convert tuple keys of dict into a new dict

Question

I have a dict like this:

{
    (\'America\', 25, \'m\', \'IT\'): 10000,
    (\'America\', 22, \'m\', \'IT\'): 8999,
    (\'Japan\',   24, \'f\', \'I         


        

Answer 1

Since I like namedtuples, here would be an alternative suggestion:

Store your dictionary as a list or set of namedtuples, e.g.,

>>> from collections import namedtuple
>>> Entry = namedtuple('entry', ('country', 'age', 'sex', 'job', 'count'))

To convert your existing dictionary dt:

>>> nt = [Entry(*list(k) + [dt[k]]) for k in dt]

Now, you could fetch the desired entries in a quite readable way, e.g.,

>>> results = [i for i in nt if (i.country=='America' and i.sex=='m' and i.job=='IT')]

Or, for example, to get the counts:

>>> [i.count for i in nt if (i.country=='America' and i.sex=='m' and i.job=='IT')]
[8999, 10000]

Edit: Performance

Was not sure if you were looking after performance since you mentioned "easier way to do it". You are right, the pure "comprehension is faster:

dt = {
    ('America', 25, 'm', 'IT'): 10000,
    ('America', 22, 'm', 'IT'): 8999,
    ('Japan',   24, 'f', 'IT'): 9999,
    ('Japan',   23, 'f', 'IT'): 9000
}

nt = [Entry(*list(k) + [dt[k]]) for k in dt]

%timeit {i.age:i.count for i in nt if (i.country=='America' and i.sex=='m' and i.job=='IT')}

100000 loops, best of 3: 3.62 µs per loop

%timeit {x: value for (w, x, y, z), value in dt.items() if w == "America" and y == "m" and z == "IT"}

100000 loops, best of 3: 2.42 µs per loop

But if you have a larger dataset and querying it over and over again I would also think about Pandas or SQLite.

df = pd.DataFrame([list(x[0]) + [x[1]] for x in dt.items()])
df.columns = ['country', 'age', 'sex', 'job', 'count']
df

df.loc[(df.country=='America') & (df.sex=='m') & (df.job=='IT')]