I have an object in the following format:
{
'af': {
'bidi': False,
'code': 'af',
'name': 'Afrikaans',
'name_local': 'Afrikaans'
},
'ar': {
'bidi': True,
'code': 'ar',
'name': 'Arabic',
'name_local': 'العربيّة'
},
...
}
This is a list of locales as found in django.conf.locale.LANG_INFO. (see here for full reference: https://github.com/django/django/blob/master/django/conf/locale/init.py).
Now, I am hoping to utilise this list in a model class:
locale = models.CharField(max_length=5, choices=get_locale_choices(), default='en')
Such that I have the following utility function:
from django.conf.locale import LANG_INFO
def get_locale_choices():
return ?
Now, that ? that gets returned, I'd like to be in the following format:
[
('af', 'Afrikaans'),
('ar', 'Arabic'),
...
]
My question is this, how would I turn the LANG_INFO dictionary into the above list of tuples?
It feels like something like this is close:
a_dict = {'color': 'blue', 'fruit': 'apple', 'pet': 'dog'}
d_items = a_dict.items()
d_items # Here d_items is a view of items
dict_items([('color', 'blue'), ('fruit', 'apple'), ('pet', 'dog')])
But...hmmm, not sure? I want to extract a sub-value from the key item...
Use list comprehension:
from django.conf.locale import LANG_INFO
def get_locale_choices():
return [(k, v['name']) for k, v in LANG_INFO.items() if 'name' in v]
The ... if 'name' in v part is necessary to ensure that the cases that have a 'fallback' but no 'name' (e.g. zh-cn, zh-my, zh-sg, etc.) are ignored.
Result:
[('af', 'Afrikaans'), ('ar', 'Arabic'), ...]
Note: django.conf.locale.LANG_INFO seems to be internal to Django, so your use of it isn't officially sanctioned by Django.
Related
I have a dictionary with some values that are type list, i need to convert each list in another dictionary and insert this new dictionary at the place of the list.
Basically, I have this dictionary
Dic = {
'name': 'P1',
'srcintf': 'IntA',
'dstintf': 'IntB',
'srcaddr': 'IP1',
'dstaddr': ['IP2', 'IP3', 'IP4'],
'service': ['P_9100', 'SNMP'],
'schedule' : 'always',
}
I need to reemplace the values that are lists
Expected output:
Dic = {
'name': 'P1',
'srcintf': 'IntA',
'dstintf': 'IntB',
'srcaddr': 'IP1',
'dstaddr': [
{'name': 'IP2'},
{'name': 'IP3'},
{'name': 'IP4'}
],
'service': [
{'name': 'P_9100'},
{'name': 'SNMP'}
],
'schedule' : 'always',
}
So far I have come up with this code:
for k,v in Dic.items():
if not isinstance(v, list):
NewDic = [k,v]
print(NewDic)
else:
values = v
keys = ["name"]*len(values)
for item in range(len(values)):
key = keys[item]
value = values[item]
SmallDic = {key : value}
liste.append(SmallDic)
NewDic = [k,liste]
which print this
['name', 'P1']
['srcintf', 'IntA']
['dstintf', 'IntB']
['srcaddr', 'IP1']
['schedule', 'always']
['schedule', 'always']
I think is a problem with the loop for, but so far I haven't been able to figure it out.
You need to re-create the dictionary. With some modifications to your existing code so that it generates a new dictionary & fixing the else clause:
NewDic = {}
for k, v in Dic.items():
if not isinstance(v, list):
NewDic[k] = v
else:
NewDic[k] = [
{"name": e} for e in v # loop through the list values & generate a dict for each
]
print(NewDic)
Result:
{'name': 'P1', 'srcintf': 'IntA', 'dstintf': 'IntB', 'srcaddr': 'IP1', 'dstaddr': [{'name': 'IP2'}, {'name': 'IP3'}, {'name': 'IP4'}], 'service': [{'name': 'P_9100'}, {'name': 'SNMP'}], 'schedule': 'always'}
I have an object in Python 3 of this format:
a = {
'events': [
{
'timestamp': 123,
'message': 'test'
},
{
'timestamp': 456,
'message': 'foo'
},
{
'timestamp': 789,
'message': 'testbar'
},
],
'first': 'abc',
'last': 'def'
}
I want to create a new object of the same format, but filtered by whether the message key's corresponding value contains a certain string, for example filtering by "test":
a = {
'events': [
{
'timestamp': 123,
'message': 'test'
},
{
'timestamp': 789,
'message': 'testbar'
},
],
'first': 'abc',
'last': 'def'
}
Can I use a nested comprehension for this? I know you can do nested list comprehensions like:
[[y*2 for y in x] for x in l]
But is there a neat way for a dict > list > dict situation?
One option would be to create a new copy of the input dict without events, and then set the filtered events as you require, like this:
copy = {k: v for k, v in a.items() if k != 'events'}
copy['events'] = [e for e in a['events'] if 'test' in e['message']]
Or if you don't mind overwriting the original input, simply do this:
a['events'] = [e for e in a['events'] if 'test' in e['message']]
I would go with a list comprehension with an if-statement like the following:
[event for event in a["events"] if event["message"] == "test" ]
Loop through the values of the "events"-key and add them to the list if the value of their "message" key equals "test".
The result is a list of dictionaries that you can assign back to a["events"] or a copy of a if you would like to preserve a["events"].
So - you can use multiple layers of comprehension, but that doesn't mean you should. I think for such an example, you'd produce cleaner code, by running it through a couple of for loops. Having that said, I think the following is technically achieves the outcome you're asking for.
>>> pprint.pprint(a)
{'events': [{'message': 'test', 'timestamp': 123},
{'message': 'foo', 'timestamp': 456},
{'message': 'testbar', 'timestamp': 789}],
'first': 'abc',
'last': 'def'}
>>> aa = copy.deepcopy(a)
>>> aa['beta'] = aa['events']
>>> pprint.pprint({k:[item for item in v if 'test' in item['message']] if isinstance(v, list) else v for k, v in aa.items()})
{'beta': [{'message': 'test', 'timestamp': 123},
{'message': 'testbar', 'timestamp': 789}],
'events': [{'message': 'test', 'timestamp': 123},
{'message': 'testbar', 'timestamp': 789}],
'first': 'abc',
'last': 'def'}
>>> pprint.pprint({k:[item for item in v if 'test' in item['message']] if isinstance(v, list) else v for k, v in a.items()})
{'events': [{'message': 'test', 'timestamp': 123},
{'message': 'testbar', 'timestamp': 789}],
'first': 'abc',
'last': 'def'}
As said, this is something you can do; I would however on behalf of everyone who's had to read other people code in their careers, respectfully request that you don't use this in production code. A couple of for loops might be more LOC, but would in most cases be much more readable and maintainable.
I have a JSON data as below.
input_list = [["Richard",[],{"children":"yes","divorced":"no","occupation":"analyst"}],
["Mary",["testing"],{"children":"no","divorced":"yes","occupation":"QA analyst","location":"Seattle"}]]
I have another list where I have the prospective keys present
list_keys = ['name', 'current_project', 'details']
I am trying to create a dic using both to make the data usable for metrics
I have summarized the both the list for the question but it goes on forever, there are multiple elements in the list. input_list is a nested list which has 500k+ elements and each list element have 70+ elements of their own (expect the details one)
list_keys also have 70+ elements in it.
I was trying to create a dict using zip but that its not helping given the size of data, also with zip I am not able to exclude the "details" element from
I am expecting output something like this.
[
{
"name": "Richard",
"current_project": "",
"children": "yes",
"divorced": "no",
"occupation": "analyst"
},
{
"name": "Mary",
"current_project" :"testing",
"children": "no",
"divorced": "yes",
"occupation": "QA analyst",
"location": "Seattle"
}
]
I have tried this so far
>>> for line in input_list:
... zipbObj = zip(list_keys, line)
... dictOfWords = dict(zipbObj)
...
>>> print dictOfWords
{'current_project': ['testing'], 'name': 'Mary', 'details': {'location': 'Seattle', 'children': 'no', 'divorced': 'yes', 'occupation': 'QA analyst'}}
but with this I am unable to to get rid of nested dict key "details". so looking for help with that
Seems like what you wanted was a list of dictionaries, here is something i coded up in the terminal and copied in here. Hope it helps.
>>> list_of_dicts = []
>>> for item in input_list:
... dict = {}
... for i in range(0, len(item)-2, 3):
... dict[list_keys[0]] = item[i]
... dict[list_keys[1]] = item[i+1]
... dict.update(item[i+2])
... list_of_dicts.append(dict)
...
>>> list_of_dicts
[{'name': 'Richard', 'current_project': [], 'children': 'yes', 'divorced': 'no', 'occupation': 'analyst'
}, {'name': 'Mary', 'current_project': ['testing'], 'children': 'no', 'divorced': 'yes', 'occupation': '
QA analyst', 'location': 'Seattle'}]
I will mention it is not the ideal method of doing this since it relies on perfectly ordered items in the input_list.
people = input_list = [["Richard",[],{"children":"yes","divorced":"no","occupation":"analyst"}],
["Mary",["testing"],{"children":"no","divorced":"yes","occupation":"QA analyst","location":"Seattle"}]]
list_keys = ['name', 'current_project', 'details']
listout = []
for person in people:
dict_p = {}
for key in list_keys:
if not key == 'details':
dict_p[key] = person[list_keys.index(key)]
else:
subdict = person[list_keys.index(key)]
for subkey in subdict.keys():
dict_p[subkey] = subdict[subkey]
listout.append(dict_p)
listout
The issue with using zip is that you have that additional dictionary in the people list. This will get the following output, and should work through a larger list of individuals:
[{'name': 'Richard',
'current_project': [],
'children': 'yes',
'divorced': 'no',
'occupation': 'analyst'},
{'name': 'Mary',
'current_project': ['testing'],
'children': 'no',
'divorced': 'yes',
'occupation': 'QA analyst',
'location': 'Seattle'}]
This script will go through every item of input_list and creates new list where there aren't any list or dictionaries:
input_list = [
["Richard",[],{"children":"yes","divorced":"no","occupation":"analyst"}],
["Mary",["testing"],{"children":"no","divorced":"yes","occupation":"QA analyst","location":"Seattle"}]
]
list_keys = ['name', 'current_project', 'details']
out = []
for item in input_list:
d = {}
out.append(d)
for value, keyname in zip(item, list_keys):
if isinstance(value, dict):
d.update(**value)
elif isinstance(value, list):
if value:
d[keyname] = value[0]
else:
d[keyname] = ''
else:
d[keyname] = value
from pprint import pprint
pprint(out)
Prints:
[{'children': 'yes',
'current_project': '',
'divorced': 'no',
'name': 'Richard',
'occupation': 'analyst'},
{'children': 'no',
'current_project': 'testing',
'divorced': 'yes',
'location': 'Seattle',
'name': 'Mary',
'occupation': 'QA analyst'}]
I have a list of dictionaries, themselves with nested lists of dictionaries. All of the nest levels have a similar structure, thankfully. I desire to sort these nested lists of dictionaries. I grasp the technique to sort a list of dictionaries by value. I'm struggling with the recursion that will sort the inner lists.
def reorder(l, sort_by):
# I have been trying to add a recursion here
# so that the function calls itself for each
# nested group of "children". So far, fail
return sorted(l, key=lambda k: k[sort_by])
l = [
{ 'name': 'steve',
'children': [
{ 'name': 'sam',
'children': [
{'name': 'sally'},
{'name': 'sabrina'}
]
},
{'name': 'sydney'},
{'name': 'sal'}
]
},
{ 'name': 'fred',
'children': [
{'name': 'fritz'},
{'name': 'frank'}
]
}
]
print(reorder(l, 'name'))
def reorder(l, sort_by):
l = sorted(l, key=lambda x: x[sort_by])
for item in l:
if "children" in item:
item["children"] = reorder(item["children"], sort_by)
return l
Since you state "I grasp the technique to sort a list of dictionaries by value" I will post some code for recursively gathering data from another SO post I made, and leave it to you to implement your sorting technique. The code:
myjson = {
'transportation': 'car',
'address': {
'driveway': 'yes',
'home_address': {
'state': 'TX',
'city': 'Houston'}
},
'work_address': {
'state': 'TX',
'city': 'Sugarland',
'location': 'office-tower',
'salary': 30000}
}
def get_keys(some_dictionary, parent=None):
for key, value in some_dictionary.items():
if '{}.{}'.format(parent, key) not in my_list:
my_list.append('{}.{}'.format(parent, key))
if isinstance(value, dict):
get_keys(value, parent='{}.{}'.format(parent, key))
else:
pass
my_list = []
get_keys(myjson, parent='myjson')
print(my_list)
Is intended to retrieve all keys recursively from the json file. It outputs:
['myjson.address',
'myjson.address.home_address',
'myjson.address.home_address.state',
'myjson.address.home_address.city',
'myjson.address.driveway',
'myjson.transportation',
'myjson.work_address',
'myjson.work_address.state',
'myjson.work_address.salary',
'myjson.work_address.location',
'myjson.work_address.city']
The main thing to note is that if isinstance(value, dict): results in get_keys() being called again, hence the recursive capabilities of it (but only for nested dictionaries in this case).
I have this data
data = [
{
'id': 'abcd738asdwe',
'name': 'John',
'mail': 'test#test.com',
},
{
'id': 'ieow83janx',
'name': 'Jane',
'mail': 'test#foobar.com',
}
]
The id's are unique, it's impossible that multiple dictonaries have the same id.
For example I want to get the item with the id "ieow83janx".
My current solution looks like this:
search_id = 'ieow83janx'
item = [x for x in data if x['id'] == search_id][0]
Do you think that's the be solution or does anyone know an alternative solution?
Since the ids are unique, you can store the items in a dictionary to achieve O(1) lookup.
lookup = {ele['id']: ele for ele in data}
then you can do
user_info = lookup[user_id]
to retrieve it
If you are going to get this kind of operations more than once on this particular object, I would recommend to translate it into a dictionary with id as a key.
data = [
{
'id': 'abcd738asdwe',
'name': 'John',
'mail': 'test#test.com',
},
{
'id': 'ieow83janx',
'name': 'Jane',
'mail': 'test#foobar.com',
}
]
data_dict = {item['id']: item for item in data}
#=> {'ieow83janx': {'mail': 'test#foobar.com', 'id': 'ieow83janx', 'name': 'Jane'}, 'abcd738asdwe': {'mail': 'test#test.com', 'id': 'abcd738asdwe', 'name': 'John'}}
data_dict['ieow83janx']
#=> {'mail': 'test#foobar.com', 'id': 'ieow83janx', 'name': 'Jane'}
In this case, this lookup operation will cost you some constant* O(1) time instead of O(N).
How about the next built-in function (docs):
>>> data = [
... {
... 'id': 'abcd738asdwe',
... 'name': 'John',
... 'mail': 'test#test.com',
... },
... {
... 'id': 'ieow83janx',
... 'name': 'Jane',
... 'mail': 'test#foobar.com',
... }
... ]
>>> search_id = 'ieow83janx'
>>> next(x for x in data if x['id'] == search_id)
{'id': 'ieow83janx', 'name': 'Jane', 'mail': 'test#foobar.com'}
EDIT:
It raises StopIteration if no match is found, which is a beautiful way to handle absence:
>>> search_id = 'does_not_exist'
>>> try:
... next(x for x in data if x['id'] == search_id)
... except StopIteration:
... print('Handled absence!')
...
Handled absence!
Without creating a new dictionary or without writing several lines of code, you can simply use the built-in filter function to get the item lazily, not checking after it finds the match.
next(filter(lambda d: d['id']==search_id, data))
should for just fine.
Would this not achieve your goal?
for i in data:
if i.get('id') == 'ieow83janx':
print(i)
(xenial)vash#localhost:~/python$ python3.7 split.py
{'id': 'ieow83janx', 'name': 'Jane', 'mail': 'test#foobar.com'}
Using comprehension:
[i for i in data if i.get('id') == 'ieow83janx']
if any(item['id']=='ieow83janx' for item in data):
#return item
As any function returns true if iterable (List of dictionaries in your case) has value present.
While using Generator Expression there will not be need of creating internal List. As there will not be duplicate values for the id in List of dictionaries, any will stop the iteration until the condition returns true. i.e the generator expression with any will stop iterating on shortcircuiting. Using List comprehension will create a entire List in the memory where as GE creates the element on the fly which will be better if you are having large items as it uses less memory.