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Get all the keys of a nested dict

With xmltodict I managed to get my code from xml in a dict and now I want to create an excel.
In this excel the header of a value is going to be all the parents (keys in the dict).
For example:
dict = {"name":"Pete", "last-name": "Pencil", "adres":{"street": "example1street", "number":"5", "roommate":{"gender":"male"}}}
The value male will have the header: adres/roommate/gender.

Here's a way to orgainze the data in the way your question asks:
d = {"name":"Pete", "last-name": "Pencil", "adres":{"street": "example1street", "number":"5", "roommate":{"gender":"male"}}}
print(d)
stack = [('', d)]
headerByValue = {}
while stack:
name, top = stack.pop()
if isinstance(top, dict):
stack += (((name + '/' if name else '') + k, v) for k, v in top.items())
else:
headerByValue[name] = top
print(headerByValue)
Output:
{'adres/roommate/gender': 'male',
'adres/number': '5',
'adres/street': 'example1street',
'last-name': 'Pencil',
'name': 'Pete'}

How to map nested list to flat values

I`m trying to parse a spreadsheet with a header that looks something like this:
My problem is those nested keys below "Контрагент". I decided to parse it like this:
['Дата',
'Номер документа',
'Дебет',
'Кредит',
['Контрагент',
['Наименование', 'ИНН', 'КПП', 'Счет', 'БИК', 'Наименование банка']],
'Назначение платежа',
'Код дебитора',
'Тип документа']
But now, I don`t really have an idea as how to map it to a flat list of values:
['21.05.2021',
'591324565436',
'0.00',
'526345428.99',
'asdasd',
'234525460140679',
'77130100123412341',
'302328105423534200000000280',
'0445252345234974',
'asdfsadfsd',
'sdfghsfgdhfdghdfgh',
'',
'dfghfgdhfdgh']
Given these variables, I want a function to return following dict:
{
"Дата": "21.05.2021",
"Номер документа": "591324565436",
"Дебет": "0.00",
"Кредит": "526345428.99",
"Контрагент": {
"Наименование": "asdasd",
"ИНН": "234525460140679",
"КПП": "77130100123412341",
"Счет": "302328105423534200000000280",
"БИК": "0445252345234974",
"Наименование банка": "asdfsadfsd"
},
"Назначение платежа": "sdfghsfgdhfdghdfgh",
"Код дебитора": "",
"Тип документа": "dfghfgdhfdgh"
}
I've gone this far without realizing it'd be raising IndexError on the 3rd line:
def map_to_schema(schema, data):
for i, elem in enumerate(data):
key = schema[i]
if isinstance(key, list):
if key[0] not in result:
result[key[0]] = {}
result[key[0]] |= {
key[1][i-len(key)]: elem
}
else:
result[key] = elem
What should I do? Maybe the structure for the schema isn't good enough? I really have no idea...

You could use a dictionary comprehension and an iterator:
headers = ['Дата', 'Номер документа', 'Дебет', 'Кредит',
['Контрагент', ['Наименование', 'ИНН', 'КПП', 'Счет', 'БИК', 'Наименование банка']],
'Назначение платежа', 'Код дебитора', 'Тип документа']
values = ['21.05.2021', '591324565436', '0.00', '526345428.99', 'asdasd', '234525460140679', '77130100123412341',
'302328105423534200000000280', '0445252345234974', 'asdfsadfsd', 'sdfghsfgdhfdghdfgh', '',
'dfghfgdhfdgh']
it = iter(values)
out = {k[0] if (islist := isinstance(k, list)) else k:
{k2: next(it) for k2 in k[1]} if islist else next(it)
for k in headers}
output:
{'Дата': '21.05.2021',
'Номер документа': '591324565436',
'Дебет': '0.00',
'Кредит': '526345428.99',
'Контрагент': {'Наименование': 'asdasd',
'ИНН': '234525460140679',
'КПП': '77130100123412341',
'Счет': '302328105423534200000000280',
'БИК': '0445252345234974',
'Наименование банка': 'asdfsadfsd'},
'Назначение платежа': 'sdfghsfgdhfdghdfgh',
'Код дебитора': '',
'Тип документа': 'dfghfgdhfdgh'}

Thanks #mozway for this solution! This is essentially the same algorithm, using a for loop.
def map(schema, s_length, row: list):
# If len(row) was less then *true* schema length, it would have thrown StopIteration.
# I ended up just extending row list by delta elements.
if (delta := s_length - len(row)) > 0:
row.extend([""] * delta)
iter_row = iter(row)
result = {}
for key in schema:
if isinstance(key, list):
result[key[0]] = {}
for sub_key in key[1]:
result[key[0]][sub_key] = next(iter_row)
else:
result[key] = next(iter_row)
return result

Python asside function returns none [duplicate]

This question already has answers here:
Recursive function does not return specified value
(2 answers)
Closed 1 year ago.
I have the orders of a platform in a csv file and I want to organize it to put in a database (probably mongodb). So the final data structure may be:
{'user1':
{'meals': {'hamburger': 2}, {'pizza': 1},
{'weekdays': {'monday': 1}},{'tuesday':1}, {'friday': 1}
}
So, I am trying to do a very simple code that organize the number of ocurrencies, but before I want to structure the dictionary keys dinamically. The data is in a csv file
import csv
import pprint
def addKeysNestedDict(dictionary, keys):
# print(dictionary)
if len(keys) > 1:
if keys[0] not in dictionary.keys():
dictionary[keys[0]] = {'meals': '', 'weekdays': ''}
print('inside addKeys IF: ')
print(dictionary)
addKeysNestedDict(dictionary, keys[1:])
else:
dictionary[keys[0]] = {'meals': '', 'weekdays': ''}
print('inside addKeys: ')
print(dictionary) # PRINTS EXPECTED VALUES
return dictionary
def organize_orders(file):
pp = pprint.PrettyPrinter(indent=4)
dict_read = csv.DictReader(file, fieldnames=['name', 'food', 'day'])
list_dict = list(dict_read)
set_names = set()
set_food = set()
set_day = set()
# people = {} # DOESN'T MATTER IF IT IS DECLARED BOFORE
for n in list_dict:
set_names.add(n['name'])
set_food.add(n['food'])
set_day.add(n['day'])
print(addKeysNestedDict({}, list(set_names))) # PRINTS NONE
people = (addKeysNestedDict({}, list(set_names)))
print('people: ')
print(people) # PRINTS NONE
# for s_n in set_names:
# for s_f in set_food:
# for s_d in set_day:
# people[s_n][s_f] = list_dict.count(s_f)
# people[s_n][s_d] = list_dict.count(s_d)
def analyze_log(path_to_file):
with open(path_to_file) as csvfile:
return(organize_orders(csvfile))
analyze_log('some csv file with path')
I culd not figure out why I get none returned from addKeysNestedDict() method, since it prints exactly what I want one line before method return

Change the function to:
def addKeysNestedDict(dictionary, keys):
# print(dictionary)
if len(keys) > 1:
if keys[0] not in dictionary.keys():
dictionary[keys[0]] = {'meals': '', 'weekdays': ''}
print('inside addKeys IF: ')
print(dictionary)
addKeysNestedDict(dictionary, keys[1:])
else:
dictionary[keys[0]] = {'meals': '', 'weekdays': ''}
print('inside addKeys: ')
print(dictionary) # PRINTS EXPECTED VALUES
return dictionary
Notice that I moved the return dictionary statement out of the specific branch, which causes both of your branches to return the dictionary you're building in your function.

multiple separator in a string python

text="Brand.*/Smart Planet.#/Color.*/Yellow.#/Type.*/Sandwich Maker.#/Power Source.*/Electrical."
I have this kind of string. I am facing the problem which splits it to 2 lists. Output will be approximately like this :
name = ['Brand','Color','Type','Power Source']
value = ['Smart Plane','Yellow','Sandwich Maker','Electrical']
Is there any solution for this.

name = []
value = []
text = text.split('.#/')
for i in text:
i = i.split('.*/')
name.append(i[0])
value.append(i[1])

This is one approach using re.split and list slicing.
Ex:
import re
text="Brand.*/Smart Planet.#/Color.*/Yellow.#/Type.*/Sandwich Maker.#/Power Source.*/Electrical."
data = [i for i in re.split("[^A-Za-z\s]+", text) if i]
name = data[::2]
value = data[1::2]
print(name)
print(value)
Output:
['Brand', 'Color', 'Type', 'Power Source']
['Smart Planet', 'Yellow', 'Sandwich Maker', 'Electrical']

You can use regex to split the text, and populate the lists in a loop.
Using regex you protect your code from invalid input.
import re
name, value = [], []
for ele in re.split(r'\.#\/', text):
k, v = ele.split('.*/')
name.append(k)
value.append(v)
>>> print(name, val)
['Brand', 'Color', 'Type', 'Power Source'] ['Smart Planet', 'Yellow', 'Sandwich Maker', 'Electrical.']

text="Brand.*/Smart Planet.#/Color.*/Yellow.#/Type.*/Sandwich Maker.#/Power Source.*/Electrical."
name=[]
value=[]
word=''
for i in range(len(text)):
temp=i
if text[i]!='.' and text[i]!='/' and text[i]!='*' and text[i]!='#':
word=word+''.join(text[i])
elif temp+1<len(text) and temp+2<=len(text):
if text[i]=='.' and text[temp+1]=='*' and text[temp+2]=='/':
name.append(word)
word=''
elif text[i]=='.' and text[temp+1]=='#' and text[temp+2]=='/':
value.append(word)
word=''
else:
value.append(word)
print(name)
print(value)
this will be work...

Categorizing sentence using dictionary

I am using below function for getting categorizing sentence in themes
def theme(x):
output =[]
category = ()
for i in x:
if 'AC' in i:
category = 'AC problem'
elif 'insects' in i:
category = 'Cleanliness'
elif 'clean' in i:
category = 'Cleanliness'
elif 'food' in i:
category = 'Food Problem'
elif 'delay' in i:
category = 'Train Delayed'
else:
category = 'None'
output.append(category)
return output
I don't want to use repeated if statements for every word in a category. Instead I want the i give a list/dictionary e.g. Cleanliness = ['Clean', 'Cleaned', 'spoilt', 'dirty'] for getting category 'Cleanliness' against the sentence if it has any of the words in list. How can i do that

You can use a dict of sets to structure your words with categories, and then generate a word-to-category lookup dict based on the said structure:
categories = {
'Cleanliness': {'insects', 'clean'},
'AC Problem': {'AC'},
'Food Problem': {'food'},
'Train Delayed': {'delay'}
}
lookup = {word: category for category, words in categories.items() for word in words}
def theme(x):
return {lookup.get(word, 'None') for word in x}
so that theme(['AC', 'clean', 'insects']) would return a set of corresponding categories:
{'Cleanliness', 'AC Problem'}

This should do what you're asking. I set all the keys to lowercase and converted i to lowercase when checking if you get a match, but with different capitalization, it still counts.
def theme(x):
output =[]
category = ()
myDict = {"ac":"AC problem", "insects":"Cleanliness", "clean":"Cleanliness", "food":"Food Problem", "delay":"Train Delayed"} #I reccomend coming up with a more suitable name for your dictionary in your actual program
for i in x:
if i.lower() in myDict: #Checks to see if i is in the dictionary before trying to print the result; prevents possible Key Errors
category = (myDict[i.lower()]) #If it is in the dictionary it category will be set to the result of the key
output.append(category)
else:
output.append("None") #If i isn't in the dictionary output will append None instead
return output
Here's some examples:
>>>print(theme(['Clean', 'Cleaned', 'spoilt', 'dirty']))
['Cleanliness', 'None', 'None', 'None']
>>>print(theme(['Delay', 'Ham', 'Cheese', 'Insects']))
['Train Delayed', 'None', 'None', 'Cleanliness']

I have worked out a another way:
def theme(x):
output = []
for i in x:
if set(cleanliness).intersection(i.lower().split()):
category = 'clean'
elif set(ac_problem).intersection(i.lower().split()):
category = 'ac problem'
else:
category = 'none'
output.append(category)
return output

Maybe you can do it like this:
def theme(x):
output = []
name_dic = {"AC": "AC problem",
"clean": "Cleanliness",
"food": "Food Problem"
}
for e in x:
output.append(name_dic.get(e))
return output
Or more exactly like this:
def theme(x):
output = []
name_list = [
("AC", "AC problem"),
("clean", "Cleanliness"),
("insects", "Cleanliness"),
("food", "Food Problem")
]
name_dic = dict(name_list)
for e in x:
output.append(name_dic.get(e))
return output
Hope it helps.