Dictionary from another dictionary-CodePudding

I have a dictionary with IDs and counts:

  items_count = {
    '1111:271': 111, 
    '1111:190': 3, 
    '1231:106': 13, 
    '1211:104': 111, 
    '1111:201': 9
    }

Key is "category":"id". I want to separate category and put it in another dictionary like:

items_count2 = {
    '1111': {'271': 111, '190': 3, '201': 0},
    '1231': {'106': 13}, 
    '1211': {'104': 111}
    }

but when I do this

items_count2 = {k.split(':')[0]: {k.split(':')[1]: v} for k, v in items_count}

I get error "ValueError: too many values to unpack (expected 2)"

please help to understand, what am I doing wrong?

CodePudding user response：

So first your suggestion should be:

items_count2 = {k.split(':')[0]: {k.split(':')[1]: v} for k, v in items_count.items()}

But this won't be working neither, cause you have multiple k, v per sub-dictionnary.

So below is a way to do, without using dict comprehension:

dct = {
    '1111:271': 111, 
    '1111:190': 3, 
    '1231:106': 13, 
    '1211:104': 111, 
    '1111:201': 9
}

# Initialize your new dict
new_dct = {k.split(':')[0]: {} for k in dct} # Or k in dct.keys()

# Loop through old dict
for k, v in dct.items():
    # Set value to top > sub key in new dict
    new_dct[k.split(':')[0]][k.split(':')[1]] = v
    
print(new_dct)

Output:

{'1111': {'271': 111, '190': 3, '201': 9}, '1231': {'106': 13}, '1211': {'104': 111}}

CodePudding user response：

items_count2  = {}
for k, v in items_count.items():  
  t = items_count2.get(k.split(':')[0],{})
  t[k.split(':')[1]] = v
  items_count2[k.split(':')[0]] = t

output:

{'1111': {'190': 3, '201': 9, '271': 111},
 '1211': {'104': 111},
 '1231': {'106': 13}}

{k.split(':')[0]: {k.split(':')[1]: v} for k, v in items_count.items()}

will not give the desired solution because if the key already exists it will be replaced by the current one. So your values will always be a directory of single keys.

CodePudding user response：

Setting up a dictionary with the needed keys is a clearer solution. But if you want to avoid looping, splitting and indexing twice you can create the needed keys on the fly with dict.setdefault.

items_count = {
    '1111:271': 111, 
    '1111:190': 3, 
    '1231:106': 13, 
    '1211:104': 111, 
    '1111:201': 9
}

items_count2 = {}
for k,v in items_count.items():
    s1, s2 = k.split(':')
    items_count2.setdefault(s1, {})[s2] = v
items_count2

Output

{'1111': {'190': 3, '201': 9, '271': 111},
 '1211': {'104': 111},
 '1231': {'106': 13}}