CODEWITHSUNDEEP
pythonpython-3.xlistdictionary
Peter Ebelsberger
Peter Ebelsberger

Reputation: 155

Performanceoptimized changing of an dictionary

I have a list of dictionaries and I have a solution, described below:

[{"A1":"B1", "C1":"D1"},{"A2":"B2", "C2":"D2"},....]

Expected output would be a dictionary like this

{
"B1":"D1",
"B2":"D2",
...
}

My current procedure is get one dictionary and create a 2 entries dict to a single one

input: 

{"A1":"B1", "C1":"D1"}

output:

{"B1":"D1"}

where the "A1" and "C1" is inside my method configureable:

def method(input_dictionary, default_key_from_dict, default_value_from_dict):
    single_dictionary[input_dictionary[default_key_from_dict]] =   input_dictionary[default_value_from_dict]

later I perform the following ones:

def function(dict_list):
    single_key_value_dictionary_list = list(map(convert_two_entries_dictionary_to_one_by_value, dict_list))
    global_dict = reduce(lambda a, b: dict(a, **b), single_key_value_dictionary_list)
    return global_dict

I ask myself if there is some better solution for this, I have the generator idea in my mind but I am not sure if it is worth to think about it? Any remarks on this?

UPDATE There are only 2 keys in each entry of the list as dictionary. 2 keys only.

BR Peter

Upvotes: 1

Views: 41

Answers (2)

user3804427
user3804427

Reputation: 442

You can use something about this:

dict1 = {"A1":"B1", "C1":"D1"}
it  = iter(dict1.values())
list1 = list(zip(it, it))
print(list1)

Upvotes: 0

Ch3steR
Ch3steR

Reputation: 20669

From the comments since you always have only 2 keys. You can use dict.update.

inp=[{"A1":"B1", "C1":"D1"},{"A2":"B2", "C2":"D2"}]
out={}
for d in inp:
    out.update((d.values(),))
out
# {'B1': 'D1', 'B2': 'D2'}

Upvotes: 1

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