I believe there must be a way to point specific key from nested dict, not in the traditional ways.
imagine dictionary like this.
dict1 = { 'level1': "value",
'unexpectable': { 'maybe': { 'onemotime': {'name': 'John'} } } }
dict2 = { 'level1': "value", 'name': 'Steve'}
dict3 = { 'find': { 'what': { 'you': { 'want': { 'in': { 'this': { 'maze': { 'I': { 'made': { 'for': { 'you': { 'which': { 'is in': { 'fact that': { 'was just': { 'bully your': { 'searching': { 'for': { 'the name': { 'even tho': { 'in fact': { 'actually': { 'there': { 'is': { 'in reality': { 'only': { 'one': { 'key': { 'named': { 'name': 'Michael' } } } } } } } } } } } } } } } } } } } } } } } } } } } } } }
in this case, if we want to point 'name' key to get 'John' and 'Steve' and the 'Michael', you should code differently against dict1 and dict2 and dict3
and the traditional way to point the key buried in nested dictionary that I know is this.
print(dict1['unexpectable']['maybe']['onemotime']['name'])
and if you don't want your code to break because of empty value of dict, you may want to use get() function.
and I'm curious that if I want to get 'name' of dict1 safely with get(), should I code like this?
print(dict1.get('unexpectable', '').get('maybe', '').get('onemotime', '').get('name', ''))
in fact, i've got error when run those get().get().get().get() thing.
And please consider if you have to print() 'name' from that horrible dict3 even it has actually only one key.
and, imagine the case you extract 'name' from unknown dict4 which you cannot imagine what nesting structure the dict4 would have.
I believe that python must have a way to deal with this.
I searched on the internet about this problem, but the solutions seems really really difficult.
I just wanted simple solution.
the solution without pointing every keys on the every level.
like just pointing that last level key, the most important key.
like, print(dict1.lastlevel('name')) --> 'John'
like, no matter what structure of nesting levels they have, no matter how many duplicates they have, even if they omitted nested key in the middle of nested dict so that dict17 has one less level of dict16, you could get what you want, the last level value of the last level key.
So Conclusion.
I want to know if there is a simple solution like
print(dict.lastlevel('name'))
without creating custom function.
I want to know if there is solution like above from the default python methods, syntax, function, logic or concept.
The solution like above can be applied to dict1, dict2, dict3, to whatever dict would come.
There is no built-in method to accomplish what you are asking for. However, you can use a recursive function to dig through a nested dictionary. The function checks if the desired key is in the dictionary and returns the value if it is. Otherwise it iterates over the dict's values for other dictionaries and scans their keys as well, repeating until it reaches the bottom.
dict1 = { 'level1': "value",
'unexpectable': { 'maybe': { 'onemotime': {'name': 'John'} } } }
dict2 = { 'level1': "value", 'name': 'Steve'}
dict3 = { 'find': { 'what': { 'you': { 'want': { 'in': { 'this': { 'maze': { 'I': {
'made': { 'for': { 'you': { 'which': { 'is in': { 'fact that': {
'was just': { 'bully your': { 'searching': { 'for': { 'the name': {
'even tho': { 'in fact': { 'actually': { 'there': { 'is': { 'in reality': {
'only': { 'one': { 'key': { 'named': { 'name': 'Michael'
} } } } } } } } } } } } } } } } } } } } } } } } } } } } } }
def get_nested_dict_key(d, key):
if key in d:
return d[key]
else:
for item in d.values():
if not isinstance(item, dict):
continue
return get_nested_dict_key(item, key)
print(get_nested_dict_key(dict1, 'name'))
print(get_nested_dict_key(dict2, 'name'))
print(get_nested_dict_key(dict3, 'name'))
# prints:
# John
# Steve
# Michael
You can make simple recursive generator function which yields value of every particular key:
def get_nested_key(source, key):
if isinstance(source, dict):
key_value = source.get(key)
if key_value:
yield key_value
for value in source.values():
yield from get_nested_key(value, key)
elif isinstance(source, (list, tuple)):
for value in source:
yield from get_nested_key(value, key)
Usage:
dictionaries = [
{'level1': 'value', 'unexpectable': {'maybe': {'onemotime': {'name': 'John'}}}},
{'level1': 'value', 'name': 'Steve'},
{'find': {'what': {'you': {'want': {'in': {'this': {'maze': {'I': {'made': {'for': {'you': {'which': {'is in': {'fact that': {'was just': {'bully your': {'searching': {'for': {'the name': {'even tho': {'in fact': {'actually': {'there': {'is': {'in reality': {'only': {'one': {'key': {'named': {'name': 'Michael'}}}}}}}}}}}}}}}}}}}}}}}}}}}}}},
{'level1': 'value', 'unexpectable': {'name': 'Alex', 'maybe': {'onemotime': {'name': 'John'}}}},
{}
]
for d in dictionaries:
print(*get_nested_key(d, 'name'), sep=', ')
Output:
John
Steve
Michael
Alex, John
Related
How am I supposed to extract the value 1234 from "password"?
account = {
'Dicky': {
'password': 1234,
'balance': {
'USD': 10,
'HKD': 10000
}
}
}
to extract values form dict:
dict["key"]
so in your case because there is a nested dict:
dict["1stkey"]["2ndkey"]
so:
account["Dicky"]["password"]
In Python, a nested dictionary is a dictionary inside a dictionary.
example:
nested_dict = { 'dictA': {'key_1': 'value_1'},
'dictB': {'key_2': 'value_2'}}
so:
nested_dict['dict1']['key_A']--->'value_A'
this question
account = {
'Dicky': {
'password': 1234,
'balance': {
'USD': 10,
'HKD': 10000
}
}
}
account["Dicky"]["password"]
print(account["Dicky"]["password"])
1234
I am working with pymongo and after writing aggregate query
db.collection.aggregate([{'$project': {'Id': '$ResultData.Id','data' : '$Results.Data'}}])
I received the object:
{'data': [{'key': 'valid', 'value': 'true'},
{'key': 'number', 'value': '543543'},
{'key': 'name', 'value': 'Saturdays cx'},
{'key': 'message', 'value': 'it is valid.'},
{'key': 'city', 'value': 'London'},
{'key': 'street', 'value': 'Bigeye'},
{'key': 'pc', 'value': '3566'}],
Is there a way that I can access the values by the key name? Like that '$Results.Data.city' and I will receive London. I would like to do that on the level of MongoDB aggregate query so it means I want to write a query in the way:
db.collection.aggregate([{'$project':
{'Id': '$ResultData.Id',
'data' : '$Results.Data',
'city' : $Results.Data.city',
'name' : $Results.Data.name',
'street' : $Results.Data.street',
'pc' : $Results.Data.pc',
}}])
And receive all the values of provided keys.
Using the $elemMatch projection operator in the following query from mongo shell:
db.collection.find(
{ _id: <some_value> },
{ _id: 0, data: { $elemMatch: { key: "city" } } }
)
The output:
{ "data" : [ { "key" : "city", "value" : "London" } ] }
Using PyMongo (gets the same output):
collection.find_one(
{ '_id': <some_value> },
{ '_id': 0, 'data': { '$elemMatch': { 'key': 'city' } } }
)
Using PyMongo aggregate method (gets the same result):
pipeline = [
{
'$project': {
'_id': 0,
'data': {
'$filter': {
'input': '$data', 'as': 'dat',
'cond': { '$eq': [ '$$dat.key', INPUT_KEY ] }
}
}
}
}
]
INPUT_KEY = 'city'
pprint.pprint(list(collection.aggregate(pipeline)))
Naming the received object "result", if result['data'] always is a list of dictionaries with 2 keys (key and value), you can convert the whole list to a dictionary using keys as keys and values as values. Given that this statement is somewhat confusing, here's the code:
data = {pair['key']: pair['value'] for pair in result['data']}
From here, data['city'] will give you 'London', data['street'] will be 'Bigeye' and so on. Obviously, this assumes that there are no conflicts amoung key values in result['data']. Note that this dictionary will (just as the original result['data']) only contain strings so don't expect data['number'] to be an integer.
Another approach would be to dynamically create an object holding each key-value pair as an attribute, allowing you to use the following syntax: data.city, data.street, ... But this would required more complicated code and is a less common and stable approach.
I am writing a function that takes 2 strings as inputs and would move a section of the dictionary to another.
def move(item_to_move, destination):
# do something....
My initial dictionary looks like this.
directories = {
'beers': {
'ipa': {
'stone': {}
}
},
'wines': {
'red': {
'cabernet': {}
}
},
'other' : {}
}
I would like to move either a subsection or section of the dictionary to another section. The sections are represented by each key of the path delimited by a '/'. For example, the inputs for my function would be:
item_to_move='beers/ipa'
destination='other'
move(directories, item_to_move,destination)
The output would be:
{
'wines': {
'red': {
'cabernet': {}
},
},
'other' :{
'beers': {
'ipa': {
'stone': {}
} }
},
}
NOTE: I am assuming all input paths for items_to_move are valid.
Find the origin's parent dictionary and the target's dictionary, then update the the target's dictionary with the origin's key and value (removing it from the origin's parent):
def move(tree,originPath,targetPath):
originKey = None
for originName in originPath.split("/"):
originParent = originParent[originKey] if originKey else tree
originKey = originName
targetDict = tree
for targetName in targetPath.split("/"):
targetDict = targetDict[targetName]
targetDict.update({originKey:originParent.pop(originKey)})
output:
directories = {
'beers': {
'ipa': {
'stone': {}
}
},
'wines': {
'red': {
'cabernet': {}
}
},
'other' : {}
}
move(directories,'beers/ipa','other')
print(directories)
{ 'beers': {},
'wines': { 'red': {'cabernet': {}} },
'other': { 'ipa': {'stone': {}} }
}
If I have a python dictionary like the following:
conf = {
'memory': {
'alarm': {
'active': 'yes',
'pagefile_error': {
'active':'no'
}
}
},
'disk': {
'alarm':{
'active':'yes',
'fixed':{
'#dev':{
'active':'yes',
'something':'else'
}
}
}
},
'cpu': {
'alarm': {
'active':'no',
'highcpu': {
'active':'yes'
}
}
}
}
how can I filter only the paths that end in 'active':'yes' and not show any other info.
In addition, for parent items that show up as active: no, I would want to disregard whatever comes after those.
conf = {
'memory': {
'alarm': {
'active': 'yes'
}
},
'disk' : {
'alarm':{
'active':'yes',
'fixed': {
'#dev': {
'active':'yes'
}
}
}
}
}
I don't have any working code for this yet as I'm not sure where to start. all I have at the moment is the starting dictionary.
Using recursion :
def keep_active_only(my_dict):
result_dict = {}
for key, value in my_dict.items():
# If there is embedded dict
if isinstance(value, dict):
# Compute the embedded dict using recursion
result_subdict = keep_active_only(value)
# Keeping result only if not empty
if result_subdict:
result_dict[key] = result_subdict
# Keep active key if value is yes
elif key == "active" and value == "yes":
result_dict[key] = value
# Returns empty dict if active is no
elif key == "active" and value == "no":
return {}
return result_dict
Output :
>>> keep_active_only(conf)
{
'disk': {
'alarm': {
'active': 'yes',
'fixed': {
'#dev': {
'active': 'yes'
}
}
}
},
'memory': {
'alarm': {
'active': 'yes'
}
}
}
You can use recursion:
def active(d):
_r, _flag = [], False
for a, b in d.items():
if a == 'active' and not _flag:
_r.append(b == 'yes')
_flag = True
if not _flag and isinstance(b, dict):
_r.append(active(b))
return all(_r)
def build(d, flag = False):
return {a:b if not isinstance(b, dict) else build(b, 'active' in b)
for a, b in d.items() if ((not isinstance(b, dict) and not flag) or a == 'active') or (isinstance(b, dict) and active(b))}
import json
print(json.dumps(build(conf), indent=4))
Output:
{
"memory": {
"alarm": {
"active": "yes"
}
},
"disk": {
"alarm": {
"active": "yes",
"fixed": {
"#dev": {
"active": "yes"
}
}
}
}
}
Not sure if I understand correctly, but here is a function that discards all data from the dict that does not take you to a particular key and value:
def filter_dict(d, key, value):
new_dict = {}
for d_key, d_value in d.items():
if d_key == key and d_value == value:
new_dict[d_key] = d_value
elif isinstance(d_value, dict):
child = filter_dict(d_value, key, value)
if child:
new_dict[d_key] = child
return new_dict
Here is how you would use it in your example:
from pprint import pprint
conf = {
'memory': {
'alarm': {
'active': 'yes',
'pagefile_error': {
'active':'no'
}
}
},
'disk': {
'alarm': {
'active': 'yes',
'fixed': {
'#dev': {
'active': 'yes',
'something': 'else'
}
}
}
}
}
pprint(filter_dict(conf, 'active', 'yes'))
# {'disk': {'alarm': {'active': 'yes', 'fixed': {'#dev': {'active': 'yes'}}}},
# 'memory': {'alarm': {'active': 'yes'}}}
I have a very large json file with several nested keys. From whaat I've read so far, if you do:
x = json.loads(data)
Python will interpret it as a dictionary (correct me if I'm wrong). The fourth level of nesting in the json file contains several elements named by an ID number and all of them contain an element called children, something like this:
{"level1":
{"level2":
{"level3":
{"ID1":
{"children": [1,2,3,4,5]}
}
{"ID2":
{"children": []}
}
{"ID3":
{"children": [6,7,8,9,10]}
}
}
}
}
What I need to do is to replace all items in all the "children" elements with nothing, meaning "children": [] if the ID number is in a list called new_ids and then convert it back to json. I've been reading on the subject for a few hours now but I haven't found anything similar to this to try to help myself.
I'm running Python 3.3.3. Any ideas are greatly appreciated!!
Thanks!!
EDIT
List:
new_ids=["ID1","ID3"]
Expected result:
{"level1":
{"level2":
{"level3":
{"ID1":
{"children": []}
}
{"ID2":
{"children": []}
}
{"ID3":
{"children": []}
}
}
}
}
First of all, your JSON is invalid. I assume you want this:
{"level1":
{"level2":
{"level3":
{
"ID1":{"children": [1,2,3,4,5]},
"ID2":{"children": []},
"ID3":{"children": [6,7,8,9,10]}
}
}
}
}
Now, load your data as a dictionary:
>>> with open('file', 'r') as f:
... x = json.load(f)
...
>>> x
{u'level1': {u'level2': {u'level3': {u'ID2': {u'children': []}, u'ID3': {u'children': [6, 7, 8, 9, 10]}, u'ID1': {u'children': [1, 2, 3, 4, 5]}}}}}
Now you can loop over the keys in x['level1']['level2']['level3'] and check whether they are in your new_ids.
>>> new_ids=["ID1","ID3"]
>>> for key in x['level1']['level2']['level3']:
... if key in new_ids:
... x['level1']['level2']['level3'][key]['children'] = []
...
>>> x
{u'level1': {u'level2': {u'level3': {u'ID2': {u'children': []}, u'ID3': {u'children': []}, u'ID1': {u'children': []}}}}}
You can now write x back to a file like this:
with open('myfile', 'w') as f:
f.write(json.dumps(x))
If your new_ids list is large, consider making it a set.
If you have simple dictionary like this
data_dict = {
"level1": {
"level2":{
"level3":{
"ID1":{"children": [1,2,3,4,5]},
"ID2":{"children": [] },
"ID3":{"children": [6,7,8,9,10]},
}
}
}
}
than you need only this:
data_dict = {
"level1": {
"level2":{
"level3":{
"ID1":{"children": [1,2,3,4,5]},
"ID2":{"children": [] },
"ID3":{"children": [6,7,8,9,10]},
}
}
}
}
new_ids=["ID1","ID3"]
for idx in new_ids:
if idx in data_dict['level1']["level2"]["level3"]:
data_dict['level1']["level2"]["level3"][idx]['children'] = []
print data_dict
'''
{
'level1': {
'level2': {
'level3': {
'ID2': {'children': []},
'ID3': {'children': []},
'ID1': {'children': []}
}
}
}
}
'''
but if you have more complicated dictionary
data_dict = {
"level1a": {
"level2a":{
"level3a":{
"ID2":{"children": [] },
"ID3":{"children": [6,7,8,9,10]},
}
}
},
"level1b": {
"level2b":{
"level3b":{
"ID1":{"children": [1,2,3,4,5]},
}
}
}
}
new_ids =["ID1","ID3"]
for level1 in data_dict.values():
for level2 in level1.values():
for level3 in level2.values():
for idx in new_ids:
if idx in level3:
level3[idx]['children'] = []
print data_dict
'''
{
'level1a': {
'level2a': {
'level3a': {
'ID2': {'children': []},
'ID3': {'children': []}
}
}
},
'level1b': {
'level2b': {
'level3b': {
'ID1': {'children': []}
}
}
}
}
'''