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Recursively sort a list of nested dictionaries by value

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).

Find item in a list of dictionaries

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.

Remove duplicates in python dictionary

I have a list of dictionaries in python and I would like to override old value with duplicate value. Please let me know how can I do.
{'message': [{'name': 'raghav', 'id': 10}, {'name': 'raghav', 'id': 11}]}
Output should be:
{'message': [ {'name': 'raghav', 'id': 11}]}

I don't know what you mean by "override old value with duplicate value". If you mean just picking the second dict from the list, you could:
print({k: [v[1]] for (k, v) in data.items()})
If the idea is to update the "name" with a newer value of "id" as you move along the list, then maybe:
def merge_records(data):
records = data['message']
users = {}
for record in records:
name = record['name']
id_ = record['id']
users[name] = id_
new_records = []
for name, id_ in users.items():
new_records.append({'name': name, 'id': id_})
return {'message': new_records}
But, if you have any control over how the data is represented, you might reconsider. You probably want a different data structure.

Here you go:
d = {'message': [{'name': 'raghav', 'id': 10}, {'name': 'raghav', 'id': 11}]}
#loop over outer dictionary
for key, value in d.items():
d[key] = [dict([t for k in value for t in k.items()])]
print(d)
Edit:
As per your requirement:
d = {'message': [ {'name': 'raghav', 'id': 11}, {'name': 'krish', 'id': 20}, {'name': 'anu', 'id': 30}]}
for key, value in d.items():
print [dict((k1,v1)) for k1,v1 in dict([tuple(i.items()) for i in value for val in i.items()]).items()]

Python: retrieve arbitrary dictionary path and amend data?

Simple Python question, but I'm scratching my head over the answer!
I have an array of strings of arbitrary length called path, like this:
path = ['country', 'city', 'items']
I also have a dictionary, data, and a string, unwanted_property. I know that the dictionary is of arbitrary depth and is dictionaries all the way down, with the exception of the items property, which is always an array.
[CLARIFICATION: The point of this question is that I don't know what the contents of path will be. They could be anything. I also don't know what the dictionary will look like. I need to walk down the dictionary as far as the path indicates, and then delete the unwanted properties from there, without knowing in advance what the path looks like, or how long it will be.]
I want to retrieve the parts of the data object (if any) that matches the path, and then delete the unwanted_property from each.
So in the example above, I would like to retrieve:
data['country']['city']['items']
and then delete unwanted_property from each of the items in the array. I want to amend the original data, not a copy. (CLARIFICATION: By this I mean, I'd like to end up with the original dict, just minus the unwanted properties.)
How can I do this in code?
I've got this far:
path = ['country', 'city', 'items']
data = {
'country': {
'city': {
'items': [
{
'name': '114th Street',
'unwanted_property': 'foo',
},
{
'name': '8th Avenue',
'unwanted_property': 'foo',
},
]
}
}
}
for p in path:
if p == 'items':
data = [i for i in data[p]]
else:
data = data[p]
if isinstance(data, list):
for d in data:
del d['unwanted_property']
else:
del data['unwanted_property']
The problem is that this doesn't amend the original data. It also relies on items always being the last string in the path, which may not always be the case.
CLARIFICATION: I mean that I'd like to end up with:
{
'country': {
'city': {
'items': [
{
'name': '114th Street'
},
{
'name': '8th Avenue'
},
]
}
}
}
Whereas what I have available in data is only [{'name': '114th Street'}, {'name': '8th Avenue'}].
I feel like I need something like XPath for the dictionary.

The problem you are overwriting the original data reference. Change your processing code to
temp = data
for p in path:
temp = temp[p]
if isinstance(temp, list):
for d in temp:
del d['unwanted_property']
else:
del temp['unwanted_property']
In this version, you set temp to point to the same object that data was referring to. temp is not a copy, so any changes you make to it will be visible in the original object. Then you step temp along itself, while data remains a reference to the root dictionary. When you find the path you are looking for, any changes made via temp will be visible in data.
I also removed the line data = [i for i in data[p]]. It creates an unnecessary copy of the list that you never need, since you are not modifying the references stored in the list, just the contents of the references.
The fact that path is not pre-determined (besides the fact that items is going to be a list) means that you may end up getting a KeyError in the first loop if the path does not exist in your dictionary. You can handle that gracefully be doing something more like:
try:
temp = data
for p in path:
temp = temp[p]
except KeyError:
print('Path {} not in data'.format(path))
else:
if isinstance(temp, list):
for d in temp:
del d['unwanted_property']
else:
del temp['unwanted_property']

The problem you are facing is that you are re-assigning the data variable to an undesired value. In the body of your for loop you are setting data to the next level down on the tree, for instance given your example data will have the following values (in order), up to when it leaves the for loop:
data == {'country': {'city': {'items': [{'name': '114th Street', 'unwanted_property': 'foo',}, {'name': '8th Avenue', 'unwanted_property': 'foo',},]}}}
data == {'city': {'items': [{'name': '114th Street', 'unwanted_property': 'foo',}, {'name': '8th Avenue', 'unwanted_property': 'foo',},]}}
data == {'items': [{'name': '114th Street', 'unwanted_property': 'foo',}, {'name': '8th Avenue', 'unwanted_property': 'foo',},]}
data == [{'name': '114th Street', 'unwanted_property': 'foo',}, {'name': '8th Avenue', 'unwanted_property': 'foo',},]
Then when you delete the items from your dictionaries at the end you are left with data being a list of those dictionaries as you have lost the higher parts of the structure. Thus if you make a backup reference for your data you can get the correct output, for example:
path = ['country', 'city', 'items']
data = {
'country': {
'city': {
'items': [
{
'name': '114th Street',
'unwanted_property': 'foo',
},
{
'name': '8th Avenue',
'unwanted_property': 'foo',
},
]
}
}
}
data_ref = data
for p in path:
if p == 'items':
data = [i for i in data[p]]
else:
data = data[p]
if isinstance(data, list):
for d in data:
del d['unwanted_property']
else:
del data['unwanted_property']
data = data_ref

def delKey(your_dict,path):
if len(path) == 1:
for item in your_dict:
del item[path[0]]
return
delKey( your_dict[path[0]],path[1:])
data
{'country': {'city': {'items': [{'name': '114th Street', 'unwanted_property': 'foo'}, {'name': '8th Avenue', 'unwanted_property': 'foo'}]}}}
path
['country', 'city', 'items', 'unwanted_property']
delKey(data,path)
data
{'country': {'city': {'items': [{'name': '114th Street'}, {'name': '8th Avenue'}]}}}

You need to remove the key unwanted_property.
names_list = []
def remove_key_from_items(data):
for d in data:
if d != 'items':
remove_key_from_items(data[d])
else:
for item in data[d]:
unwanted_prop = item.pop('unwanted_property', None)
names_list.append(item)
This will remove the key. The second parameter None is returned if the key unwanted_property does not exist.
EDIT:
You can use pop even without the second parameter. It will raise KeyError if the key does not exist.
EDIT 2: Updated to recursively go into depth of data dict until it finds the items key, where it pops the unwanted_property as desired and append into the names_list list to get the desired output.

Using operator.itemgetter you can compose a function to return the final key's value.
import operator, functools
def compose(*functions):
'''returns a callable composed of the functions
compose(f, g, h, k) -> f(g(h(k())))
'''
def compose2(f, g):
return lambda x: f(g(x))
return functools.reduce(compose2, functions, lambda x: x)
get_items = compose(*[operator.itemgetter(key) for key in path[::-1]])
Then use it like this:
path = ['country', 'city', 'items']
unwanted_property = 'unwanted_property'
for thing in get_items(data):
del thing[unwanted_property]
Of course if the path contains non-existent keys it will throw a KeyError - you probably should account for that:
path = ['country', 'foo', 'items']
get_items = compose(*[operator.itemgetter(key) for key in path[::-1]])
try:
for thing in get_items(data):
del thing[unwanted_property]
except KeyError as e:
print('missing key:', e)

You can try this:
path = ['country', 'city', 'items']
previous_data = data[path[0]]
previous_key = path[0]
for i in path:
previous_data = previous_data[i]
previous_key = i
if isinstance(previous_data, list):
for c, b in enumerate(previous_data):
if "unwanted_property" in b:
del previous_data[c]["unwanted_property"]
current_dict = {}
previous_data_dict = {}
for i, a in enumerate(path):
if i == 0:
current_dict[a] = data[a]
previous_data_dict = data[a]
else:
if a == previous_key:
current_dict[a] = previous_data
else:
current_dict[a] = previous_data_dict[a]
previous_data_dict = previous_data_dict[a]
data = current_dict
print(data)
Output:
{'country': {'city': {'items': [{'name': '114th Street'}, {'name': '8th Avenue'}]}}, 'items': [{'name': '114th Street'}, {'name': '8th Avenue'}], 'city': {'items': [{'name': '114th Street'}, {'name': '8th Avenue'}]}}

parse multilevel json to string with condition

I have this nested json item that I just want to flatten out to a comma separated string (i.e. parkinson:5, billy mays:4)so I can store in a database if needed for future analysis. I wrote out the function below but am wondering if there's a more elegant way using list comprehension (or something else). I found this post but I'm not sure how to adapt it for my needs (Python - parse JSON values by multilevel keys).
Data looks like this:
{'persons':
[{'name': 'parkinson', 'sentiment': '5'},
{'name': 'knott david', 'sentiment': 'none'},
{'name': 'billy mays', 'sentiment': '4'}],
'organizations':
[{'name': 'piper jaffray companies', 'sentiment': 'none'},
{'name': 'marketbeat.com', 'sentiment': 'none'},
{'name': 'zacks investment research', 'sentiment': 'none'}]
'locations': []
}
Here's my code:
def parse_entities(data):
results = ''
for category in data.keys():
# for c_id, category in enumerate(data.keys()):
entity_data = data[category]
for e_id, entity in enumerate(entity_data):
if not entity_data[e_id]['sentiment'] == 'none':
results = results + (data[category][e_id]['name'] + ":" +
data[category][e_id]['sentiment'] + ",")
return results

Firstly, the most important thing to make your code shorter and nicer to look at is to use your own variables. Be aware that entity_data = data[category] and entity = entity_data[e_id]. So you can write entity['name'] instead of data[category][e_id]['name'].
Secondly, if you want something like
for category in data.keys():
entity_data = data[category]
you can make it shorter and easier to read by changing it to
for category, entity_data in data.items():
But you don't even need that here, you can just use the data.values() iterator to get the values. When combining these improvements your code looks like this:
def parse_entities(data):
results = ''
for entity_data in data.values():
for entity in entity_data:
if entity['sentiment'] != 'none':
results += entity['name'] + ":" + entity['sentiment'] + ","
return results
(I have also changed results = results + ... to results += ... and if not entity['sentiment'] == 'none' to if entity['sentiment'] != 'none', because it is shorter and doesn't lower the readability)
When you have this it is much easier to make it even shorter and more elegant by using list comprehension:
def parse_entities(data):
return ",".join([entity['name'] + ":" + entity['sentiment']
for entity_data in data.values()
for entity in entity_data
if not entity['sentiment'] == 'none'])

Maybe something like this will work?
def parse_entities(data):
results = []
for category in data.keys():
results += list(map(lambda x: '{0}:{1}'.format(x['name'], x['sentiment']),
filter(lambda i: i['sentiment'] != 'none', data[category])))
return ','.join(results)
if __name__ == '__main__':
print(parse_entities(data))
With the output looking like this
parkinson:5,billy mays:4

This might be a way to do it. Even though using a 'proper library' (depending on your actual use case) makes more sense.
data = {
'persons':
[{'name': 'parkinson', 'sentiment': '5'},
{'name': 'knott david', 'sentiment': 'none'},
{'name': 'billy mays', 'sentiment': '4'}],
'organizations':
[{'name': 'piper jaffray companies', 'sentiment': 'none'},
{'name': 'marketbeat.com', 'sentiment': 'none'},
{'name': 'zacks investment research', 'sentiment': 'none'}],
'locations': []
}
import itertools
# eq. = itertools.chain.from_iterable(data.values())
dicts = itertools.chain(*data.values())
pairs = [":".join([d['name'], d['sentiment']])
for d in dicts if d['sentiment'] != 'none']
result = ",".join(pairs)
print(result)
# parkinson:5,billy mays:4
# short, but less readable version
result = ",".join([":".join([d['name'], d['sentiment']])
for d in itertools.chain(*data.values())
if d['sentiment'] != 'none'])

This is a problem where we need to perform the 3 separate tasks:
Filter out unqualified rows of data
Flatten the dict of lists into a simple list
Transform each dictionary object into a simple tuple, ready for formatting
Here is the code:
def parse_entities(data):
new_data = [
(row['name'], row['sentiment']) # 3. Transform
for rows in data.values() # 2. Flatten
for row in rows # 2. Flatten
if row['sentiment'] != 'none' # 1. Filter
]
# e.g, new_data = [('parkinson', '5'), ('billy mays', '4')]
return ','.join('{}:{}'.format(*row) for row in new_data)
#
# test code
#
data = {
'locations': [],
'organizations': [
{'name': 'piper jaffray companies', 'sentiment': 'none'},
{'name': 'marketbeat.com', 'sentiment': 'none'},
{'name': 'zacks investment research', 'sentiment': 'none'}
],
'persons': [
{'name': 'parkinson', 'sentiment': '5'},
{'name': 'knott david', 'sentiment': 'none'},
{'name': 'billy mays', 'sentiment': '4'}
],
}
print parse_entities(data)
Output:
parkinson:5,billy mays:4

Here's a generator expression that does it:
data = {'persons': [
{'name': 'parkinson', 'sentiment': '5'},
{'name': 'knott david', 'sentiment': 'none'},
{'name': 'billy mays', 'sentiment': '4'}],
'organizations': [
{'name': 'piper jaffray companies', 'sentiment': 'none'},
{'name': 'marketbeat.com', 'sentiment': '99'},
{'name': 'zacks investment research', 'sentiment': 'none'}],
'locations': []
}
results = ','.join(entity['name'] + ':' + entity['sentiment']
for category, entity_data in data.items()
for entity in entity_data if entity['sentiment'] is not 'none')
print(results) # -> parkinson:5,billy mays:4,marketbeat.com:99
Note: I changed the sample data slightly to make sure it handled data in more than one category the same as your code.