name': 'In Stock', 'Price': 'In Stock', 'Status': 'In Stock'},
'Item 3': {'Name': 'In Stock', 'Price': 'In Stock', 'Status': 'In Stock'},
'Item 4': {'Name': 'In Stock', 'Price': 'In Stock', 'Status': 'Out of Stock'},
'Item 5': {'Name': 'In Stock', 'Price': 'In Stock', 'Status': 'In Stock'}}
Expected Output:
{'Item 1': {'Name': 'Wireless Power Bank',
'Price': '$50',
'Status': 'Out of Stock'},
'Item 2': {'Name': 'Case - iPhone 12',
'Price': '$50',
'Status': 'In Stock'},
'Item 3': {'Name': 'The Case - iPhone 12',
'Price': '$50',
'Status': 'In Stock'},
'Item 4': {'Name': 'Case - iPhone 11',
'Price': '$50',
'Status': 'Out of Stock'},
'Item 5': {'Name': 'The Card Case - iPhone X/XS',
'Price': '$50',
'Status': 'In Stock'}}
Solved thanks to #luk2302!
for k, v in d.items():
if "sold-out" in d[k]["Status"]:
d[k]["Status"] = 'Out of Stock'
else:
d[k]["Status"] = 'In Stock'
Output:
{'Item 1': {'Name': 'Wireless Power Bank', 'Price': '$50', 'Status': 'Out of Stock'},
'Item 2': {'Name': 'Case - iPhone 12', 'Price': '$50', 'Status': 'In Stock'},
'Item 3': {'Name': 'The Case - iPhone 12', 'Price': '$50', 'Status': 'In Stock'},
'Item 4': {'Name': 'Case - iPhone 11', 'Price': '$50', 'Status': 'Out of Stock'},
'Item 5': {'Name': 'The Card Case - iPhone X/XS', 'Price': '$50', 'Status': 'In Stock'}}
Looking into translating the following nested dictionary which is an API pull from Yelp into a pandas dataframe to run visualization on:
Top 50 Pizzerias in Chicago
{'businesses': [{'alias': 'pequods-pizzeria-chicago',
'categories': [{'alias': 'pizza', 'title': 'Pizza'}],
'coordinates': {'latitude': 41.92187, 'longitude': -87.664486},
'display_phone': '(773) 327-1512',
'distance': 2158.7084581522413,
'id': 'DXwSYgiXqIVNdO9dazel6w',
'image_url': 'https://s3-media1.fl.yelpcdn.com/bphoto/8QJUNblfCI0EDhOjuIWJ4A/o.jpg',
'is_closed': False,
'location': {'address1': '2207 N Clybourn Ave',
'address2': '',
'address3': '',
'city': 'Chicago',
'country': 'US',
'display_address': ['2207 N Clybourn Ave',
'Chicago, IL 60614'],
'state': 'IL',
'zip_code': '60614'},
'name': "Pequod's Pizzeria",
'phone': '+17733271512',
'price': '$$',
'rating': 4.0,
'review_count': 6586,
'transactions': ['restaurant_reservation', 'delivery'],
'url': 'https://www.yelp.com/biz/pequods-pizzeria-chicago?adjust_creative=wt2WY5Ii_urZB8YeHggW2g&utm_campaign=yelp_api_v3&utm_medium=api_v3_business_search&utm_source=wt2WY5Ii_urZB8YeHggW2g'},
{'alias': 'lou-malnatis-pizzeria-chicago',
'categories': [{'alias': 'pizza', 'title': 'Pizza'},
{'alias': 'italian', 'title': 'Italian'},
{'alias': 'sandwiches', 'title': 'Sandwiches'}],
'coordinates': {'latitude': 41.890357,
'longitude': -87.633704},
'display_phone': '(312) 828-9800',
'distance': 4000.9990531720227,
'id': '8vFJH_paXsMocmEO_KAa3w',
'image_url': 'https://s3-media3.fl.yelpcdn.com/bphoto/9FiL-9Pbytyg6usOE02lYg/o.jpg',
'is_closed': False,
'location': {'address1': '439 N Wells St',
'address2': '',
'address3': '',
'city': 'Chicago',
'country': 'US',
'display_address': ['439 N Wells St',
'Chicago, IL 60654'],
'state': 'IL',
'zip_code': '60654'},
'name': "Lou Malnati's Pizzeria",
'phone': '+13128289800',
'price': '$$',
'rating': 4.0,
'review_count': 6368,
'transactions': ['pickup', 'delivery'],
'url': 'https://www.yelp.com/biz/lou-malnatis-pizzeria-chicago?adjust_creative=wt2WY5Ii_urZB8YeHggW2g&utm_campaign=yelp_api_v3&utm_medium=api_v3_business_search&utm_source=wt2WY5Ii_urZB8YeHggW2g'},
....]
I've tried the below and iterations of it but haven't had any luck.
df = pd.DataFrame.from_dict(topresponse)
Im really new to coding so any advice would be helpful
response["businesses"] is a list of records, so:
df = pd.DataFrame.from_records(response["businesses"])
Lets say I have the following list in python. It is ordered first by Equip, then by Date:
my_list = [
{'Equip': 'A-1', 'Job': 'Job 1', 'Date': '2018-01-01'},
{'Equip': 'A-1', 'Job': 'Job 1', 'Date': '2018-01-02'},
{'Equip': 'A-1', 'Job': 'Job 1', 'Date': '2018-01-03'},
{'Equip': 'A-1', 'Job': 'Job 2', 'Date': '2018-01-04'},
{'Equip': 'A-1', 'Job': 'Job 2', 'Date': '2018-01-05'},
{'Equip': 'A-2', 'Job': 'Job 1', 'Date': '2018-01-03'},
{'Equip': 'A-2', 'Job': 'Job 3', 'Date': '2018-01-04'},
{'Equip': 'A-2', 'Job': 'Job 3', 'Date': '2018-01-05'}
]
What I want to do is collapse the list by each set where a given piece of Equipment's job does not change, and grab the first and last date the equipment was there. E.g., this simple example should change to:
list_by_job = [
{'Equip': 'A-1', 'Job': 'Job 1', 'First': '2018-01-01', 'Last': '2018-01-03'},
{'Equip': 'A-1', 'Job': 'Job 2', 'First': '2018-01-04', 'Last': '2018-01-05'},
{'Equip': 'A-2', 'Job': 'Job 1', 'First': '2018-01-03', 'Last': '2018-01-03'},
{'Equip': 'A-2', 'Job': 'Job 3', 'First': '2018-01-04', 'Last': '2018-01-05'}
]
A couple of things to note:
A-2 on Job 1 is only there for a single day, thus its First and Last Date should be the same.
A piece of equipment could be on a job, leave that job, and come back. In this case, I'd need to see an entry for each time it was on the job, not just one single summary.
As stated before, the list is already sorted first by Equip, then by Date, so that ordering can be assumed. (If there is a better way to sort to accomplish this, I am all ears)
For point 3, the list
my_list = [
{'Equip': 'A-1', 'Job': 'Job 1', 'Date': '2018-01-01'},
{'Equip': 'A-1', 'Job': 'Job 2', 'Date': '2018-01-02'},
{'Equip': 'A-1', 'Job': 'Job 1', 'Date': '2018-01-03'}
]
should yield
list_by_job = [
{'Equip': 'A-1', 'Job': 'Job 1', 'First': '2018-01-01', 'Last': '2018-01-01'},
{'Equip': 'A-2', 'Job': 'Job 2', 'First': '2018-01-02', 'Last': '2018-01-02'},
{'Equip': 'A-1', 'Job': 'Job 1', 'First': '2018-01-03', 'Last': '2018-01-03'}
]
Currently I am doing so in a simple loop/non-pythonic way:
list_by_job = []
last_entry = None
for entry in my_list:
if last_entry is None or last_entry['Equip'] != entry['Equip'] or last_entry['Job'] != entry['Job']:
list_by_job.append({'Equip': entry['Equip'], 'Job': entry['Job'], 'First': entry['Date'], 'Last': entry['Date']})
else:
list_by_job[-1]['Last'] = entry['Date']
last_entry = entry
Is there a more pythonic way to do this using Python's list comprehension, etc?
You can use itertools.groupby:
import itertools
def _key(d):
return (d['Equip'], d['Job'])
my_list = [{'Date': '2018-01-01', 'Equip': 'A-1', 'Job': 'Job 1'}, {'Date': '2018-01-02', 'Equip': 'A-1', 'Job': 'Job 1'}, {'Date': '2018-01-03', 'Equip': 'A-1', 'Job': 'Job 1'}, {'Date': '2018-01-04', 'Equip': 'A-1', 'Job': 'Job 2'}, {'Date': '2018-01-05', 'Equip': 'A-1', 'Job': 'Job 2'}, {'Date': '2018-01-03', 'Equip': 'A-2', 'Job': 'Job 1'}, {'Date': '2018-01-04', 'Equip': 'A-2', 'Job': 'Job 3'}, {'Date': '2018-01-05', 'Equip': 'A-2', 'Job': 'Job 3'}]
new_data = [[a, list(b)] for a, b in itertools.groupby(my_list, key=_key)]
final_result = [{"Equip":c, 'Job':d, 'First':b[0]['Date'], 'Last':b[-1]['Date']} for [c, d], b in new_data]
Output:
[{'Equip': 'A-1', 'Job': 'Job 1', 'Last': '2018-01-03', 'First': '2018-01-01'},
{'Equip': 'A-1', 'Job': 'Job 2', 'Last': '2018-01-05', 'First': '2018-01-04'},
{'Equip': 'A-2', 'Job': 'Job 1', 'Last': '2018-01-03', 'First': '2018-01-03'},
{'Equip': 'A-2', 'Job': 'Job 3', 'Last': '2018-01-05', 'First': '2018-01-04'}]
Edit:
Using data as suggested in your comment:
my_list = [{'Date': '2018-01-01', 'Equip': 'A-1', 'Job': 'Job 1'}, {'Date': '2018-01-02', 'Equip': 'A-1', 'Job': 'Job 2'}, {'Date': '2018-01-03', 'Equip': 'A-1', 'Job': 'Job 1'}, {'Date': '2018-01-04', 'Equip': 'A-1', 'Job': 'Job 2'}, {'Date': '2018-01-05', 'Equip': 'A-1', 'Job': 'Job 2'}, {'Date': '2018-01-03', 'Equip': 'A-2', 'Job': 'Job 1'}, {'Date': '2018-01-04', 'Equip': 'A-2', 'Job': 'Job 3'}, {'Date': '2018-01-05', 'Equip': 'A-2', 'Job': 'Job 3'}]
Output:
[{'Equip': 'A-1', 'Job': 'Job 1', 'Last': '2018-01-01', 'First': '2018-01-01'},
{'Equip': 'A-1', 'Job': 'Job 2', 'Last': '2018-01-02', 'First': '2018-01-02'},
{'Equip': 'A-1', 'Job': 'Job 1', 'Last': '2018-01-03', 'First': '2018-01-03'},
{'Equip': 'A-1', 'Job': 'Job 2', 'Last': '2018-01-05', 'First': '2018-01-04'},
{'Equip': 'A-2', 'Job': 'Job 1', 'Last': '2018-01-03', 'First': '2018-01-03'},
{'Equip': 'A-2', 'Job': 'Job 3', 'Last': '2018-01-05', 'First': '2018-01-04'}]
I suggest using pandas for this.
itertools.groupby is cool but IMO a bit harder to comprehend.
>>> import pandas as pd
>>>
>>> my_list = [
...: {'Equip': 'A-1', 'Job': 'Job 1', 'Date': '2018-01-01'},
...: {'Equip': 'A-1', 'Job': 'Job 1', 'Date': '2018-01-02'},
...: {'Equip': 'A-1', 'Job': 'Job 1', 'Date': '2018-01-03'},
...: {'Equip': 'A-1', 'Job': 'Job 2', 'Date': '2018-01-04'},
...: {'Equip': 'A-1', 'Job': 'Job 2', 'Date': '2018-01-05'},
...: {'Equip': 'A-2', 'Job': 'Job 1', 'Date': '2018-01-03'},
...: {'Equip': 'A-2', 'Job': 'Job 3', 'Date': '2018-01-04'},
...: {'Equip': 'A-2', 'Job': 'Job 3', 'Date': '2018-01-05'}
...:]
>>>
>>> df = pd.DataFrame(my_list)
>>> df['Date'] = pd.to_datetime(df['Date'])
>>> groups = df.groupby(['Equip', 'Job']).agg({'Date': [min, max]}).reset_index()
>>> groups.columns = ['Equip', 'Job', 'First', 'Last']
>>> groups
>>>
Equip Job First Last
0 A-1 Job 1 2018-01-01 2018-01-03
1 A-1 Job 2 2018-01-04 2018-01-05
2 A-2 Job 1 2018-01-03 2018-01-03
3 A-2 Job 3 2018-01-04 2018-01-05
>>>
>>> groups.to_dict(orient='records')
>>>
[{'Equip': 'A-1',
'First': Timestamp('2018-01-01 00:00:00'),
'Job': 'Job 1',
'Last': Timestamp('2018-01-03 00:00:00')},
{'Equip': 'A-1',
'First': Timestamp('2018-01-04 00:00:00'),
'Job': 'Job 2',
'Last': Timestamp('2018-01-05 00:00:00')},
{'Equip': 'A-2',
'First': Timestamp('2018-01-03 00:00:00'),
'Job': 'Job 1',
'Last': Timestamp('2018-01-03 00:00:00')},
{'Equip': 'A-2',
'First': Timestamp('2018-01-04 00:00:00'),
'Job': 'Job 3',
'Last': Timestamp('2018-01-05 00:00:00')}]
I suggest keeping the dates as time stamps.
You can use pandas here, which is some sort of "database interface" for data:
import pandas as pd
df = pd.DataFrame(my_list)
df2 = df.groupby(['Equip', 'Job']).agg(['min', 'max']).rename(columns={'min': 'First', 'max': 'Last'})
df2.columns = df2.columns.droplevel()
df2 = df2.reset_index()
result = df2.to_dict('records')
for the given sample input, this gives:
>>> df2.to_dict('records')
[{'Equip': 'A-1', 'Job': 'Job 1', 'First': '2018-01-01', 'Last': '2018-01-03'},
{'Equip': 'A-1', 'Job': 'Job 2', 'First': '2018-01-04', 'Last': '2018-01-05'},
{'Equip': 'A-2', 'Job': 'Job 1', 'First': '2018-01-03', 'Last': '2018-01-03'},
{'Equip': 'A-2', 'Job': 'Job 3', 'First': '2018-01-04', 'Last': '2018-01-05'}]
In case the date format is not '%Y-%m-%d', then one first needs to convert it with pd.to_datetime(..) like:
import pandas as pd
df = pd.DataFrame(my_list)
df['Date'] = pd.to_datetime(df['Date'])
df2 = df.groupby(['Equip', 'Job']).agg(['min', 'max']).rename(columns={'min': 'First', 'max': 'Last'})
df2.columns = df2.columns.droplevel()
df2 = df2.reset_index()
result = df2.to_dict('records')
I have 2 lists with the following data structure:
array1:
[{'student': {'name': 'abc'}, 'address': 'add_abc'},
{'student': {'name': 'xyz'}, 'address': 'add_xyz'}]
array2:
[{'student': {'name': 'abc'}, 'address': 'add_abc'},
{'student': {'name': 'rst'}, 'address': 'add_rst'}]
I want to have an array3 with union of the above 2 lists
array3:
[{'student': {'name': 'abc'}, 'address': 'add_abc'},
{'student': {'name': 'rst'}, 'address': 'add_rst'},
{'student': {'name': 'xyz'}, 'address': 'add_xyz'}]
How can I do that in Python?
They are lists, not arrays, but here is a solution:
a1 = [{'student': {'name': 'abc'}, 'address': 'add_abc'},
{'student': {'name': 'xyz'}, 'address': 'add_xyz'}]
a2 = [{'student': {'name': 'abc'}, 'address': 'add_abc'},
{'student': {'name': 'rst'}, 'address': 'add_rst'}]
a3 = a1 + [i for i in a2 if i not in a1]
the value of a3 will be
[{'student': {'name': 'abc'}, 'address': 'add_abc'},
{'student': {'name': 'xyz'}, 'address': 'add_xyz'},
{'student': {'name': 'rst'}, 'address': 'add_rst'}]
If you want your resulting list to be sorted (by student name, for example), simply use sort method:
a3.sort(key=lambda x: x["student"]["name"])
and the resulting a3 would be:
[{'student': {'name': 'abc'}, 'address': 'add_abc'},
{'student': {'name': 'rst'}, 'address': 'add_rst'},
{'student': {'name': 'xyz'}, 'address': 'add_xyz'}]
Normally you would simply use set(a1 + a2) but the problem here is that you have dicts inside your lists and they are unhashable, therefore they cannot be set items.
I'm looking for a concise way of taking two dicts that have a common key/value, and copying a key and value into one of the dicts. Example:
d1 = [{'name': 'john', 'uid': 'ax01', 'phone': '555-555-5555'},
{'name': 'jane', 'uid': 'ax02', 'phone': '555-555-5555'},
{'name': 'jimmy', 'uid': 'ax03', 'phone': '555-555-5555'}]
d2 = [{'uid': 'ax01', 'orderid': '9999', 'note': 'testing this'},
{'uid': 'ax02', 'orderid': '6666', 'note': 'testing this'},
{'uid': 'ax03', 'orderid': '7777', 'note': 'testing this'}]
Here uid is the key that I want to use to then copy the orderid key and the value for that matching data point. In the end I'd get something like:
output = [
{'name': 'john', 'uid': 'ax01', 'phone': '555-555-5555', 'orderid': '9999'},
{'name': 'jane', 'uid': 'ax02', 'phone': '555-555-5555', 'orderid': '6666'},
{'name': 'jimmy', 'uid': 'ax03', 'phone': '555-555-5555', 'orderid': '7777'}
]
Where the orderid is pulled into d1. I'm looking for the pythonic way if at all possible.
You can use dict() to copy one dictionary and pass in the extra key. You need to create a mapping from uid to orderid first though:
uid_to_orderid = {d['uid']: d['orderid'] for d in d2}
output = [dict(d, orderid=uid_to_orderid[d['uid']]) for d in d1]
This assumes that you want to leave the dictionaries in d1 untouched otherwise. Other assumptions made are that uid values are unique, and that all uid values in d1 are present in d2.
Demo:
>>> d1 = [{'name': 'john', 'uid': 'ax01', 'phone': '555-555-5555'},
... {'name': 'jane', 'uid': 'ax02', 'phone': '555-555-5555'},
... {'name': 'jimmy', 'uid': 'ax03', 'phone': '555-555-5555'}]
>>> d2 = [{'uid': 'ax01', 'orderid': '9999', 'note': 'testing this'},
... {'uid': 'ax02', 'orderid': '6666', 'note': 'testing this'},
... {'uid': 'ax03', 'orderid': '7777', 'note': 'testing this'}]
>>> uid_to_orderid = {d['uid']: d['orderid'] for d in d2}
>>> [dict(d, orderid=uid_to_orderid[d['uid']]) for d in d1]
[{'orderid': '9999', 'phone': '555-555-5555', 'name': 'john', 'uid': 'ax01'}, {'orderid': '6666', 'phone': '555-555-5555', 'name': 'jane', 'uid': 'ax02'}, {'orderid': '7777', 'phone': '555-555-5555', 'name': 'jimmy', 'uid': 'ax03'}]
Great answer #Martijn Pieters
I can't think of a clever way to protect against missing data other something brutish like this:
d1 = [{'name': 'john', 'uid': 'ax01', 'phone': '555-555-5555'},
{'name': 'jane', 'uid': 'ax02', 'phone': '555-555-5555'},
{'name': 'jimmy', 'uid': 'ax03', 'phone': '555-555-5555'},
{'name': 'jack', 'uid': 'ax04', 'phone': '555-555-5555'},
{'name': 'joan', 'uid': 'ax05', 'phone': '555-555-5555'}]
d2 = [{'uid': 'ax01', 'orderid': '9999', 'note': 'testing this'},
{'uid': 'ax02', 'orderid': '6666', 'note': 'testing this'},
{'uid': 'ax03', 'orderid': '6666', 'note': 'testing this'},
{'uid': 'ax05', 'orderid-not-here': '7777', 'note': 'testing this'}]
def get_orderid(search):
match = [x for x in d2 if x['uid']==search]
if len(match) == 0:
return 'None'
else:
return match[0].get('orderid','None')
[dict(d, orderid=get_orderid(d['uid'])) for d in d1]
[ {'name': 'john', 'orderid': '9999', 'phone': '555-555-5555', 'uid': 'ax01'},
{'name': 'jane', 'orderid': '6666', 'phone': '555-555-5555', 'uid': 'ax02'},
{'name': 'jimmy', 'orderid': '6666', 'phone': '555-555-5555', 'uid': 'ax03'},
{'name': 'jack', 'orderid': 'None', 'phone': '555-555-5555', 'uid': 'ax04'},
{'name': 'joan', 'orderid': 'None', 'phone': '555-555-5555', 'uid': 'ax05'}]
`