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I have a list of tuples. This could look like this:
tuple_list = [
('species', 'flower'),
('flower', 'dorsal flower'),
('dorsal flower', 'pink'),
('pink', 'white'),
('pink', 'greenish'),
('species', 'branch'),
]
Note: The tuples are not in order and in this example, they could also vary in order. The 'deepness' can also vary.
I would like to create a dict of dict that would look like this:
dod = {'species': {'branch':{},'flower': {'dorsal flower':{'pink': {'white':{}}, 'greenish':{}}}}}
In this case I want the species at top level, as it has no items that 'contain' species'. E.g. species contains 'flower' and 'branch' and so on.
I feel this entire process can be wrapped in a simple recursive function (e.g. yield from) instead of writing an elaborative for loop that iterates over all values.
In the end, I want to use this function to create a list of lists that contains the proper values as a list (Kudos to #Stef for this function):
def undict_to_lists(d, acc = []):
if d == {}:
yield acc
else:
for k, v in d.items():
yield from undict_to_tuples(v, acc + [k,])
This would result in the following:
print(list(undict_to_lists(dod)))
[['species', 'branch'],
['species', 'flower', 'dorsal flower', 'pink', 'white'],
['species', 'flower', 'dorsal flower', 'greenish']]
Thanks for thinking along! All suggestions are welcome.
You could first create a dictionary key (with {} as value) for each key that occurs in the input. Then iterate those tuples to find the value that corresponds to the start key, and populate the sub dictionary with the end key, and the subdictionary that corresponds to that end key.
Finally, derive which is the root by excluding all those nodes that are children.
tuple_list = [('species', 'flower'), ('flower', 'dorsal flower'), ('dorsal flower', 'pink'),('pink', 'white'),('pink', 'greenish'),('species', 'branch')]
d = { key: {} for pair in tuple_list for key in pair }
for start, end in tuple_list:
d[start][end] = d[end]
root = None
for key in set(d.keys()).difference(end for _, end in tuple_list):
root = d[key]
print(root)
tuple_list = [
('species', 'flower'),
('flower', 'dorsal flower'),
('dorsal flower', 'pink'),
('pink', 'white'),
('pink', 'greenish'),
('species', 'branch'),
]
# Create the nested dict, using a "master" dict
# to quickly look up nodes in the nested dict.
nested_dict, master_dict = {}, {}
for a, b in tuple_list:
if a not in master_dict:
nested_dict[a] = master_dict[a] = {}
master_dict[a][b] = master_dict[b] = {}
# Flatten into lists.
def flatten_dict(d):
if not d:
return [[]]
return [[k] + f for k, v in d.items() for f in flatten_dict(v)]
print(flatten_dict(nested_dict))
#[['species', 'flower', 'dorsal flower', 'pink', 'white'],
# ['species', 'flower', 'dorsal flower', 'pink', 'greenish'],
# ['species', 'branch']]
Here's another alternative (loosely based on #trincot answer) that uses a defaultdict to simplify the code slightly and which figures out the root of the tree as it goes through the list of tuples:
from collections import defaultdict
d = defaultdict(dict)
root = tuple_list[0][0] # first parent value
for parent, child in tuple_list:
d[parent][child] = d[child]
if root == child:
root = parent
result = { root : d[root] }
Output:
{
"species": {
"branch": {},
"flower": {
"dorsal flower": {
"pink": {
"greenish": {},
"white": {}
}
}
}
}
}
Alternative :
def find_node( tree, parent, child ):
if parent in tree:
tree[parent][child] = {}
return True
for node in tree.values():
if find_node( node, parent, child ):
return True
# new node
tree[parent] = { child : {} }
root = {}
for parent, child in tuple_list:
find_node( root, parent, child )
I'm looking for any suggestions to resolve an issue I'm facing. It might seem as a simple problem, but after a few days trying to find an answer - I think it is not anymore.
I'm receiving data (StringType) in a following JSON-like format, and there is a requirement to turn it into flat key-value pair dictionary. Here is a payload sample:
s = """{"status": "active", "name": "{\"first\": \"John\", \"last\": \"Smith\"}", "street_address": "100 \"Y\" Street"}"""
and the desired output should look like this:
{'status': 'active', 'name_first': 'John', 'name_last': 'Smith', 'street_address': '100 "Y" Street'}
The issue is I can't find a way to turn original string (s) into a dictionary. If I can achieve that the flattening part is working perfectly fine.
import json
import collections
import ast
#############################################################
# Flatten complex structure into a flat dictionary
#############################################################
def flatten_dictionary(dictionary, parent_key=False, separator='_', value_to_str=True):
"""
Turn a nested complex json into a flattened dictionary
:param dictionary: The dictionary to flatten
:param parent_key: The string to prepend to dictionary's keys
:param separator: The string used to separate flattened keys
:param value_to_str: Force all returned values to string type
:return: A flattened dictionary
"""
items = []
for key, value in dictionary.items():
new_key = str(parent_key) + separator + key if parent_key else key
try:
value = json.loads(value)
except BaseException:
value = value
if isinstance(value, collections.MutableMapping):
if not value.items():
items.append((new_key,None))
else:
items.extend(flatten_dictionary(value, new_key, separator).items())
elif isinstance(value, list):
if len(value):
for k, v in enumerate(value):
items.extend(flatten_dictionary({str(k): (str(v) if value_to_str else v)}, new_key).items())
else:
items.append((new_key,None))
else:
items.append((new_key, (str(value) if value_to_str else value)))
return dict(items)
# Data sample; sting and dictionary
s = """{"status": "active", "name": "{\"first\": \"John\", \"last\": \"Smith\"}", "street_address": "100 \"Y\" Street"}"""
d = {"status": "active", "name": "{\"first\": \"John\", \"last\": \"Smith\"}", "street_address": "100 \"Y\" Street"}
# Works for dictionary type
print(flatten_dictionary(d))
# Doesn't work for string type, for any of the below methods
e = eval(s)
# a = ast.literal_eval(s)
# j = json.loads(s)
Try:
import json
import re
def jsonify(s):
s = s.replace('"{','{').replace('}"','}')
s = re.sub(r'street_address":\s+"(.+)"(.+)"(.+)"', r'street_address": "\1\2\3"',s)
return json.loads(s)
If you must keep the quotes around Y, try:
def jsonify(s):
s = s.replace('"{','{').replace('}"','}')
search = re.search(r'street_address":\s+"(.+)"(.+)"(.+)"',s)
if search:
s = re.sub(r'street_address":\s+"(.+)"(.+)"(.+)"', r'street_address": "\1\2\3"',s)
dict_version = json.loads(s)
dict_version['street_address'] = dict_version['street_address'].replace(search.group(2),'"'+search.group(2)+'"')
return dict_version
A more generalized attempt:
def jsonify(s):
pattern = r'(?<=[,}])\s*"(.[^\{\}:,]+?)":\s+"([^\{\}:,]+?)"([^\{\}:,]+?)"([^\{\}:,]+?)"([,\}])'
s = s.replace('"{','{').replace('}"','}')
search = re.search(pattern,s)
matches = []
if search:
matches = re.findall(pattern,s)
s = re.sub(pattern, r'"\1": "\2\3\4"\5',s)
dict_version = json.loads(s)
for match in matches:
dict_version[match[0]] = dict_version[match[0]].replace(match[2],'"'+match[2]+'"')
return dict_version
I have been struggling to extract sub dictionaries where the value is "0" from a list of dictionaries and add them to temporary dictionaries.
I tried this:
new_users = [{'user1':{'book1':'0', 'book2':'4', 'book3':'1'}},{'user2':{'book1':'5', 'book2':'1', 'book3':'0'}}]
def approachA():
for data in new_users: # new_users is a list of nested dictionaries
if data == '0':
print("found 0")
keys = data.keys()
for key in keys:
if key == '0':
key.pop() # tried to deleted delete the elements at first
It did not work for some reason, and I have been trying to do it for 2 hours so please do not ask questions not related to the problem.
This is a simple version of what I am trying to do:
[{'user1':{'book1':'0', 'book2':'4', 'book3':'1'}},{'user2':{'book1':'5', 'book2':'1', 'book3':'0'}}] -> [{'user1':{'book1':'0'}}, {'user2':{'book3':'0'}}]
So basically the keys with value "0" get copied to a temp list of dictionaries.
As I mentioned in a comment I'm not sure this will solve your problem, but based on the sample transformation you gave, here's one way to achieve it:
LIST_IN = [{"name": {"book1": "0", "book2": "1", "book3": "0"}}]
def proccess(list_in):
result = []
for this_dict in list_in:
for key in this_dict["name"]:
if this_dict["name"][key] == "0":
result.append({key: this_dict["name"][key]})
return result
print(proccess(LIST_IN))
This should accomplish the "sample" snippet. Please give some additional details if this is not what you were trying.
def solve_nested(nested_dict: list) -> list:
result = list()
for dictionary in nested_dict:
for k, v in dictionary.items():
for _k, _v in v.items():
if _v == '0':
result.append({_k: _v})
return result
if __name__ == '__main__':
print(solve_nested([{"name": {"book1": "0", "book2": "1", "book3": "0"}}]))
If you're interested in recursion, the below example ought to get you going:
# d_xy => x = nest level
# y = dictionary item in nest level x
d = {'d_00' : {'d_10' : {'d_20' : 0,
'd_21' : 43,
'd_22' : 12},
'd_11' : 4,
'd_12' : 0,
'd_13' : 1},
'd_01' : 0,
'd_02' : {'d_14' : {'d_23' : 0,
'd_24' : 1,
'd_25' : 0},
'd_15' : 4,
'd_16' : {'d_26' : 0,
'd_27' : {'d_30' : 3,
'd_31' : 0,
'd_32' : {'d_40' : 0},
'd_33' : 0},
'd_28' : 1},
'd_17' : 0}}
important_items = []
def get_0_key_values(dic):
for key in dic:
if type(dic[key]) == dict:
get_0_key_values(dic[key])
else:
if dic[key] == 0:
important_items.append({key : dic[key]})
get_0_key_values(d)
print(important_items)
I have a json in below format:
{"MainName":[{"col1":"12345","col2":"False","col3":"190809","SubName1":{"col4":30.00,"SubName2":{"col5":"19703","col6":"USD"}},"col7":"7372267","SubName3":[{"col8":"345337","col9":"PC"}],"col10":"10265","col11":"29889004","col12":"calculated","col13":"9218","SubName4":{"col14":1,"SubName5":{"col15":"1970324","col16":"integer"}},"col17":"434628","col18":"2020-02-06T13:47:40.000-0800","col19":"754878037","SubName6":{"col20":30.00,"SubName7":{"col21":"19703248","col22":"USD"}}},{"col1":"12345","col2":"False","col3":"190809","SubName1":{"col4":30.00,"SubName2":{"col5":"19703","col6":"USD"}},"col7":"7372267","SubName3":[{"col8":"345337","col9":"PC"}],"col10":"10265","col11":"29889004","col12":"calculated","col13":"9218","SubName4":{"col14":1,"SubName5":{"col15":"1970324","col16":"integer"}},"col17":"434628","col18":"2020-02-06T13:47:40.000-0800","col19":"754878037","SubName6":{"col20":30.00,"SubName7":{"col21":"19703248","col22":"USD"}}}],"skip":0,"top":2,"next":"/v1/APIName?skip=2&top=2"}
I want to convert it into csv with below format:
MainName_col1,MainName_col2,MainName_col3,MainName_SubName1_col4,MainName_SubName1_SubName2_col5,MainName_SubName1_SubName2_col6,MainName_col7,MainName_SubName3_col8,MainName_SubName3_col9,MainName_col10,MainName_col11,MainName_col12,MainName_col13,MainName_SubName4_col14,MainName_SubName4_SubName5_col15,MainName_SubName4_SubName5_col16,MainName_col17,MainName_col18,MainName_col19,MainName_SubName6_col20,MainName_SubName6_SubName7_col21,MainName_SubName6_SubName7_col22
12345,False,190809,30.0,19703,USD,7372267,345337,PC,10265,29889004,calculated,9218,1,1970324,integer,434628,2020-02-06T13:47:40.000-0800,754878037,30.0,19703248,USD
12345,False,190809,30.0,19703,USD,7372267,345337,PC,10265,29889004,calculated,9218,2,123453,integer,434628,2020-02-06T13:47:40.000-0800,754878037,30.0,19703248,USD
Kindly help me out in this.
Use below function to flatten your JSON data.
dc = {"MainName":[{"col1":"12345","col2":False,"col3":"190809","SubName1":{"col4":30.00,"SubName2":{"col5":"19703","col6":"USD"}},"col7":"7372267","SubName3":[{"col8":"345337","col9":"PC"}],"col10":"10265","col11":"29889004","col12":"calculated","col13":"9218","SubName4":{"col14":1,"SubName5":{"col15":"1970324","col16":"integer"}},"col17":"434628","col18":"2020-02-06T13:47:40.000-0800","col19":"754878037","SubName6":{"col20":30.00,"SubName7":{"col21":"19703248","col22":"USD"}}}],"skip":0,"top":1,"next":"/v1/APIName?skip=1&top=1"}
def flatten(root: str, dict_obj: dict):
flat = {}
for i in dict_obj.keys():
val = dict_obj[i]
if not isinstance(val, dict) and not isinstance(val, list):
flat[f'{root}_{i}'] = val
else:
if isinstance(val, list):
val = val[-1]
flat.update(flatten(f'{root}_{i}', val))
return flat
flatten('MainName', dc['MainName'][0])
It will give you expected output. Then use it the way you want.
{'MainName_col1': '12345',
'MainName_col2': False,
'MainName_col3': '190809',
'MainName_SubName1_col4': 30.0,
'MainName_SubName1_SubName2_col5': '19703',
'MainName_SubName1_SubName2_col6': 'USD',
'MainName_col7': '7372267',
'MainName_SubName3_col8': '345337',
'MainName_SubName3_col9': 'PC',
'MainName_col10': '10265',
'MainName_col11': '29889004',
'MainName_col12': 'calculated',
'MainName_col13': '9218',
'MainName_SubName4_col14': 1,
'MainName_SubName4_SubName5_col15': '1970324',
'MainName_SubName4_SubName5_col16': 'integer',
'MainName_col17': '434628',
'MainName_col18': '2020-02-06T13:47:40.000-0800',
'MainName_col19': '754878037',
'MainName_SubName6_col20': 30.0,
'MainName_SubName6_SubName7_col21': '19703248',
'MainName_SubName6_SubName7_col22': 'USD'}
As of my understanding, your dc will look like below
dc = {"MainName":[{"col1":"12345","col2":"False","col3":"190809","SubName1":{"col4":30.00,"SubName2":{"col5":"19703","col6":"USD"}},"col7":"7372267","SubName3":[{"col8":"345337","col9":"PC"}],"col10":"10265","col11":"29889004","col12":"calculated","col13":"9218","SubName4":{"col14":1,"SubName5":{"col15":"1970324","col16":"integer"}},"col17":"434628","col18":"2020-02-06T13:47:40.000-0800","col19":"754878037","SubName6":{"col20":30.00,"SubName7":{"col21":"19703248","col22":"USD"}}},{"col1_a":"12345XX","col2_b":"False","col3_c":"190809","SubName1":{"col4_d":30.00,"SubName2":{"col5_e":"19703","col6_f":"USD"}},"col7_g":"7372267","SubName3":[{"col8_h":"345337","col9":"PC"}],"col10_i":"10265","col11_j":"29889004","col12_k":"calculated","col13_l":"9218","SubName4":{"col14_m":1,"SubName5":{"col15_n":"1970324","col16_o":"integer"}},"col17_p":"434628","col18_q":"2020-02-06T13:47:40.000-0800","col19_r":"754878037","SubName6":{"col20_s":30.00,"SubName7":{"col21_t":"19703248","col22_u":"USDZZ"}}}],"skip":0,"top":2,"next":"/v1/APIName?skip=2&top=2"}
I used the above answer to flatten everything into single object
def flatten(root: str, dict_obj: dict):
flat = {}
for i in dict_obj.keys():
val = dict_obj[i]
if not isinstance(val, dict) and not isinstance(val, list):
flat[f'{root}_{i}'] = val
else:
if isinstance(val, list):
val = val[-1]
flat.update(flatten(f'{root}_{i}', val))
return flat
keys_list = []
values_list = []
for i in range(len(dc['MainName'])):
result = flatten('MainName', dc['MainName'][i])
keys_list.append(list(result.keys()))
values_list.append(list(result.values()))
for k in keys_list:
for res in k:
guestFile = open("sample.csv","a")
guestFile.write(res)
guestFile.write(",")
guestFile.close()
for v in values_list:
for res in v:
guestFile = open("sample.csv","a")
guestFile.write(str(res))
guestFile.write(",")
guestFile.close()
Checkout my code at https://repl.it/#TamilselvanLaks/jsontocsvmul
Note: Use the 'run' button to run the program, left side you can see sample.csv
there you can see all keys as like you want
Please let me know my answer meets your expectation
Search for a value and get the parent dictionary names (keys):
Dictionary = {dict1:{
'part1': {
'.wbxml': 'application/vnd.wap.wbxml',
'.rl': 'application/resource-lists+xml',
},
'part2':
{'.wsdl': 'application/wsdl+xml',
'.rs': 'application/rls-services+xml',
'.xop': 'application/xop+xml',
'.svg': 'image/svg+xml',
},
'part3':{...}, ...
dict2:{
'part1': { '.dotx': 'application/vnd.openxmlformats-..'
'.zaz': 'application/vnd.zzazz.deck+xml',
'.xer': 'application/patch-ops-error+xml',}
},
'part2':{...},
'part3':{...},...
},...
In above dictionary I need to search values like: "image/svg+xml". Where, none of the values are repeated in the dictionary. How to search the "image/svg+xml"? so that it should return the parent keys in a dictionary { dict1:"part2" }.
Please note: Solutions should work unmodified for both Python 2.7 and Python 3.3.
Here's a simple recursive version:
def getpath(nested_dict, value, prepath=()):
for k, v in nested_dict.items():
path = prepath + (k,)
if v == value: # found value
return path
elif hasattr(v, 'items'): # v is a dict
p = getpath(v, value, path) # recursive call
if p is not None:
return p
Example:
print(getpath(dictionary, 'image/svg+xml'))
# -> ('dict1', 'part2', '.svg')
To yield multiple paths (Python 3 only solution):
def find_paths(nested_dict, value, prepath=()):
for k, v in nested_dict.items():
path = prepath + (k,)
if v == value: # found value
yield path
elif hasattr(v, 'items'): # v is a dict
yield from find_paths(v, value, path)
print(*find_paths(dictionary, 'image/svg+xml'))
This is an iterative traversal of your nested dicts that additionally keeps track of all the keys leading up to a particular point. Therefore as soon as you find the correct value inside your dicts, you also already have the keys needed to get to that value.
The code below will run as-is if you put it in a .py file. The find_mime_type(...) function returns the sequence of keys that will get you from the original dictionary to the value you want. The demo() function shows how to use it.
d = {'dict1':
{'part1':
{'.wbxml': 'application/vnd.wap.wbxml',
'.rl': 'application/resource-lists+xml'},
'part2':
{'.wsdl': 'application/wsdl+xml',
'.rs': 'application/rls-services+xml',
'.xop': 'application/xop+xml',
'.svg': 'image/svg+xml'}},
'dict2':
{'part1':
{'.dotx': 'application/vnd.openxmlformats-..',
'.zaz': 'application/vnd.zzazz.deck+xml',
'.xer': 'application/patch-ops-error+xml'}}}
def demo():
mime_type = 'image/svg+xml'
try:
key_chain = find_mime_type(d, mime_type)
except KeyError:
print ('Could not find this mime type: {0}'.format(mime_type))
exit()
print ('Found {0} mime type here: {1}'.format(mime_type, key_chain))
nested = d
for key in key_chain:
nested = nested[key]
print ('Confirmation lookup: {0}'.format(nested))
def find_mime_type(d, mime_type):
reverse_linked_q = list()
reverse_linked_q.append((list(), d))
while reverse_linked_q:
this_key_chain, this_v = reverse_linked_q.pop()
# finish search if found the mime type
if this_v == mime_type:
return this_key_chain
# not found. keep searching
# queue dicts for checking / ignore anything that's not a dict
try:
items = this_v.items()
except AttributeError:
continue # this was not a nested dict. ignore it
for k, v in items:
reverse_linked_q.append((this_key_chain + [k], v))
# if we haven't returned by this point, we've exhausted all the contents
raise KeyError
if __name__ == '__main__':
demo()
Output:
Found image/svg+xml mime type here: ['dict1', 'part2', '.svg']
Confirmation lookup: image/svg+xml
Here is a solution that works for a complex data structure of nested lists and dicts
import pprint
def search(d, search_pattern, prev_datapoint_path=''):
output = []
current_datapoint = d
current_datapoint_path = prev_datapoint_path
if type(current_datapoint) is dict:
for dkey in current_datapoint:
if search_pattern in str(dkey):
c = current_datapoint_path
c+="['"+dkey+"']"
output.append(c)
c = current_datapoint_path
c+="['"+dkey+"']"
for i in search(current_datapoint[dkey], search_pattern, c):
output.append(i)
elif type(current_datapoint) is list:
for i in range(0, len(current_datapoint)):
if search_pattern in str(i):
c = current_datapoint_path
c += "[" + str(i) + "]"
output.append(i)
c = current_datapoint_path
c+="["+ str(i) +"]"
for i in search(current_datapoint[i], search_pattern, c):
output.append(i)
elif search_pattern in str(current_datapoint):
c = current_datapoint_path
output.append(c)
output = filter(None, output)
return list(output)
if __name__ == "__main__":
d = {'dict1':
{'part1':
{'.wbxml': 'application/vnd.wap.wbxml',
'.rl': 'application/resource-lists+xml'},
'part2':
{'.wsdl': 'application/wsdl+xml',
'.rs': 'application/rls-services+xml',
'.xop': 'application/xop+xml',
'.svg': 'image/svg+xml'}},
'dict2':
{'part1':
{'.dotx': 'application/vnd.openxmlformats-..',
'.zaz': 'application/vnd.zzazz.deck+xml',
'.xer': 'application/patch-ops-error+xml'}}}
d2 = {
"items":
{
"item":
[
{
"id": "0001",
"type": "donut",
"name": "Cake",
"ppu": 0.55,
"batters":
{
"batter":
[
{"id": "1001", "type": "Regular"},
{"id": "1002", "type": "Chocolate"},
{"id": "1003", "type": "Blueberry"},
{"id": "1004", "type": "Devil's Food"}
]
},
"topping":
[
{"id": "5001", "type": "None"},
{"id": "5002", "type": "Glazed"},
{"id": "5005", "type": "Sugar"},
{"id": "5007", "type": "Powdered Sugar"},
{"id": "5006", "type": "Chocolate with Sprinkles"},
{"id": "5003", "type": "Chocolate"},
{"id": "5004", "type": "Maple"}
]
},
...
]
}
}
pprint.pprint(search(d,'svg+xml','d'))
>> ["d['dict1']['part2']['.svg']"]
pprint.pprint(search(d2,'500','d2'))
>> ["d2['items']['item'][0]['topping'][0]['id']",
"d2['items']['item'][0]['topping'][1]['id']",
"d2['items']['item'][0]['topping'][2]['id']",
"d2['items']['item'][0]['topping'][3]['id']",
"d2['items']['item'][0]['topping'][4]['id']",
"d2['items']['item'][0]['topping'][5]['id']",
"d2['items']['item'][0]['topping'][6]['id']"]
Here are two similar quick and dirty ways of doing this type of operation. The function find_parent_dict1 uses list comprehension but if you are uncomfortable with that then find_parent_dict2 uses the infamous nested for loops.
Dictionary = {'dict1':{'part1':{'.wbxml':'1','.rl':'2'},'part2':{'.wbdl':'3','.rs':'4'}},'dict2':{'part3':{'.wbxml':'5','.rl':'6'},'part4':{'.wbdl':'1','.rs':'10'}}}
value = '3'
def find_parent_dict1(Dictionary):
for key1 in Dictionary.keys():
item = {key1:key2 for key2 in Dictionary[key1].keys() if value in Dictionary[key1][key2].values()}
if len(item)>0:
return item
find_parent_dict1(Dictionary)
def find_parent_dict2(Dictionary):
for key1 in Dictionary.keys():
for key2 in Dictionary[key1].keys():
if value in Dictionary[key1][key2].values():
print {key1:key2}
find_parent_dict2(Dictionary)
Traverses a nested dict looking for a particular value. When success is achieved the full key path to the value is printed. I left all the comments and print statements for pedagogical purposes (this isn't production code!)
#!/usr/bin/env python3
# -*- coding: utf-8 -*-
"""
Created on Mon Jan 24 17:16:46 2022
#author: wellington
"""
class Tree(dict):
"""
allows autovivification as in Perl hashes
"""
def __missing__(self, key):
value = self[key] = type(self)()
return value
# tracking the key sequence when seeking the target
key_list = Tree()
# dict storing the target success result
success = Tree()
# example nested dict of dicts and lists
E = {
'AA':
{
'BB':
{'CC':
{
'DD':
{
'ZZ':'YY',
'WW':'PP'
},
'QQ':
{
'RR':'SS'
},
},
'II':
{
'JJ':'KK'
},
'LL':['MM', 'GG', 'TT']
}
}
}
def find_keys_from_value(data, target):
"""
recursive function -
given a value it returns all the keys in the path to that value within
the dict "data"
there are many paths and many false routes
at the end of a given path if success has not been achieved
the function discards keys to get back to the next possible path junction
"""
print(f"the number of keys in the local dict is {len(data)}")
key_counter = 0
for key in data:
key_counter += 1
# if target has been located stop iterating through keys
if success[target] == 1:
break
else:
# eliminate prior key from path that did not lead to success
if key_counter > 1:
k_list.pop()
# add key to new path
k_list.append(key)
print(f"printing k_list after append{k_list}")
# if target located set success[target] = 1 and exit
if key == target or data[key] == target:
key_list[target] = k_list
success[target] = 1
break
# if the target has not been located check to see if the value
# associated with the new key is a dict and if so return to the
# recursive function with the new dict as "data"
elif isinstance(data[key], dict):
print(f"\nvalue is dict\n {data[key]}")
find_keys_from_value(data[key], target)
# check to see if the value associated with the new key is a list
elif isinstance(data[key], list):
# print("\nv is list\n")
# search through the list
for i in data[key]:
# check to see if the list element is a dict
# and if so return to the recursive function with
# the new dict as "data
if isinstance(i, dict):
find_keys_from_value(i, target)
# check to see if each list element is the target
elif i == target:
print(f"list entry {i} is target")
success[target] = 1
key_list[target] = k_list
elif i != target:
print(f"list entry {i} is not target")
print(f"printing k_list before pop_b {k_list}")
print(f"popping off key_b {key}")
# so if value is not a key and not a list and not the target then
# discard the key from the key list
elif data[key] != target:
print(f"value {data[key]} is not target")
print(f"printing k_list before removing key_before {k_list}")
print(f"removing key_c {key}")
k_list.remove(key)
# select target values
values = ["PP", "SS", "KK", "TT"]
success = {}
for target in values:
print(f"\nlooking for target {target}")
success[target] = 0
k_list = []
find_keys_from_value(E, target)
print(f"\nprinting key_list for target {target}")
print(f"{key_list[target]}\n")
print("\n****************\n\n")