Apply A Function Pairwise On A Pandas Series

I've a pandas series whose elements constitute frozensets: data = {0: frozenset({'apple', 'banana'}), 1: frozenset({'apple', 'orange'}), 2: frozenset({'banana'}), 3:

Solution 1:

Couple of ways

Option 1] list comprehension

In [3631]: pd.Series([x[0].union(x[1])
                      for x in zip(tokens, tokens.shift(-1).fillna(''))],
                     index=tokens.index)
Out[3631]:
0              (orange, banana, apple)
1              (orange, apple, banana)
2            (orange, kumquat, banana)
3                    (orange, kumquat)
4                       (orange, pear)
5                       (orange, pear)
6        (orange, pear, banana, apple)
7     (persimmon, pear, banana, apple)
8           (apple, persimmon, banana)
9                      (apple, banana)
10                     (banana, apple)
11                             (apple)
dtype: object

Option 2]map

In [3632]: pd.Series(map(lambda x: x[0].union(x[1]), 
                         zip(tokens, tokens.shift(-1).fillna(''))),
                     index=tokens.index)
Out[3632]:
0              (orange, banana, apple)
1              (orange, apple, banana)
2            (orange, kumquat, banana)
3                    (orange, kumquat)
4                       (orange, pear)
5                       (orange, pear)
6        (orange, pear, banana, apple)
7     (persimmon, pear, banana, apple)
8           (apple, persimmon, banana)
9                      (apple, banana)
10                     (banana, apple)
11                             (apple)
dtype: object

Option 3] Using concat and apply

In [3633]: pd.concat([tokens, tokens.shift(-1).fillna('')],
                     axis=1).apply(lambda x: x[0].union(x[1]), axis=1)
Out[3633]:
0              (orange, banana, apple)
1              (orange, apple, banana)
2            (orange, kumquat, banana)
3                    (orange, kumquat)
4                       (orange, pear)
5                       (orange, pear)
6        (orange, pear, banana, apple)
7     (persimmon, pear, banana, apple)
8           (apple, persimmon, banana)
9                      (apple, banana)
10                     (banana, apple)
11                             (apple)
dtype: object

---

Timings

In [3647]: tokens.shape
Out[3647]: (60000L,)

In [3648]: %timeit pd.Series([x[0].union(x[1]) for x in zip(tokens, tokens.shift(-1).fillna(''))], index=tokens.index)
10 loops, best of 3: 35 ms per loop

In [3649]: %timeit pd.Series(map(lambda x: x[0].union(x[1]), zip(tokens, tokens.shift(-1).fillna(''))), index=tokens.index)
10 loops, best of 3: 40.9 ms per loop

In [3650]: %timeit pd.concat([tokens, tokens.shift(-1).fillna('')], axis=1).apply(lambda x: x[0].union(x[1]), axis=1)
1 loop, best of 3: 2.2 s per loop

Unrelated and for sake of a number on diff

In [3653]: %timeit tokens.diff()
10 loops, best of 3: 10.8 ms per loop