Develop Reference

How do extract query-like results from a nested dictionary in Python?

I have a relatively simple nested dictionary as below:
emp_details = {
'Employee': {
'jim': {'ID':'001', 'Sales':'75000', 'Title': 'Lead'},
'eva': {'ID':'002', 'Sales': '50000', 'Title': 'Associate'},
'tony': {'ID':'003', 'Sales': '150000', 'Title': 'Manager'}
}
}
I can get the sales info of 'eva' easily by:
print(emp_details['Employee']['eva']['Sales'])
but I'm having difficulty writing a statement to extract information on all employees whose sales are over 50000.

You can't use one statement because the list initializer expression can't have an if without an else.
Use a for loop:
result = {} # dict expression
result_list = [] # list expression using (key, value)
for key, value in list(emp_details['Employee'].items())): # iterate from all items in dictionary
if int(value['Sales']) > 50000: # your judgement
result[key] = value # add to dict
result_list.append((key, value)) # add to list
print(result)
print(result_list)
# should say:
'''
{'jim': {'ID':'001', 'Sales':'75000', 'Title': 'Lead'}, 'tony': {'ID':'003', 'Sales': '150000', 'Title': 'Manager'}}
[('jim', {'ID':'001', 'Sales':'75000', 'Title': 'Lead'}), ('tony', {'ID':'003', 'Sales': '150000', 'Title': 'Manager'})]
'''

Your Sales is of String type.
Therefore, we can do something like this to get the information of employees whose sales are over 50000 : -
Method1 :
If you just want to get the information : -
emp_details={'Employee':{'jim':{'ID':'001', 'Sales':'75000', 'Title': 'Lead'}, \
'eva':{'ID':'002', 'Sales': '50000', 'Title': 'Associate'}, \
'tony':{'ID':'003', 'Sales': '150000', 'Title': 'Manager'}
}}
for emp in emp_details['Employee']:
if int(emp_details['Employee'][emp]['Sales']) > 50000:
print(emp_details['Employee'][emp])
It print outs to -:
{'ID': '001', 'Sales': '75000', 'Title': 'Lead'}
{'ID': '003', 'Sales': '150000', 'Title': 'Manager'}
Method2 : You can use Dict and List comprehension to get complete information : -
emp_details={'Employee':{'jim':{'ID':'001', 'Sales':'75000', 'Title': 'Lead'}, \
'eva':{'ID':'002', 'Sales': '50000', 'Title': 'Associate'}, \
'tony':{'ID':'003', 'Sales': '150000', 'Title': 'Manager'}
}}
emp_details_dictComp = {k:v for k,v in list(emp_details['Employee'].items()) if int(v['Sales']) > 50000}
print(emp_details_dictComp)
emp_details_listComp = [(k,v) for k,v in list(emp_details['Employee'].items()) if int(v['Sales']) > 50000]
print(emp_details_listComp)
Result : -
{'jim': {'ID': '001', 'Sales': '75000', 'Title': 'Lead'}, 'tony': {'ID': '003', 'Sales': '150000', 'Title': 'Manager'}}
[('jim', {'ID': '001', 'Sales': '75000', 'Title': 'Lead'}), ('tony', {'ID': '003', 'Sales': '150000', 'Title': 'Manager'})]

Merge dictionaries in a list of dictionaries by combining the values

Help me please.
I have query set from Model.objects.value('name', 'language__name') and it give me the list of dictionary:
list = [{'id': 1, 'name': 'Adel', 'language': 'С#'},
{'id': 1, 'name': 'Adel', 'language': 'Python'},
{'id': 5, 'name': 'Dora', 'language': 'С#'},
{'id': 5, 'name': 'Dora', 'language': 'Java'},
{'id': 6, 'name': 'Dars', 'language': 'Python'}];
how can I to do a list of dictionary but to unit key and value?
I want to get like this:
list = [{'id': 1, 'name': 'Adel', 'language':['С#','Python']},
{'id': 5, 'name': 'Dora', 'language': ['С#','Java']},
{'id': 6, 'name': 'Dars', 'language': 'Python'}];
I tried this:
mapping = {}
for d in qs:
try:
entry = mapping[d['id']] # raises KeyError
entry['language__name'].append(d['language__name']) # raises AttributeError
except KeyError:
mapping[d['id']] = d
except AttributeError:
entry['language__name'] = [entry['language__name'], d['language__name']]
print(list(mapping.values()))

You can iterate over lst and create a dictionary out where the keys correspond to "id" values in the dicts in lst and the values are dicts. In each iteration, check if the value under "language" key is a list or not and append to the list if it's a list, create a list, if not. Finally, pass the values of out to a list constructor for the final outcome.
out = {}
for d in lst:
if d['id'] in out:
if isinstance(out[d['id']]['language'], list):
out[d['id']]['language'].append(d['language'])
else:
out[d['id']]['language'] = [out[d['id']]['language'], d['language']]
else:
out[d['id']] = d
out = list(out.values())
Output:
[{'id': 1, 'name': 'Adel', 'language': ['С#', 'Python']},
{'id': 5, 'name': 'Dora', 'language': ['С#', 'Java']},
{'id': 6, 'name': 'Dars', 'language': 'Python'}]

How can I print a list without the first square brackets?

How can I print a list without the first square brackets?
[[{'id': '5b0cecebc15ae360393438c2', 'owner': {'id': '58ebd0b12c593e5253505ad4'}, 'linked_id': None, 'type': 'product', 'name': 'Трансформаторное масло Т-1500у'}]]
At the same time, it is necessary to preserve its essence of the list.
So that the parser can read the id.
items = response['data']
usr_ids = items[i]["owner"]["id"]

Simply print the first item of the list
lst = [[{'id': '5b0cecebc15ae360393438c2', 'owner': {'id': '58ebd0b12c593e5253505ad4'}, 'linked_id': None, 'type': 'product', 'name': 'Трансформаторное масло Т-1500у'}]]
print(lst[0])

Retrieve only certain keys and values from a dictionary, nested inside a list

I've been stuck on this for hours.. I want to retrieve only ONE individuals keys and values from a dictionary that is nested inside of a list.
GAMERS = [{
'name': 'Fatboi',
'parent': 'Dick Van Dyke',
'game': 'Dark Souls 3',
'weight': '420 lbs'
},
{
'name': 'Justin',
'parent': 'Heather Blueberry',
'game': 'Tetris',
'weight': '180 lbs'
},
{
'name': 'jerkhead',
'parent': 'none',
'games': 'Hello Kitty',
'weight': '240 lbs'
},{
'name': 'Tumor',
'parent': 'Jack Black',
'games': 'Trying to live',
'weight': '150 lbs'
}]
So for instance I want to get Justins information printed only, nobody elses. Any insights?

You can pass the key which you want and push it to separate list.
GAMERS = [{
'name': 'Fatboi',
'parent': 'Dick Van Dyke',
'game': 'Dark Souls 3',
'weight': '420 lbs'
},
{
'name': 'Justin',
'parent': 'Heather Blueberry',
'game': 'Tetris',
'weight': '180 lbs'
},{
'name': 'jerkhead',
'parent': 'none',
'games': 'Hello Kitty',
'weight': '240 lbs'
}]
def get_key_pair_list(input_dict, key):
new_list = []
for item in input_dict:
my_dict = {}
if key in item.keys():
my_dict[key] = item[key]
new_list.append(my_dict)
return new_list
print(get_key_pair_list(GAMERS, 'name'))
Output:
[{'name': 'Fatboi'}, {'name': 'Justin'}, {'name': 'jerkhead'}]
Comprehensive way:
key = 'name'
my_list = [{key, item[key]} for item in GAMERS if key in item.keys() ]
print(my_list)
output:
[{'name', 'Fatboi'}, {'name', 'Justin'}, {'name', 'jerkhead'}]

You want to filter the list and grab the first value that matches a predicate. Make sure to handle the case where the item doesnt exist!
filtered_info = (
item for item in GAMERS if item['name'] == 'Justin'
)
justin_info = next(filtered_info, None)
if justin_info is not None:
print(justin_info)

How to split string values across a list of dictionaries, so that each part of the split string is in it's own dictionary?

How to split string values across a list of dictionaries, so that each part of the split string is in it's own dictionary?
For example:
lst = [{item:'321,121,343','name':'abby','amount':'400'},{item:'111,222,333','name':'chris','amount':'500'}]
I'd like each part of the item key-value pair to be in its own dictionary
Desired_List = [{item:'321','name':'abby','amount':'400'},{item:'121','name':'abby','amount':'400'},{item:'343','name':'abby','amount':'400'},{item:'111','name':'chris','amount':'500'},{item:'222','name':'chris','amount':'500'},{item:'333','name':'chris','amount':'500'}]
I've tried this with no luck:
[li.update({li['item']:spl}) for li in lst for spl in li['item'].split(',')]
return Li

def unpack_dict(d, field, unpack):
packed = d.pop(field)
for item in unpack(packed):
result = d.copy()
result[field] = item
yield result
lst = [
{'item':'321,121,343','name':'abby','amount':'400'},
{'item':'111,222,333','name':'chris','amount':'500'}
]
new_lst = [
ud
for d in lst
for ud in unpack_dict(d, "item", lambda item: item.split(","))
]
gives new_lst =
[
{'amount': '400', 'item': '321', 'name': 'abby'},
{'amount': '400', 'item': '121', 'name': 'abby'},
{'amount': '400', 'item': '343', 'name': 'abby'},
{'amount': '500', 'item': '111', 'name': 'chris'},
{'amount': '500', 'item': '222', 'name': 'chris'},
{'amount': '500', 'item': '333', 'name': 'chris'}
]

Here is a working script:
lst = [{'item':'321,121,343','name':'abby','amount':'400'}, {'item':'321,121,343','name':'chris','amount':'500'}]
new_list = [{'item':j, 'name': i['name'], 'amount': i['amount']} for i in lst for j in i['item'].split(',')]
>>> print new_list
[{item:'321','name':'abby','amount':'400'},{item:'121','name':'abby','amount':'400'},{item:'343','name':'abby','amount':'400'},{item:'111','name':'chris','amount':'500'},{item:'222','name':'chris','amount':'500'},{item:'333','name':'chris','amount':'500'}]
Demo: http://repl.it/RaE