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Selecting dictionary nested in a list by index

Here is am example of the list:
{'Item': 'milk', 'Price': '2.0', 'Quantity': '2'}, {'Item': 'egg', 'Price': '12.0', 'Quantity': '1'}]
Here is my code:
def edit_items(info):
xy = info
print('Index | Orders')
for x in enumerate(xy):
print('\n')
print(x)
choice = int(input('Which entry would you like to edit? Choose by index. :'))
print(x[choice])
Id like the user to able to chose an entry by index, and allow them to edit information inside the dictionary.
So far my code prints out:
Index | Orders
(0, {'Item': 'milk', 'Price': '2.0', 'Quantity': '2'})
(1, {'Item': 'egg', 'Price': '12.0', 'Quantity': '1'})
But i have no idea how to choose one, assign It to a variable and carry out the ability to edit whats inside.
Cheers. Nalpak_

def edit_items(info):
xy = info
# to make multiple edits.
while True:
print('Index | Orders')
for x in range(len(xy)):
print(x,xy[x])
choice = int(input('Which entry would you like to edit?\nChoose by index: '))
print(xy[choice])
edit = input('What you want to edit: ') # Key val of dict
value = input("Enter: ") # Value for the specific key in dict
xy[choice][edit] = value
print('list updated.')
print(xy[choice])
more_edits = input('\nDo you want to make more edits?(y/n): ')
if more_edits == 'n':
break
edit_items(info)
this will help you make multiple edits.

If you want to edit an item in a dictionary, you can easily do it by accessing it by the key.
First, we set up the data
xy = [{'Item': 'milk', 'Price': '2.0', 'Quantity': '2'}, {'Item': 'egg', 'Price': '12.0', 'Quantity': '1'}]
Then if I understood you correctly, this edit_items method should do exactly what you need:
def edit_items(i):
name = input('Type in a new item name: ')
xy[i]['Item'] = name # 'Item' is the key.
Everything else is pretty much the same:
print('Index | Orders')
for x in enumerate(xy):
print('\n')
print(x)
choice = int(input('Which entry would you like to edit? Choose by index. :'))
print(xy[choice])
edit_items(choice)
print(xy)
If you want, you can also use input for getting a key (property) of an item you want to edit.

How to insert data into an array of dictionaries in an order without missing data via regex

This is my code:
I'm trying to use the following code to insert data into an array of dictionaries but unable to insert properly.
Code:
test_list = {'module_serial-1': 'PSUXA12345680', 'module_name-1': 'CH1.FM5', 'module_name-2': 'CH1.FM6', 'module_serial-2': 'PSUXA12345681'}
def parse_subdevice_modules(row):
modules = []
module = {}
for k, v in row.items():
if v:
if re.match("module_name", k):
module['name'] = v
if re.match("module_serial", k):
module['serial'] = v
modules.append(module)
module = {}
return modules
print(parse_subdevice_modules(test_list))
Expected output:
[{'name':'CH1.FM5', serial': 'PSUXA12345680'}, {'name': 'CH1.FM6', 'serial': 'PSUXA12345681'}]
Actual output:
['serial': 'PSUXA12345680'}, {'name': 'CH1.FM6', 'serial': 'PSUXA12345681'}]
Run it here: https://repl.it/repls/WetSteelblueRange
Please note that the order of the data test_list cannot be altered as it comes via an external API so I used regex. Any ideas would be appreciated.

Your code relies on the wrong assumption that keys are ordered and that the serial will always follow the name. The proper solution here is to use a dict (actually a collections.defaultdict to make things easier) to collect and regroup the values you're interested in based on the module number (the final '-N' in the key). Note that you don't need regexps here - Python string already provide the necessary operations for this task:
from collections import defaultdict
def parse_subdevice_modules(row):
modules = defaultdict(dict)
for k, v in row.items():
# first get rid of what we're not interested in
if not v:
continue
if not k.startswith("module_"):
continue
# retrieve the key number (last char) with
# negative string indexing:
key_num = k[-1]
# retrieve the useful part of the key ("name" or "serial")
# by splitting the string:
key_name = k.split("_")[1].split("-")[0]
# and now we just have to store this in our defaultdict
modules[key_num][key_name] = v
# and return only the values.
# NB: in py2.x you don't need the call to `list`,
# you can just return `modules.values()` directly
modules = list(modules.values())
return modules
test_list = {
'profile': '', 'chassis_name': '123', 'supplier_order_num': '',
'device_type': 'mass_storage', 'device_subtype': 'flashblade',
'module_serial-1': 'PSUXA12345680', 'module_name-1': 'CH1.FM5',
'module_name-2': 'CH1.FM6', 'rack_total_pos': '',
'asset_tag': '002000027493', 'module_serial-2': 'PSUXA12345681',
'purchase_order': '0004530869', 'build': 'Test_Build_for_SNOW',
'po_line_num': '00190', 'mac_address': '', 'position': '7',
'model': 'FB-528TB-10X52.8TB', 'manufacturer': 'PureStorage',
'rack': 'Test_Rack_2', 'serial': 'PMPAM1842147D', 'name': 'FB02'
}
print(parse_subdevice_modules(test_list))

You can do somthing like this also.
test_list = {'module_serial-1': 'PSUXA12345680', 'module_name-1': 'CH1.FM5', 'module_name-2': 'CH1.FM6',
'module_serial-2': 'PSUXA12345681'}
def parse_subdevice_modules(row):
modules_list = []
for key, value in row.items():
if not value or key.startswith('module_name'):
continue
if key.startswith('module_serial'):
module_name_key = f'module_name-{key.split("-")[-1]}'
modules_list.append({'serial': value, 'name': row[module_name_key]})
return modules_list
print(parse_subdevice_modules(test_list))
Output:
[{'serial': 'PSUXA12345680', 'name': 'CH1.FM5'}, {'serial': 'PSUXA12345681', 'name': 'CH1.FM6'}]

You would need to check if module contains 2 elements and append it to modules:
test_list = {'module_serial-1': 'PSUXA12345680', 'module_name-1': 'CH1.FM5', 'module_name-2': 'CH1.FM6', 'module_serial-2': 'PSUXA12345681'}
def parse_subdevice_modules(row):
modules = []
module = {}
for k, v in row.items():
if v:
if k.startswith('module_name'):
module['name'] = v
elif k.startswith("module_serial"):
module['serial'] = v
if len(module) == 2:
modules.append(module)
module = {}
return modules
print(parse_subdevice_modules(test_list))
Returns:
[{'serial': 'PSUXA12345680'}, {'name': 'CH1.FM5'}, {'name': 'CH1.FM6'}, {'serial': 'PSUXA12345681'}]

Dynamically changing key value in dictionary

I am checking the key in dictionary, if it contains space remove it.
def query_combination(sentence,mydict):
for key in mydict.keys():
if key == 'key':
pass
else:
print 'key is : ',key
if " " in key:
temp = key
key = key.replace(' ',"")
print 'new key : ',key
sentence = sentence.replace(temp ,key)
print 'new sentence : ',sentence
print mydict
mydict = {'films': {'match': ['Space', 'Movie', 'six', 'two', 'one']}, u'Popeye Doyle': {'score': 100, 'match': [u'People', 'heaven', 'released']}}
sentence ='What films featured the character Popeye Doyle'
combination = query_combination(sentence,mydict)
I could not dynamically change the new key value to mydict. Any suggestion much appreciable

If you get a string out of the dictionary, and then change it and make a new string, the dictionary won't know about it; you can add a new entry to the dictionary and remove the old one:
if " " in key:
newkey = key.replace(' ',"")
mydict[newkey] = mydict[key]
del mydict[key]
print 'new key : ', newkey

You could try this
def query_combination(sentence,mydict):
for key in mydict.iterkeys():
if " " in key:
temp = key
mydict[key.replace(" ","")] = mydict[key] # create new key
del mydict[key] # delete old key
sentence = sentence.replace(temp ,key)
Another solution in one line would be
mydict[key.replace(" ","")] = mydict.pop(key)

key = key.replace(' ',"") does not affect the actual key in the dictionary, it is changing a copy of that key. You need to add the value to the dictionary with the new key and remove the old key. Here's one way to do it:
def query_combination(sentence, mydict):
for old_key, new_key in [(key, key.replace(' ', '')) for key in mydict if ' ' in key]:
mydict[new_key] = mydict.pop(old_key)
sentence = sentence.replace(old_key, new_key)
Note, however, that you are replacing the key in the string sentence, but sentence is local to function query_combination(), so the outer scope sentence is unaffected by the replacement. I am not sure if that was what you hoped your code would do, but if it was you could simply return the revised sentence from the function, or include it as an item in the dictionary.
Given that sentence is not actually updated by your function, you can simplify the whole function to a mere dictionary comprehension:
>>> mydict = {'films': {'match': ['Space', 'Movie', 'six', 'two', 'one']}, u'Popeye Doyle': {'score': 100, 'match': [u'People', 'heaven', 'released']}}
>>> mydict = {key.replace(' ', '') : value for key, value in mydict.items()}
>>> mydict
{'films': {'match': ['Space', 'Movie', 'six', 'two', 'one']}, u'PopeyeDoyle': {'score': 100, 'match': [u'People', 'heaven', 'released']}}

Failing to append to dictionary. Python

I am experiencing a strange faulty behaviour, where a dictionary is only appended once and I can not add more key value pairs to it.
My code reads in a multi-line string and extracts substrings via split(), to be added to a dictionary. I make use of conditional statements. Strangely only the key:value pairs under the first conditional statement are added.
Therefore I can not complete the dictionary.
How can I solve this issue?
Minimal code:
#I hope the '\n' is sufficient or use '\r\n'
example = "Name: Bugs Bunny\nDOB: 01/04/1900\nAddress: 111 Jokes Drive, Hollywood Hills, CA 11111, United States"
def format(data):
dic = {}
for line in data.splitlines():
#print('Line:', line)
if ':' in line:
info = line.split(': ', 1)[1].rstrip() #does not work with files
#print('Info: ', info)
if ' Name:' in info: #middle name problems! /maiden name
dic['F_NAME'] = info.split(' ', 1)[0].rstrip()
dic['L_NAME'] = info.split(' ', 1)[1].rstrip()
elif 'DOB' in info: #overhang
dic['DD'] = info.split('/', 2)[0].rstrip()
dic['MM'] = info.split('/', 2)[1].rstrip()
dic['YY'] = info.split('/', 2)[2].rstrip()
elif 'Address' in info:
dic['STREET'] = info.split(', ', 2)[0].rstrip()
dic['CITY'] = info.split(', ', 2)[1].rstrip()
dic['ZIP'] = info.split(', ', 2)[2].rstrip()
return dic
if __name__ == '__main__':
x = format(example)
for v, k in x.iteritems():
print v, k

Your code doesn't work, at all. You split off the name before the colon and discard it, looking only at the value after the colon, stored in info. That value never contains the names you are looking for; Name, DOB and Address all are part of the line before the :.
Python lets you assign to multiple names at once; make use of this when splitting:
def format(data):
dic = {}
for line in data.splitlines():
if ':' not in line:
continue
name, _, value = line.partition(':')
name = name.strip()
if name == 'Name':
dic['F_NAME'], dic['L_NAME'] = value.split(None, 1) # strips whitespace for us
elif name == 'DOB':
dic['DD'], dic['MM'], dic['YY'] = (v.strip() for v in value.split('/', 2))
elif name == 'Address':
dic['STREET'], dic['CITY'], dic['ZIP'] = (v.strip() for v in value.split(', ', 2))
return dic
I used str.partition() here rather than limit str.split() to just one split; it is slightly faster that way.
For your sample input this produces:
>>> format(example)
{'CITY': 'Hollywood Hills', 'ZIP': 'CA 11111, United States', 'L_NAME': 'Bunny', 'F_NAME': 'Bugs', 'YY': '1900', 'MM': '04', 'STREET': '111 Jokes Drive', 'DD': '01'}
>>> from pprint import pprint
>>> pprint(format(example))
{'CITY': 'Hollywood Hills',
'DD': '01',
'F_NAME': 'Bugs',
'L_NAME': 'Bunny',
'MM': '04',
'STREET': '111 Jokes Drive',
'YY': '1900',
'ZIP': 'CA 11111, United States'}

issue in list of dict

class MyOwnClass:
# list who contains the queries
queries = []
# a template dict
template_query = {}
template_query['name'] = 'mat'
template_query['age'] = '12'
obj = MyOwnClass()
query = obj.template_query
query['name'] = 'sam'
query['age'] = '23'
obj.queries.append(query)
query2 = obj.template_query
query2['name'] = 'dj'
query2['age'] = '19'
obj.queries.append(query2)
print obj.queries
It gives me
[{'age': '19', 'name': 'dj'}, {'age': '19', 'name': 'dj'}]
while I expect to have
[{'age': '23' , 'name': 'sam'}, {'age': '19', 'name': 'dj'}]
I thought to use a template for this list because I'm gonna to use it very often and there are some default variable who does not need to be changed.
Why does doing it the template_query itself changes? I'm new to python and I'm getting pretty confused.

this is because you are pointing to the same dictionary each time ... and overwriting the keys ...
# query = obj.template_query - dont need this
query = {}
query['name'] = 'sam'
query['age'] = '23'
obj.queries.append(query)
query2 = {} #obj.template_query-dont need this
query2['name'] = 'dj'
query2['age'] = '19'
obj.queries.append(query2)
this should demonstrate your problem
>>> q = {'a':1}
>>> lst = []
>>> lst.append(q)
>>> q['a']=2
>>> lst
[{'a': 2}]
>>> lst.append(q)
>>> lst
[{'a': 2}, {'a': 2}]
you could implement your class differently
class MyOwnClass:
# a template dict
#property
def template_query():
return {'name':'default','age':-1}
this will make obj.template_query return a new dict each time

This is because query and query2 are both referring to the same object. obj.template_query, in this case.
Better to make a template factory:
def template_query(**kwargs):
template = {'name': 'some default value',
'age': 'another default value',
'car': 'generic car name'}
template.update(**kwargs)
return template
That creates a new dictionary every time it's called. So you can do:
>>> my_query = template_query(name="sam")
>>> my_query
{'name': 'sam', 'age': 'another default value', 'car': 'generic car name'}

You're copying the same dict into query2. Instead, you might want to create the needed dict by creating a function template_query() and constructing a new dict each time:
class MyOwnClass:
# a template dict
def template_query():
d = {}
d['name'] = 'mat'
d['age'] = '12'
d['car'] = 'ferrari'
return d