Create new dictionaries based on names of keys of another dictionary-CodePudding

I have a dictionary "A",

A = {'Industry1':1, 'Industry2':1, 'Industry3':1, 'Customer1':1, 'Customer2':1, 'LocalShop1':1, 'LolcalShop2':1 }

A = {'Industry1': 1,
 'Industry2': 1,
 'Industry3': 1,
 'Customer1': 1,
 'Customer2': 1,
 'LocalShop1': 1,
 'LolcalShop2': 1}

I want to group by key names and create new dictionaries for each "category", the names should be generated automatically

Expected Output

Industry = {'Industry1': 1,
 'Industry2': 1,
 'Industry3': 1,
}

Customer = { 'Customer1': 1,
 'Customer2': 1,
}

LocalShop = { 'LocalShop1': 1,
 'LolcalShop2': 1}

Can you guys give me a hint to achieve this output, please?

CodePudding user response：

You can use set() to identify the unique keys in your input dict, removing last char.

Then, for each key, you can retrieve all entries from input_dict and and create a new dictionary with the previously retrieved unique keys and those that were initially present as values.

Then simply compare the string of the new key with the old keys (minus the last character).

Full code:

A = {
    'Industry1': 1,
    'Industry2': 1,
    'Industry3': 1,
    'Customer1': 1,
    'Customer2': 1,
    'LocalShop1': 1,
    'LocalShop2': 1
}

unique_keys = set([x[:-1] for x in A.keys()])  # {'Customer', 'LocalShop', 'Industry'}

new_dict = {}
for k in unique_keys:
    new_dict[k] = {key:val for key,val in A.items() if key[:-1]==k}

# new_dict = {'LocalShop': {'LocalShop1': 1, 'LocalShop2': 1}, 'Industry': {'Industry1': 1, 'Industry2': 1, 'Industry3': 1}, 'Customer': {'Customer1': 1, 'Customer2': 1}}

Industry = new_dict['Industry']
Customer = new_dict['Customer']
LocalShop = new_dict['LocalShop']

CodePudding user response：

For the purposes of this answer I've assumed that every key in the dictionary A has either the word "Industry", "Customer", or "Shop" in it. This allows us to detect what category each entry needs to be in by checking if each key contains a certain substring (i.e. "Industry"). If this assumption doesn't hold for your specific circumstances, you'll have to find a different way to write the if/elif statements in the solutions below that fits your situation better.

Here's one way to do it. You make a new dictionary for each category and check if "Industry", "Customer", or "Shop" is in each key.

industries = {}
customers = {}
shops = {}

for key, value in A.items():
    if "Industry" in key:
        industries[key] = value
    elif "Customer" in key:
        customers[key] = value
    elif "Shop" in key:
        shops[key] = value

Another, cleaner version would be where you have a nested dictionary that stores all of your categories, and each category would have its own dictionary inside the main one. This would help in the future if you needed to add more categories. You'd only have to add them in one place (in the dictionary definition) and the code would automatically adjust.

categories = {
    "Industry": {},
    "Customer": {},
    "Shop": {},
}

for key, value in A.items():
    for category_name, category_dict in categories.items():
        if category_name in key:
            category_dict[key] = value

If you can't detect the category from the string of an entry, then you may have to store that categorical information in the key or the value of each entry in A, so that you can detect the category when trying to filter everything.

CodePudding user response：

You can use itertools.groupby with a key that only extracts the word without the number. I wouldn't recommend to make variables of them, this is not scalable if there are more than 3 keys... just put them in a new dictionary or in a list.

A = {'Industry1': 1,
 'Industry2': 1,
 'Industry3': 1,
 'Customer1': 1,
 'Customer2': 1,
 'LocalShop1': 1,
 'LocalShop2': 1}

grouped = [dict(val) for k, val in itertools.groupby(A.items(), lambda x: re.match('(. )\d{1,}', x[0]).group(1))]

Output grouped:

[
    {'Industry1': 1, 'Industry2': 1, 'Industry3': 1}, 
    {'Customer1': 1, 'Customer2': 1}, 
    {'LocalShop1': 1, 'LocalShop2': 1}
]

If you are sure, that there are exactly 3 elements in that list and you really want them as variables, you can do it with tuple unpacking:

Industry, Customer, LocalShop = [dict(val) for k, val in itertools.groupby(A.items(), lambda x: re.match('(. )\d{1,}', x[0]).group(1))]

I think I would save the results in a new dictionary with the grouped key as new key and the list as value:

grouped_dict = {k: dict(val) for k, val in itertools.groupby(A.items(), lambda x: re.match('(. )\d{1,}', x[0]).group(1))}

Output grouped_dict:

{'Industry': {'Industry1': 1, 'Industry2': 1, 'Industry3': 1},
 'Customer': {'Customer1': 1, 'Customer2': 1},
 'LocalShop': {'LocalShop1': 1, 'LocalShop2': 1}}