CODEWITHSUNDEEP
pythonpandas
Danny
Danny

Reputation: 403

Python series where values are lists, get another series with list of indexes correspond to each item list

Sorry the title is hard to understand—not sure how to phrase this one. Say I have a series that looks like this 

s = pd.Series(index = ['a','b','c'], data = [['x','y','z'], ['y','z'], ['x','z']]).

I would want something like this 

{'x':['a','c'], 'y':['a','b'], 'z':['a','b','c']}

I.e. I can see which keys correspond to each element from the series of lists. Any ideas how I could do this as efficiently as possible? Thanks!

Upvotes: 3

Views: 652

Answers (3)

Scott Boston
Scott Boston

Reputation: 153570

Another solution using default dict for speed:

from collections import defaultdict

d = defaultdict(list)
q = s.explode()
for k, v in q.items():
    d[v].append(k)

dict(d)

Output:

{'x': ['a', 'c'], 'y': ['a', 'b'], 'z': ['a', 'b', 'c']}

Timings:

%timeit s.explode().reset_index().groupby(0)['index'].agg(list).to_dict()
3.94 ms ± 119 µs per loop (mean ± std. dev. of 7 runs, 100 loops each)

%%timeit d = defaultdict(list) method
300 µs ± 33.4 µs per l0op (mean ± std. dev. of 7 runs, 1000 loops each)

Upvotes: 1

oppressionslayer
oppressionslayer

Reputation: 7222

Here's a second solution as well:

x = s.explode() 

pd.DataFrame({'X':x.index, 'Y':x.values}).groupby('Y')['X'].apply(list).to_dict()


# {'x': ['a', 'c'], 'y': ['a', 'b'], 'z': ['a', 'b', 'c']}

Upvotes: 0

BENY
BENY

Reputation: 323396

Let us use explode

s.explode().reset_index().groupby(0)['index'].agg(list).to_dict()
{'x': ['a', 'c'], 'y': ['a', 'b'], 'z': ['a', 'b', 'c']}

Upvotes: 4

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