How to convert dict of dict to dict of specified format?-CodePudding

I have a dictionary of dictionaries as shown below:

d = {0: {1: ["hello"], 2: ["How are you"]}, 1: {1: ["!"], 2: ["?"]}}

and I would want it be in required format:

result = {1:["hello", "!"], 2: ["How are you", "?"]}

However, I get this in the following format using the code below:

new_d = {}
for sub in d.values():
    for key, value in sub.items():
        new_d.setdefault(key, []).append(value)

The result is not of required structure and it causes a list of lists.

{1: [['hello'], ['!']], 2: [['How are you'], ['?']]}

Any help here would be highly appreciated. Thanks.

CodePudding user response：

use extend instead of append:

new_d.setdefault(key, []).extend(value)

The extend() method adds all the elements of an iterable (list, tuple, string etc.) to the end of the list.

CodePudding user response：

If you want to solve this problem with using append() function try this code:

new_d = {}
for sub in d.values():
    for key, value in sub.items():

        # Control key exist...
        if(key in new_d.keys()):
            
            new_d[key].append(value[0])
        else:
            new_d[key] = value

CodePudding user response：

You can either use .extend(value) instead of .append(value) or you can add a basic for loop to flatten the list of all dictionary values as shown below.

new_d = {}
for sub in d.values():
    for key, value in sub.items():
        new_d.setdefault(key, []).extend(value)


for i in range (0,len(d)):
    new_d[i 1] = [item for sublist in new_d.get(i 1) for item in sublist]
print(new_d)

CodePudding user response：

The accepted answer by @Gabip correctly identifies that your only mistake was using append instead of extend.

That mistake being corrected, I'd also like to suggest a slightly different approach using dict comprehensions:

d = {0: {1: ["hello"], 2: ["How are you"]}, 1: {1: ["!"], 2: ["?"]}}

new_d = {key: d[0].get(key, [])   d[1].get(key, []) for key in d[0]}
# {1: ['hello', '!'], 2: ['How are you', '?']}

Or a more robust version that takes keys from both d[0] and d[1], in case some keys are in d[1] but not in d[0]:

d = {0: {1: ["hello"], 2: ["How are you"]}, 1: {1: ["!"], 2: ["?"], 3: ['>>>']}}

new_d = {key: d[0].get(key, [])   d[1].get(key, []) for key in set(d[0].keys()) | set(d[1].keys())}
# {1: ['hello', '!'], 2: ['How are you', '?'], 3: ['>>>']}

Finally, this wasn't explicitly part of your question, but I suggest using str.join to join the strings:

d = {0: {1: ["hello"], 2: ["How are you"]}, 1: {1: ["!"], 2: ["?"]}}

new_d = {key: ''.join(d[0].get(key, [])   d[1].get(key, [])) for key in d[0]}
# {1: 'hello!', 2: 'How are you?'}