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pythonperformancepandasvectorizationreduce

"Reduce" function for Series


Is there an analog for reduce for a pandas Series?

For example, the analog for map is pd.Series.apply, but I can't find any analog for reduce.


My application is, I have a pandas Series of lists:

>>> business["categories"].head()

0                      ['Doctors', 'Health & Medical']
1                                        ['Nightlife']
2                 ['Active Life', 'Mini Golf', 'Golf']
3    ['Shopping', 'Home Services', 'Internet Servic...
4    ['Bars', 'American (New)', 'Nightlife', 'Loung...
Name: categories, dtype: object

I'd like to merge the Series of lists together using reduce, like so:

categories = reduce(lambda l1, l2: l1 + l2, categories)

but this takes a horrific time because merging two lists together is O(n) time in Python. I'm hoping that pd.Series has a vectorized way to perform this faster.


Solution

  • With itertools.chain() on the values

    This could be faster:

    from itertools import chain
    categories = list(chain.from_iterable(categories.values))
    

    Performance

    from functools import reduce
    from itertools import chain
    
    categories = pd.Series([['a', 'b'], ['c', 'd', 'e']] * 1000)
    
    %timeit list(chain.from_iterable(categories.values))
    1000 loops, best of 3: 231 µs per loop
    
    %timeit list(chain(*categories.values.flat))
    1000 loops, best of 3: 237 µs per loop
    
    %timeit reduce(lambda l1, l2: l1 + l2, categories)
    100 loops, best of 3: 15.8 ms per loop
    

    For this data set the chaining is about 68x faster.

    Vectorization?

    Vectorization works when you have native NumPy data types (pandas uses NumPy for its data after all). Since we have lists in the Series already and want a list as result, it is rather unlikely that vectorization will speed things up. The conversion between standard Python objects and pandas/NumPy data types will likely eat up all the performance you might get from the vectorization. I made one attempt to vectorize the algorithm in another answer.