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在熊猫系列上成对应用函数

frozenset set dataframe pandas python

3

cs95 abhishek58g · 技术社区 · 8 年前

我有一个熊猫系列,它的元素构成了冻结集:

data = {0: frozenset({'apple', 'banana'}),
     1: frozenset({'apple', 'orange'}),
     2: frozenset({'banana'}),
     3: frozenset({'kumquat', 'orange'}),
     4: frozenset({'orange'}),
     5: frozenset({'orange', 'pear'}),
     6: frozenset({'orange', 'pear'}),
     7: frozenset({'apple', 'banana', 'pear'}),
     8: frozenset({'banana', 'persimmon'}),
     9: frozenset({'apple'}),
     10: frozenset({'banana'}),
     11: frozenset({'apple'})}

tokens = pd.Series(data); tokens

0           (apple, banana)
1           (orange, apple)
2                  (banana)
3         (orange, kumquat)
4                  (orange)
5            (orange, pear)
6            (orange, pear)
7     (apple, banana, pear)
8       (persimmon, banana)
9                   (apple)
10                 (banana)
11                  (apple)
Name: Tokens, dtype: object

我想成对应用一个函数。例如 tokens.diff

0                   NaN
1              (orange)
2              (banana)
3     (orange, kumquat)
4                    ()
5                (pear)
6                    ()
7       (apple, banana)
8           (persimmon)
9               (apple)
10             (banana)
11              (apple)
Name: Tokens, dtype: object

我想做同样的事情,但不是集差,我想要连续行上的集并集。因此,我希望:

0                                 NaN
1             (orange, apple, banana)
2             (banana, orange, apply)
3           (orange, kumquat, banana)
4                   (orange, kumquat)
                                  ...

zip

1 回复 | 直到 8 年前

1

4

Zero 8 年前

选项1]

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

选项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

选项3] 使用 concat 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

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

无关,为了一个数字 diff

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