find all dictionary matching key=value pairs in dynamic dictionary - python

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'}}}

Related

point specific key and value from highly nested dictionary

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

Conditional Parameter inside dictionary Python

From my dataframe here:
OrigC OrigZ OrigN Weigh DestC DestZ DestN Mvt
0 PL 97 TP 59 DE 63 SN DD
1 TR 23 GH 66 SN 65 US DP
I want to pass conditional parameter in my dictionary based on the value of a column from my dataframe.
My code looks like this without the condition :
dic = {}
dic['section'] = []
for ix, row in df.iterrows():
in_dict = {
'location': {
'zip_code': {
'OrigC': row['OrigC'],
'OrigZ': row['OrigZ'],
},
'location': {'id': 1},
'OrigN': 'TP',
},
'CarriageParameter': {
'road': {
'truckLoad': 'Auto'}
},
'load': {
'Weigh': str(row['Weigh']),
}
}
dic['section'].append(in_dict)
I want to pass a condition inside my dictionary like this somehow wont work:
dic = {}
dic['section'] = []
for ix, row in df.iterrows():
in_dict = {
'location': {
if row['Mvt'] = 'DP':
return 'zip_code': {
'OrigC': row['OrigC'],
'OrigZ': row['OrigZ'],
}
elif row['Mvt'] = 'DD':
return 'iata_code': {
'OrigC': row['OrigN'],
}
'location': {'id':1},
'OrigN': 'TP',
},
'CarriageParameter': {
'road': {
'truckLoad': 'Auto'}
},
'load': {
'Weigh': str(row['Weigh']),
}
}
dic['section'].append(in_dict)
Have the common key value pairs in the in_dict dictionary setup initially, and later update the dictionary according to the condition.
dic = {}
dic['section'] = []
for ix, row in df.iterrows():
in_dict = {
'location': {
'CarriageParameter': {
'road': {
'truckLoad': 'Auto'
}
},
'load': {
'Weigh': str(row['Weigh']),
}
}
if row['Mvt'] == 'DP':
in_dict['location']['zip_code'] = {
'OrigC': row['OrigC'],
'OrigZ': row['OrigZ'],
}
elif row['Mvt'] == 'DD':
in_dict['location']['iata_code'] = {
'OrigC': row['OrigN']
}
in_dict['location']['location'] = {'id':1}
in_dict['location']['OrigN'] = 'TP'
dic['section'].append(in_dict)

How to move sub-dictionary to another dictionary?

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': {}} }
}

Comparing Nested Python dict with list and dict

I've seen similar questions but none that exactly match what I'm doing and I believe other developers might face same issue if they are working with MongoDB.
I'm looking to compare two nested dict objects with dict and arrays and return a dict with additions and deletion (like you would git diff two files)
Here is what I have so far:
def dict_diff(alpha, beta, recurse_adds=False, recurse_dels=False):
"""
:return: differences between two python dict with adds and dels
example:
(This is the expected output)
{
'adds':
{
'specific_hours': [{'ends_at': '2015-12-25'}],
}
'dels':
{
'specific_hours': [{'ends_at': '2015-12-24'}],
'subscription_products': {'review_management': {'thiswillbedeleted': 'deleteme'}}
}
}
"""
if type(alpha) is dict and type(beta) is dict:
a_keys = alpha.keys()
b_keys = beta.keys()
dels = {}
adds = {}
for key in a_keys:
if type(alpha[key]) is list:
if alpha[key] != beta[key]:
adds[key] = dict_diff(alpha[key], beta[key], recurse_adds=True)
dels[key] = dict_diff(alpha[key], beta[key], recurse_dels=True)
elif type(alpha[key]) is dict:
if alpha[key] != beta[key]:
adds[key] = dict_diff(alpha[key], beta[key], recurse_adds=True)
dels[key] = dict_diff(alpha[key], beta[key], recurse_dels=True)
elif key not in b_keys:
dels[key] = alpha[key]
elif alpha[key] != beta[key]:
adds[key] = beta[key]
dels[key] = alpha[key]
for key in b_keys:
if key not in a_keys:
adds[key] = beta[key]
elif type(alpha) is list and type(beta) is list:
index = 0
adds=[]
dels=[]
for elem in alpha:
if alpha[index] != beta[index]:
dels.append(alpha[index])
adds.append(beta[index])
# print('update', adds, dels)
index+=1
else:
raise Exception("dict_diff function can only get dict objects")
if recurse_adds:
if bool(adds):
return adds
return {}
if recurse_dels:
if bool(dels):
return dels
return {}
return {'adds': adds, 'dels': dels}
The result I'm getting now is:
{'adds': {'specific_hours': [{'ends_at': '2015-12-24',
'open_hours': ['07:30-11:30', '12:30-21:30'],
'starts_at': '2015-12-22'},
{'ends_at': '2015-01-03',
'open_hours': ['07:30-11:30'],
'starts_at': '2015-01-0'}],
'subscription_products': {'review_management': {}}},
'dels': {'specific_hours': [{'ends_at': '2015-12-24',
'open_hours': ['07:30-11:30', '12:30-21:30'],
'starts_at': '2015-12-2'},
{'ends_at': '2015-01-03',
'open_hours': ['07:30-11:30'],
'starts_at': '2015-01-0'}],
'subscription_products': {'review_management': {'thiswillbedeleted': 'deleteme'}}}}
And this is the two objects I'm trying to compare:
alpha = {
'specific_hours': [
{
"starts_at": "2015-12-2",
"ends_at": "2015-12-24",
"open_hours": [
"07:30-11:30",
"12:30-21:30"
]
},
{
"starts_at": "2015-01-0",
"ends_at": "2015-01-03",
"open_hours": [
"07:30-11:30"
]
}
],
'subscription_products': {'presence_management':
{'expiration_date': 1953291600,
'payment_type': {
'free': 'iamfree',
'test': "test",
},
},
'review_management':
{'expiration_date': 1511799660,
'payment_type': {
'free': 'iamfree',
'test': "test",
},
'thiswillbedeleted': "deleteme",
}
},
}
beta = {
'specific_hours': [
{
"starts_at": "2015-12-22",
"ends_at": "2015-12-24",
"open_hours": [
"07:30-11:30",
"12:30-21:30"
]
},
{
"starts_at": "2015-01-0",
"ends_at": "2015-01-03",
"open_hours": [
"07:30-11:30"
]
}
],
'subscription_products': {'presence_management':
{'expiration_date': 1953291600,
'payment_type': {
'free': 'iamfree',
'test': "test",
},
},
'review_management':
{'expiration_date': 1511799660,
'payment_type': {
'free': 'iamfree',
'test': "test",
},
}
},
}

How to dive into stacked dict gracefully

Let's say I have a dict like
{
"key_a": "value",
"key_b": {
"key_b_a": "value",
"key_b_b": {
"key_b_b_a": "value"
}
}
}
What I want is to create a method to delete the given key or change its value.
def del_key(key):
my_dict = <dictionary described above>
keys = key.split(':')
if len(keys) == 1:
del my_dict[keys[0]]
elif len(keys) == 2:
del my_dict[keys[0]][keys[1]]
elif len(keys) == 3:
del my_dict[keys[0]][keys[1]][keys[2]]
. . .
del_key('key_b:key_b_b:key_b_b_a')
del_key('key_b:key_b_b')
del_key('key_a')
How can I do this gracefully?
It assumes your input is valid key,otherwise you have to check.
data = {
"key_a": "value",
"key_b": {
"key_b_a": "value",
"key_b_b": {
"key_b_b_a": "value"
}
}
}
def del_key(key):
key = key.split(':')
temp = data
for i in key[:-1]:
temp = temp[i]
del temp[key[-1]]
return data
print del_key('key_b:key_b_b:key_b_b_a')
print del_key('key_b:key_b_b')
print del_key('key_a')
output:
{'key_a': 'value', 'key_b': {'key_b_a': 'value', 'key_b_b': {}}}
{'key_a': 'value', 'key_b': {'key_b_a': 'value'}}
{'key_b': {'key_b_a': 'value'}}

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